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Rút gọn căn thức – Các biểu thức số

Trong bài này ta tổng hợp các kĩ năng thực hiện các phép tính toán, khai căn, phân tích thành tích, trục căn thức ở mẫu để làm các bài toán phức tạp hơn.

Chú ý khi làm bài. Trong các bài này ta có thể rút gọn các phân thức riêng lẻ trước nếu được bằng cách phân tích thành tích, tiếp theo thì trục căn thức và rút gọn các biểu thức trong ngoặc, không nên qui đồng vì tính toán sẽ rất phức tạp.

Ví dụ 1. Rút gọn

a) $\dfrac{6-6\sqrt{3}}{1-\sqrt{3}}+\dfrac{3\sqrt{3}+3}{\sqrt{3}+1}$.
b) $\dfrac{2-\sqrt{2}}{1-\sqrt{2}}+\dfrac{\sqrt{2}-\sqrt{6}}{\sqrt{3}-1}$.
c) $\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}+\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}$.
d) $\dfrac{3\sqrt{2}-6}{\sqrt{2}-1}+\dfrac{6\sqrt{2}-4}{\sqrt{2}-3}$.

Giải

a)  $\dfrac{6-6\sqrt{3}}{1-\sqrt{3}}+\dfrac{3\sqrt{3}+3}{\sqrt{3}+1}$.\\
Ta có:\\
$\begin{aligned}
&\dfrac{6-6\sqrt{3}}{1-\sqrt{3}}+\dfrac{3\sqrt{3}+3}{\sqrt{3}+1}\\
&=\dfrac{6\left(1-\sqrt{3}\right)}{1-\sqrt{3}}+\dfrac{3\left(\sqrt{3}+1\right)}{\sqrt{3}+1}\\
&=6+3\\
&=9
\end{aligned}$
b) $\dfrac{2-\sqrt{2}}{1-\sqrt{2}}+\dfrac{\sqrt{2}-\sqrt{6}}{\sqrt{3}-1}$.\\
Ta có:\\
$\begin{aligned}
&\dfrac{2-\sqrt{2}}{1-\sqrt{2}}+\dfrac{\sqrt{2}-\sqrt{6}}{\sqrt{3}-1}\\
&=\dfrac{\sqrt{2}\left(\sqrt{2}-1\right)}{-\left(\sqrt{2}-1\right)}+\dfrac{\sqrt{2}\left(1-\sqrt{3}\right)}{-\left(1-\sqrt{3}\right)}\\
&=-\sqrt{2}-\sqrt{2}\\
&=-2\sqrt{2}
\end{aligned}$
c) $\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}+\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}$.\\
Ta có:\\
$\begin{aligned}
&\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}+\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}\\
&=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{2\left(\sqrt{5}-2\right)}\\
&=\sqrt{5}+\dfrac{\sqrt{5}}{2}\\
&=\dfrac{3\sqrt{5}}{2}
\end{aligned}$
d) $\dfrac{3\sqrt{2}-6}{\sqrt{2}-1}+\dfrac{6\sqrt{2}-4}{\sqrt{2}-3}$.\\
Ta có:\\
$\begin{aligned}
&\dfrac{3\sqrt{2}-6}{\sqrt{2}-1}+\dfrac{6\sqrt{2}-4}{\sqrt{2}-3}\\
&=\dfrac{3\sqrt{2}\left(1-\sqrt{2}\right)}{-\left(1-\sqrt{2}\right)}+\dfrac{2\sqrt{2}\left(3-\sqrt{2}\right)}{-\left(3-\sqrt{2}\right)}\\
&=-3\sqrt{2}-2\sqrt{2}\\
&=-5\sqrt{2}
\end{aligned}$

Ví dụ 2. Rút gọn

a) $\dfrac{6}{\sqrt{5}-1}+\dfrac{7}{1-\sqrt{3}}-\dfrac{2}{\sqrt{3}-\sqrt{5}}$.
b) $\dfrac{\sqrt{12}-6}{\sqrt{8}-\sqrt{24}}-\dfrac{3+\sqrt{3}}{\sqrt{3}}+\dfrac{4}{1-\sqrt{7}}$.
c) $\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{1}{\sqrt{3}-\sqrt{5}}+\dfrac{1}{\sqrt{7}-\sqrt{5}}$.
d) $\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}$.

