1. Ví dụ

Ví dụ 1: Giải các phương trình sau:

a/ $(x-1)^{2}=2\left(x^{2}-1\right)$
b/ $2(x+2)^{2}-x^{3}-8=0$
c/ $(x-1)\left(x^{2}+5 x-2\right)-x^{3}+1=0$
d/ $(x-3)^{2}=(2 x+7)^{2}$

Giải

a/ $ (x-1)^{2}=2\left(x^{2}-1\right) $
$\Leftrightarrow (x-1)^{2}-2\left(x^{2}-1\right) = 0 $
$\Leftrightarrow (x-1)[x-1-2(x+1)] = 0 $
$\Leftrightarrow (x-1)(-x-3) = 0 $
$\Leftrightarrow \left[
\begin{array}{l}
x-1=0 \\
-x-3=0
\end{array} \right.
\Leftrightarrow \left[
\begin{array}{l}
x = 1 \\
x=-3
\end{array} \right. $
Vậy $ S = \{ 1; -3 \} $

b/ $2(x+2)^{2}-x^{3}-8=0 $
$\Leftrightarrow 2(x+2)^2 -(x+2)(x^2-2x+4) = 0 $
$\Leftrightarrow (x+2)[2(x+2)-(x^2-2x+4)] = 0 $
$\Leftrightarrow (x+2)(-x^2+4x) = 0 $
$\Leftrightarrow -x(x+2)(x-4) = 0 $
$\Leftrightarrow \left[
\begin{array}{l}
x = 0 \\
x+2=0 \\
x-4 = 0
\end{array} \right.
\Leftrightarrow \left[
\begin{array}{l}
x = 0 \\
x=-2 \\
x=4
\end{array} \right. $
Vậy $ S = \{ 0; -2; 4 \} $

c/ $(x-1)\left(x^{2}+5 x-2\right)-x^{3}+1=0 $
$ \Leftrightarrow (x-1)(x^2+5x-2)-(x-1)(x^2+x+1) = 0 $
$\Leftrightarrow (x-1)[x^2+5x-2-(x^2+x+1)]= 0 $
$\Leftrightarrow (x-1)(4x-3) = 0 $
$\Leftrightarrow \left[
\begin{array}{l}
x-1=0 \\
4x-3 = 0
\end{array} \right.
\Leftrightarrow \left[
\begin{array}{l}
x=1 \\
x=\dfrac{3}{4}
\end{array} \right. $
Vậy $ S = \left \{ 1; \dfrac{3}{4} \right \} $

d/ $(x-3)^{2}=(2 x+7)^{2}$
$ \Leftrightarrow (x-3)^2 – (2x+7)^2 = 0 $
$\Leftrightarrow [(x-3)+(2x+7)][(x-3)-(2x+7)] = 0 $
$\Leftrightarrow (3x+4)(-x-10) = 0 $
$\Leftrightarrow \left[
\begin{array}{l}
3x+4 =0 \\
-x-10= 0
\end{array} \right.
\Leftrightarrow \left[
\begin{array}{l}
x= \dfrac{-4}{3} \\
x= -10
\end{array} \right. $
Vậy $ S = \left \{ \dfrac{-4}{3}; -10 \right \} $

Ví dụ 2: Giải các phương trình sau:

a/ $(2 x-5)^{2}-(x+2)^{2}=0$
b/ $\left(3 x^{2}+10 x-8\right)^{2}=\left(5 x^{2}-2 x+10\right)^{2}$
c/ $\left(x^{2}-2 x+1\right)-4=0$
d/ $\left(x^{2}-9\right)^{2}-9(x-3)^{2}=0$

Giải

a/ $(2 x-5)^{2}-(x+2)^{2}=0 $
$ \Leftrightarrow [(2x-5)+(x+2)][(2x-5)-(x+2)] = 0 $
$\Leftrightarrow (3x-3)(x-7) = 0 $
$\Leftrightarrow \left[
\begin{array}{l}
3x-3=0 \\
x-7=0
\end{array} \right.
\Leftrightarrow \left[
\begin{array}{l}
x=1 \\
x=7
\end{array} \right. $
Vậy $ S = \left \{ 1; 7 \right \} $

b/ $\left(3 x^{2}+10 x-8\right)^{2}=\left(5 x^{2}-2 x+10\right)^{2}$
$\Leftrightarrow \left(3 x^{2}+10 x-8\right)^{2}-\left(5 x^{2}-2 x+10\right)^{2} = 0 $
$ \Leftrightarrow [(3x^2+10x-8)+(5x^2-2x+10)][(3x^2+10x-8)-(5x^2-2x+10)] = 0 $
$\Leftrightarrow (8x^2+8x+2)(-2x^2+12x-18)= 0 $
$\Leftrightarrow -4(2x+1)^2(x-3)^2 = 0 $
$\Leftrightarrow \left[
\begin{array}{l}
2x+1 = 0 \\
x-3=0
\end{array} \right.
\Leftrightarrow \left[
\begin{array}{l}
x=\dfrac{-1}{2} \\
x=3
\end{array} \right. $
Vậy $ S = \left \{ \dfrac{-1}{2}; 3 \right \} $