Giải

a)$\dfrac{6}{\sqrt{5}-1}+\dfrac{7}{1-\sqrt{3}}-\dfrac{2}{\sqrt{3}-\sqrt{5}}$.
Ta có:
$\begin{aligned}
&\dfrac{6}{\sqrt{5}-1}+\dfrac{7}{1-\sqrt{3}}-\dfrac{2}{\sqrt{3}-\sqrt{5}}\\
&=\dfrac{6}{5-1}\left(\sqrt{5}+1\right)+\dfrac{7}{1-3}\left(1+\sqrt{3}\right)-\dfrac{2}{3-5}\left(\sqrt{3}+\sqrt{5}\right)\\
&=\dfrac{3}{2}\left(\sqrt{5}+1\right)-\dfrac{7}{2}\left(1+\sqrt{3}\right)+\sqrt{3}+\sqrt{5}\\
&=\dfrac{3\sqrt{5}}{2}+\dfrac{3}{2}-\dfrac{7}{2}-\dfrac{7\sqrt{3}}{2}+\sqrt{3}+\sqrt{5}\\
&=\dfrac{5\sqrt{5}}{2}-\dfrac{5\sqrt{3}}{2}-2\\
&=\dfrac{5}{2}\left(\sqrt{5}-\sqrt{3}\right)-2
\end{aligned}$
b) $\dfrac{\sqrt{12}-6}{\sqrt{8}-\sqrt{24}}-\dfrac{3+\sqrt{3}}{\sqrt{3}}+\dfrac{4}{1-\sqrt{7}}$.
Ta có:
$\begin{aligned}
&\dfrac{\sqrt{12}-6}{\sqrt{8}-\sqrt{24}}-\dfrac{3+\sqrt{3}}{\sqrt{3}}+\dfrac{4}{1-\sqrt{7}}\\
&=\dfrac{\sqrt{6}\left(\sqrt{2}-\sqrt{6}\right)}{2\left(\sqrt{2}-\sqrt{6}\right)}-\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}}+\dfrac{4}{1-7}\left(1+\sqrt{7}\right)\\
&=\dfrac{\sqrt{6}}{2}-\left(\sqrt{3}+1\right)-\dfrac{2}{3}\left(1+\sqrt{7}\right)\\
&=-\dfrac{2}{3}\sqrt{7}+\dfrac{\sqrt{6}}{2}-\sqrt{3}-\dfrac{5}{3}
\end{aligned}$
c) $\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{1}{\sqrt{3}-\sqrt{5}}+\dfrac{1}{\sqrt{7}-\sqrt{5}}$.\\
Ta có:\\
$\begin{aligned}
&\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{1}{\sqrt{3}-\sqrt{5}}+\dfrac{1}{\sqrt{7}-\sqrt{5}}\\
&=\dfrac{1}{2-3}\left(\sqrt{2}+\sqrt{3}\right)-\dfrac{1}{3-5}\left(\sqrt{3}+\sqrt{5}\right)+\dfrac{1}{7-5}\left(\sqrt{7}+\sqrt{5}\right)\\
&=-\left(\sqrt{2}+\sqrt{3}\right)+\dfrac{1}{2}\left(\sqrt{3}+\sqrt{5}\right)+\dfrac{1}{2}\left(\sqrt{7}+\sqrt{5}\right)\\
&=-\sqrt{2}-\sqrt{3}+\dfrac{1}{2}\sqrt{3}+\dfrac{1}{2}\sqrt{5}+\dfrac{1}{2}\sqrt{7}+\dfrac{1}{2}\sqrt{5}\\
&=\dfrac{1}{2}\sqrt{7}+\sqrt{5}-\dfrac{1}{2}\sqrt{3}-\sqrt{2}
\end{aligned}$
d) $\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}$.\\
Ta có:
$\begin{aligned}
&\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\\
&=\left[\dfrac{\sqrt{7}\left(\sqrt{2}-1\right)}{-\left(\sqrt{2}-1\right)}+\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{-\left(\sqrt{3}-1\right)}\right].\left(\sqrt{7}-\sqrt{5}\right)\\
&=\left(-\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\\
&=-(7-5)\\
&=-2
\end{aligned}$

Ví dụ 3. Rút gọn

a) $\left(\dfrac{12}{\sqrt{5}+1}-\dfrac{4}{\sqrt{5}+2}+\dfrac{20}{3+\sqrt{5}}\right)(10+3\sqrt{5})$.
b) $\left(\dfrac{24}{\sqrt{7}+1}+\dfrac{4}{3+\sqrt{7}}-\dfrac{3}{\sqrt{7}+2}\right)(4-\sqrt{7})$.
c) $\left(\dfrac{8}{\sqrt{3}-1}-\dfrac{4}{\sqrt{3}+1}+\dfrac{4}{\sqrt{5}+\sqrt{3}}\right):\sqrt{14+6\sqrt{5}}$.
d) $\left(\dfrac{7}{\sqrt{2}-1}+\dfrac{56}{\sqrt{2}-4}+\dfrac{3}{\sqrt{3}+\sqrt{2}}\right):\sqrt{12-6\sqrt{3}}$.

Giải

a) $\left(\dfrac{12}{\sqrt{5}+1}-\dfrac{4}{\sqrt{5}+2}+\dfrac{20}{3+\sqrt{5}}\right)(10+3\sqrt{5})$.
Ta có:
$\begin{aligned}
&\left(\dfrac{12}{\sqrt{5}+1}-\dfrac{4}{\sqrt{5}+2}+\dfrac{20}{3+\sqrt{5}}\right)(10+3\sqrt{5})\\
&=\left[\dfrac{12}{5-1}\left(\sqrt{5}-1\right)-\dfrac{4}{5-4}\left(\sqrt{5}-2\right)+\dfrac{20}{9-5}\left(3-\sqrt{5}\right)\right]\left(10+3\sqrt{5}\right)\\
&=\left[3\left(\sqrt{5}-1\right)-4\left(\sqrt{5}-2\right)+5\left(3-\sqrt{5}\right)\right]\left(10+3\sqrt{5}\right)\\
&=\left[3\sqrt{5}-3-4\sqrt{5}+8+15-5\sqrt{5}\right]\left(10+3\sqrt{5}\right)\\
&=\left(-6\sqrt{5}+20\right)\left(10+3\sqrt{5}\right)\\
&=2\left(10-3\sqrt{5}\right)\left(10+3\sqrt{5}\right)\\
&=2(100-45)\\
&=110
\end{aligned}$
b) $\left(\dfrac{24}{\sqrt{7}+1}+\dfrac{4}{3+\sqrt{7}}-\dfrac{3}{\sqrt{7}+2}\right)(4-\sqrt{7})$.
Ta có:
$\begin{aligned}
&\left(\dfrac{24}{\sqrt{7}+1}+\dfrac{4}{3+\sqrt{7}}-\dfrac{3}{\sqrt{7}+2}\right)(4-\sqrt{7})\\
&=\left[\dfrac{24}{7-1}\left(\sqrt{7}-1\right)+\dfrac{4}{9-7}\left(3-\sqrt{7}\right)-\dfrac{3}{7-4}\left(\sqrt{7}-2\right)\right]\left(4-\sqrt{7}\right)\\
&=\left[4\left(\sqrt{7}-1\right)+2\left(3-\sqrt{7}\right)-\left(\sqrt{7}-2\right)\right]\left(4-\sqrt{7}\right)\\
&=\left(4\sqrt{7}-4+6-2\sqrt{7}-\sqrt{7}+2\right)\left(4-\sqrt{7}\right)\\
&=\left(\sqrt{7}+4\right)\left(4-\sqrt{7}\right)\\
&=16-7
&=9
\end{aligned}$
c) $\left(\dfrac{8}{\sqrt{3}-1}-\dfrac{4}{\sqrt{3}+1}+\dfrac{4}{\sqrt{5}+\sqrt{3}}\right):\sqrt{14+6\sqrt{5}}$.
Ta có:
$\begin{aligned}
&\left(\dfrac{8}{\sqrt{3}-1}-\dfrac{4}{\sqrt{3}+1}+\dfrac{4}{\sqrt{5}+\sqrt{3}}\right):\sqrt{14+6\sqrt{5}}\\
&=\left[\dfrac{8}{3-1}\left(\sqrt{3}+1\right)-\dfrac{4}{3-1}\left(\sqrt{3}-1\right)+\dfrac{4}{5-3}\left(\sqrt{5}-\sqrt{3}\right)\right]:\left(3+\sqrt{5}\right)\\
&=\left[4\left(\sqrt{3}+1\right)-2\left(\sqrt{3}-1\right)+2\left(\sqrt{5}-\sqrt{3}\right)\right]:\left(3+\sqrt{5}\right)\\
&=\left(4\sqrt{3}+4-2\sqrt{3}+2+2\sqrt{5}-2\sqrt{3}\right):\left(3+\sqrt{5}\right)\\
&=\left(6+2\sqrt{5}\right):\left(3+\sqrt{5}\right)\\
&=2
\end{aligned}$
d) $\left(\dfrac{7}{\sqrt{2}-1}+\dfrac{56}{\sqrt{2}-4}+\dfrac{3}{\sqrt{3}+\sqrt{2}}\right):\sqrt{12-6\sqrt{3}}$.\\
Ta có:
$\begin{aligned}
&\left(\dfrac{7}{\sqrt{2}-1}+\dfrac{56}{\sqrt{2}-4}+\dfrac{3}{\sqrt{3}+\sqrt{2}}\right):\sqrt{12-6\sqrt{3}}\\
&=\left[\dfrac{7}{2-1}\left(\sqrt{2}+1\right)+\dfrac{56}{2-16}\left(\sqrt{2}+4\right)+\dfrac{3}{3-2}\left(\sqrt{3}-\sqrt{2}\right)\right]:\left(3-\sqrt{3}\right)\\
&=\left[7\left(\sqrt{2}+1\right)-4\left(\sqrt{2}+4\right)+3\left(\sqrt{3}-\sqrt{2}\right)\right]:\left(3-\sqrt{3}\right)\\
&=\left(7\sqrt{2}+7-4\sqrt{2}-16+3\sqrt{3}-3\sqrt{2}\right):\left(3-\sqrt{3}\right)\\
&=\left(-9+3\sqrt{3}\right):\left(3-\sqrt{3}\right)\\
&=-3
\end{aligned}$