c/ $\left(x^{2}-2 x+1\right)-4=0 $
$ \Leftrightarrow (x-1)^2-2^2 = 0 $
$\Leftrightarrow (x-1+2)(x-1-2) = 0 $
$\Leftrightarrow (x+1)(x-3) = 0 $
$\Leftrightarrow \left[
\begin{array}{l}
x+1 =0 \\
x-3=0
\end{array} \right.
\Leftrightarrow \left[
\begin{array}{l}
x=-1 \\
x=3
\end{array} \right. $
Vậy $ S = \left \{ -1; 3 \right \} $

d/ $\left(x^{2}-9\right)^{2}-9(x-3)^{2}=0$
$\Leftrightarrow [(x^2-9)+3(x-3)][(x^2-9)-3(x-3)] = 0 $
$\Leftrightarrow (x^2+3x-18)(x^2-3x) =0 $
$\Leftrightarrow (x+6)(x-3)x(x-3) = 0 $
$\Leftrightarrow x(x+6)(x-3)^2 = 0 $
$\Leftrightarrow \left[
\begin{array}{l}
x=0 \\
x+6=0\\
x-3=0
\end{array} \right.
\Leftrightarrow \left[
\begin{array}{l}
x=0 \\
x=-6 \\
x=3
\end{array} \right. $
Vậy $ S = \left \{ 0; -6; 3 \right \} $

Ví dụ 3: Giải các phương trình sau:

a/ $x^{2}-3 x+2=0$
b/ $x^{2}+7 x+12=0$
c/ $x^{2}-3 x-10=0$
d/ $x^{3}-3 x^{2}-3 x+9=0$

Giải

a/ $x^{2}-3 x+2=0$
$ \Leftrightarrow x^2-2x-x+2 =0$
$\Leftrightarrow x(x-2)-(x-2) = 0 $
$\Leftrightarrow (x-2)(x-1) = 0$
$\Leftrightarrow \left[
\begin{array}{l}
x-2 = 0 \\
x-1=0
\end{array} \right.
\Leftrightarrow \left[
\begin{array}{l}
x=2 \\
x=1
\end{array} \right. $
Vậy $ S = \left \{ 2; 1 \right \} $

b/ $x^{2}+7 x+12=0$
$\Leftrightarrow x^2+3x+4x+12 = 0 $
$\Leftrightarrow x(x+3)+4(x+3) = 0$
$\Leftrightarrow (x+3)(x+4) = 0 $
$\Leftrightarrow \left[
\begin{array}{l}
x+3=0\\
x+4=0
\end{array} \right.
\Leftrightarrow \left[
\begin{array}{l}
x=-3 \\
x=-4
\end{array} \right. $
Vậy $ S = \left \{ -3; -4 \right \} $

c/ $x^{2}-3 x-10=0$
$\Leftrightarrow x^2-5x+2x-10 = 0 $
$\Leftrightarrow x(x-5)+2(x-5)=0 $
$\Leftrightarrow (x-5)(x+2)=0 $
$\Leftrightarrow \left[
\begin{array}{l}
x-5=0\\
x+2=0
\end{array} \right.
\Leftrightarrow \left[
\begin{array}{l}
x=5 \\
x=-2
\end{array} \right. $
Vậy $ S = \left \{ -2; 5 \right \} $

d/ $x^{3}-3 x^{2}-3 x+9=0$
$\Leftrightarrow x^2(x-3)-3(x-3) =0 $
$\Leftrightarrow (x-3)(x^2-3) = 0 $
$\Leftrightarrow (x-3)(x+\sqrt{3})(x-\sqrt{3}) = 0 $
$\Leftrightarrow \left[
\begin{array}{l}
x-3=0\\
x+\sqrt{3}=0\\
x-\sqrt{3}=0
\end{array} \right.
\Leftrightarrow \left[
\begin{array}{l}
x=3 \\
x=-\sqrt{3} \\
x=\sqrt{3}
\end{array} \right. $
Vậy $ S = \left \{ 3; -\sqrt{3}; \sqrt{3} \right \} $

2. Bài tập tự luyện

Bài 1: Giải các phương trình sau:

a/ $9(x-3)^{2}=4(x+2)^{2}$
b/ $\left(4 x^{2}-3 x-18\right)^{2}=\left(4 x^{2}+3 x\right)^{2}$
c/ $(2 x-1)^{2}=49$
d/ $(5 x-3)^{2}-(4 x-7)^{2}=0$
e/ $(2 x+7)^{2}=9(x+2)^{2}$
f/ $4(2 x+7)^{2}=9(x+3)^{2}$

Bài 2: Giải các phương trình sau:

a/ $3 x^{2}+2 x-1=0$
b/ $x^{2}-5 x+6=0$
c/ $x^{2}-3 x+2=0$
d/ $2 x^{2}-6 x+1=0$
e/ $4 x^{2}-12 x+5=0$
f/ $2 x^{2}+5 x+3=0$

Bài 3: Giải các phương trình sau:

a/ $3 x^{2}+12 x-66=0$
b/ $9 x^{2}-30 x+25=0$
c/ $x^{2}+3 x-10=0$
d/ $3 x^{2}-7 x+1=0$
e/ $3 x^{2}-7 x+8=0$
f/ $4 x^{2}-12 x+9=0$

Bài 4: Giải các phương trình sau:

a/ $2 x^{2}-6 x+1=0$
b/ $3 x^{2}+4 x-4=0$
c/ $x^{3}-8 x^{2}+21 x-18=0$
d/ $x^{4}+x^{2}+6 x-8=0$
e/ $ x^4 +2x^3-4x^2-5x-6 = 0 $
f/ $x^4-10x^3+15x^2-50x+24 = 0 $