Bài tập rèn luyện

Bài 1. Rút gọn

a) $\dfrac{\sqrt{160}-\sqrt{80}}{\sqrt{8}-\sqrt{2}}-\dfrac{\sqrt{40}-\sqrt{15}}{2\sqrt{2}-\sqrt{3}}$.
b) $\left(\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}-2\right)\left(\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}-2\right)$.
c) $\left(\dfrac{\sqrt{216}}{3}-\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}\right)\dfrac{1}{\sqrt{6}}$.
d) $\left(\dfrac{\sqrt{343}}{21}-\dfrac{28+4\sqrt{7}}{\sqrt{63}+3}\right)\dfrac{\sqrt{7}}{7}$.

Bài 2. Rút gọn

a) $\dfrac{5\sqrt{2}-2\sqrt{5}}{\sqrt{5}-\sqrt{2}}+\dfrac{6}{2-\sqrt{10}}$.
b) $\dfrac{3}{\sqrt{5}-\sqrt{2}}-\dfrac{2}{2-\sqrt{2}}+\dfrac{1}{\sqrt{3}+\sqrt{2}}$.
c) $\dfrac{-4}{\sqrt{7}-\sqrt{5}}+\dfrac{1}{\sqrt{3}-1}+\dfrac{4-2\sqrt{5}}{\sqrt{5}-2}$.
d) $\dfrac{5}{3-\sqrt{7}}-\dfrac{2}{\sqrt{2}+\sqrt{3}}+\dfrac{-1}{\sqrt{2}-1}$.

Bài 3. Rút gọn

a) $\dfrac{(\sqrt{3}-\sqrt{5})^2+4\sqrt{15}}{\sqrt{3}+\sqrt{5}}$.
b) $(\sqrt{5}+2)\dfrac{(\sqrt{5}+2)^2-8\sqrt{5}}{\sqrt{5}-2}$.
c) $\dfrac{(\sqrt{2}+1)^2-4\sqrt{2}}{\sqrt{2}-1}\cdot(\sqrt{2}+1)$.
d) $\dfrac{(\sqrt{3}-\sqrt{2})^2+4\sqrt{6}}{(\sqrt{3}+\sqrt{2})^2}\cdot(\sqrt{3}-\sqrt{2})$.

Căn bậc hai – Tính chất cơ bản phần 2

Bài 1. Khai triển các biểu thức sau

a) $(\sqrt{x}-1)^2+(\sqrt{x}+1)^2$.
b) $(\sqrt{x}+2)(\sqrt{x}-3)-(\sqrt{x}+1)(2\sqrt{x}-5)$.
c) $(2\sqrt{x}-3)^2+3(\sqrt{x}-1)(\sqrt{x}+2)$.
d) $(3-\sqrt{x})(3+\sqrt{x})+(\sqrt{x}-2)^2$.

Giải

a) $(\sqrt{x}-1)^2+(\sqrt{x}+1)^2$

$= {{(\sqrt{x}-1)}^2}+{{(\sqrt{x}+1)}^2}$

$=x-2\sqrt{x}+1+x+2\sqrt{x}+1=2x+2$.
b) $(\sqrt{x}+2)(\sqrt{x}-3)-(\sqrt{x}+1)(2\sqrt{x}-5)$
$=(\sqrt{x}+2)(\sqrt{x}-3)-(\sqrt{x}+1)(2\sqrt{x}-5)$

$=x-\sqrt{x}-6-2x+3\sqrt{x}+5$

$=-x+2\sqrt{x}-1=-{{\left(\sqrt{x}-1\right)}^2}$.
c) $(2\sqrt{x}-3)^2+3(\sqrt{x}-1)(\sqrt{x}+2)$
$={{(2\sqrt{x}-3)}^2}+3(\sqrt{x}-1)(\sqrt{x}+2)$

$=4x-12\sqrt{x}+9+3\left(x+\sqrt{x}-2\right)$

$=7x-9\sqrt{x}+3$.
d) $(3-\sqrt{x})(3+\sqrt{x})+(\sqrt{x}-2)^2$

$=(3-\sqrt{x})(3+\sqrt{x})+{{(\sqrt{x}-2)}^2}$

$=9-x+x-4\sqrt{x}+4$

$=13-4\sqrt{x}$.

Bài 2. Rút gọn các biểu thức sau:
a) $A=(\sqrt{x}+2)(5-\sqrt{x})-(\sqrt{x}+3)(\sqrt{x}+1)-(3x+4\sqrt{x}+5)$. $(x \geq 0)$
b) $B=(2\sqrt{a}+\sqrt{b})(\sqrt{a}+1)-(2-\sqrt{a b})(\sqrt{a}-1)$. ($a, b \geq 0$)

Giải

a) $A=(\sqrt{x}+2)(5-\sqrt{x})-(\sqrt{x}+3)(\sqrt{x}+1)-(3x+4\sqrt{x}+5)$
$A=(\sqrt{x}+2)(5-\sqrt{x})-(\sqrt{x}+3)(\sqrt{x}+1)-(3x+4\sqrt{x}+5)$
$A=x+3\sqrt{x}+10-\left(x+4\sqrt{x}+3\right)-3x-4\sqrt{x}-5$
$A=x+3\sqrt{x}+10-x-4\sqrt{x}-3-3x-4\sqrt{x}-5$
$A=-3x-5\sqrt{x}+2$

b) $B=(2\sqrt{a}+\sqrt{b})(\sqrt{a}+1)-(2-\sqrt{a b})(\sqrt{a}-1)$
$B=(2\sqrt{a}+\sqrt{b})(\sqrt{a}+1)-(2-\sqrt{ab})(\sqrt{a}-1)$
$B=2a+2\sqrt{a}+\sqrt{ab}+\sqrt{b}-\left(2\sqrt{a}-2-a \sqrt{b}+\sqrt{ab}\right)$

$B=2a+2\sqrt{a}+\sqrt{ab}+\sqrt{b}-2\sqrt{a}+2+a \sqrt{b}-\sqrt{ab}$
$B=2a+\sqrt{b}+2+a \sqrt{b}$

Bài 3. Phân tích các đa thức sau thành nhân tử:

a) $A=x-\sqrt{x}-2$.
b) $B=x-y+3\sqrt{x}-3\sqrt{y}$.
c) $C=\sqrt{a b}+2\sqrt{a}-\sqrt{b}-2$.
d) $D=x\sqrt{x}+x-2\sqrt{x}$.

Giải

a)  $A=x-\sqrt{x}-2={{\left(\sqrt{x}\right)}^2}-1 \left(\sqrt{x}+1\right)$

$=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)$

$=\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)$.
b) $B=x-y+3\sqrt{x}-3\sqrt{y}=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)+3\left(\sqrt{x}-\sqrt{y}\right)$

$=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}+3\right)$.

c)$C=\sqrt{ab}+2\sqrt{a}-\sqrt{b}-2=\sqrt{a}.\sqrt{b}+2\sqrt{a}-\sqrt{b}-2$

$=\sqrt{b}\left(\sqrt{a}-1\right)+2\left(\sqrt{a}-1\right)$

$=\left(\sqrt{a}-1\right)\left(\sqrt{b}+2\right)$.
d)
$D=x\sqrt{x}+x-2\sqrt{x}$
$=x\sqrt{x}-\sqrt{x}+x-\sqrt{x}$
$=\sqrt{x}(x-1)+\sqrt{x}\left(\sqrt{x}-1\right)$
$=\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+\sqrt{x}\left(\sqrt{x}-1\right)$
$=\sqrt{x}\left(\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)$

Bài 4. Rút gọn các biểu thức sau:
a) $\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}$.
b) $\dfrac{x-4\sqrt{x}+4}{x-2\sqrt{x}}$.
c) $\dfrac{x\sqrt{x}+8}{\sqrt{x}+2}-x-4$.
d) $\dfrac{x-4\sqrt{x}-5}{\sqrt{x}+1}$.

Giải

a)Ta có $\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{{{\left(\sqrt{x}-1\right)}^2}}{\left(\sqrt{x}-1\right)}=\sqrt{x}-1$.
b) Ta có $\dfrac{x-4\sqrt{x}+4}{x-2\sqrt{x}}=\dfrac{{{\left(\sqrt{x}-2\right)}^2}}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}}=1-\dfrac{2}{\sqrt{x}}$.
c) Ta có $\dfrac{x\sqrt{x}+8}{\sqrt{x}+2}-x-4=\dfrac{x\sqrt{x}+8-x\sqrt{x}-2\sqrt{x}-4\sqrt{x}-8}{\sqrt{x}+2}=\dfrac{-6\sqrt{x}}{\sqrt{x}+2}$.
d) Ta có $\dfrac{x-4\sqrt{x}-5}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}=\sqrt{x}-5$.

Bài tập rèn luyện

Bài 1. Khai triển

a) $(\sqrt{a}+2)^2 – (\sqrt(a)-1)^2$.

b) $\sqrt{b}(\sqrt{b}+1)^2 – 2b(\sqrt{b}+3)$.

c) $(\sqrt{x}-1)(\sqrt{y}+4)- 2(2\sqrt{x}+1)(2-\sqrt{y})$.

d) $(\sqrt{x}-1)^3 – 3(\sqrt{x}+2)(\sqrt{x}-1) – 2x(\sqrt{x}-1)$.

Bài 2. Cho $x = \sqrt{3} – \sqrt{2}$.
a) Tính giá trị của biểu thức $A = x^2 -4x+1$.
b) Tính giá trị của biểu thức $B = x^4 -x^2+1$.
Bài 3. Rút gọn các biểu thức sau:
a) $\dfrac{{a\sqrt a – 1}}{{\sqrt a – 1}} – \sqrt a $
b) $\dfrac{{x\sqrt x + 8}}{{\sqrt x + 2}} – 2\sqrt x $
Bài 4. Rút gọn các biểu thức sau:

a)  $\dfrac{{a – 1}}{{\sqrt a + 1}} + \dfrac{{4 – a}}{{\sqrt a + 2}}$.
b) $\dfrac{x-3\sqrt{x}+2}{\sqrt{x}-2}+\dfrac{x-5\sqrt{x}+4}{\sqrt{x}-1}$.

 

Khai phương một biểu thức

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Tính chất 1: Với mọi $A$ ta có hằng đẳng thức:

$\sqrt {A^2}=\left| A \right| $

Ví dụ 1: Tính:

a) $\sqrt {(-7)^2}$.

b) $\sqrt {\left ( \sqrt 5 -2 \right )^2}$.

c)$\sqrt {\left ( 3-2\sqrt 3 \right )^2}$.

Giải

a) $\sqrt {(-7)^2}=\left | -7 \right |=7$.

b) $\sqrt {\left ( \sqrt 5 -2 \right )^2}=\left | \sqrt 5 -2 \right |=\sqrt 5-2$.

c) $\sqrt {\left ( 3-2\sqrt 3 \right )^2}=\left | 3-2\sqrt 3 \right |=2\sqrt 3-3$.

Ví dụ 2: Khai căn các biểu thức sau:

a) $\sqrt {4-2\sqrt 3}$, $\sqrt {4+2\sqrt 3}$.

b) $\sqrt {7+2\sqrt 6}$, $\sqrt {13-2\sqrt {12}}$.

Giải

a) $\sqrt {4-2\sqrt 3}=\sqrt {3-2\sqrt 3+1}=\sqrt {\left ( \sqrt 3-1 \right )^2}=\left | \sqrt 3-1 \right
|=\sqrt 3-1$.

$\sqrt {4+2\sqrt 3}=\sqrt {3+2\sqrt 3+1}=\sqrt {\left ( \sqrt 3 +1 \right )^2}=\left | \sqrt 3+1 \right |=\sqrt 3+1$.

b) $\sqrt {7+2\sqrt 6}=\sqrt {6+2\sqrt 6+1}=\sqrt {\left ( \sqrt 6+1 \right )^2}=\left | \sqrt 6+1 \right |=\sqrt 6+1$.

$\sqrt {13-2\sqrt {12}}=\sqrt {12-2\sqrt {12}+1}=\sqrt {\left ( \sqrt {12}-1 \right )^2}=\left | \sqrt {12}-1\right |=\sqrt {12} -1$.

Tính chất 2: Cho $A$, $B$ là các số không âm. Khi đó ta có các đẳng thức sau:

  • $\sqrt {AB}=\sqrt A \sqrt B$.
  • $\sqrt {\dfrac{A}{B} }=\dfrac{\sqrt A}{\sqrt B}$ $(B>0)$.
  • $\sqrt {A^2B}=\left | A \right | \sqrt B$.

Ví dụ 3:  Tính:

a) $\sqrt {25.169}$.

b) $\sqrt {\dfrac {49}{81} }$.

c) $\sqrt {\dfrac {0,16.0,49}{1,21} }$.

Giải

a) $\sqrt {25.169}=\sqrt {25} .\sqrt {169}=5.13=65$

b) $\sqrt {\dfrac {49}{81} }=\dfrac{\sqrt {49}}{\sqrt {81}}=\dfrac{7}{8}$

c) $\sqrt {\dfrac {0,16.0,49}{1,21} }=\dfrac{\sqrt {0,16.0,49}}{\sqrt {1,21}}=\dfrac{\sqrt {0,16}.\sqrt {0,49}}{\sqrt {1,21}}=\dfrac{0,4.0,7}{1.1}=\dfrac{14}{55}$.

Bài tập:

Bài 1:  Rút gọn các biểu thức sau:

a) $3\sqrt 8-4\sqrt {18} $.

b) $\sqrt {125} -2\sqrt {20} -3\sqrt {80}$.

c) $\sqrt {48} -4\sqrt {27} -2\sqrt {75} +\sqrt {108}$.

Bài 2:  Thực hiện các phép tính:

a) $A=\left ( \sqrt 2-1\right )^2+\left ( \sqrt 2+3 \right )^2$.

b) $B=\left (\sqrt 3+2\sqrt 2 \right )^2-\left ( \sqrt 3-\sqrt 2 \right )^2$.

c) $C=\left ( \sqrt 2 +1 \right )^3-\left ( \sqrt 2 -2 \right )^3$.

d) $D=\left ( \sqrt 2 -\sqrt 3 \right )\left ( \sqrt 6+1 \right )-\sqrt 2 \left ( \sqrt 6+3\sqrt 2 \right )$.

Bài 3:  Khai căn các biểu thức sau:

a) $\sqrt {12+2\sqrt {35}}$, $\sqrt {18-2\sqrt  {65}}$.

b) $\sqrt {16+6\sqrt 7}$, $\sqrt {14-6\sqrt 5}$.

c) $\sqrt {27+10\sqrt 2}$, $\sqrt {9+4\sqrt 5}$.

d) $\sqrt {21-2\sqrt {108}}$, $\sqrt {17-2\sqrt {72}}$.

Bài 4: Rút gọn các biểu thức sau:

a) $\sqrt {x+2\sqrt {x-1}}-\sqrt {x-2\sqrt {x-1}}$  với $x \ge 2$.

b) $\sqrt {2m+2\sqrt {2m-1}}-\sqrt {2m-2\sqrt {2m-1}}$.

c) $\sqrt {x+3+4\sqrt {x-1}}+\sqrt {x+8-6\sqrt {x-1}}$.