1. Kiến thức cần nhớ
Tìm điều kiện xác định (ĐKXĐ) của phương trình chứa ẩn ở mẫu là tìm các giá trị của ẩn để tất cả các mẫu của phương trình đều khác $0$.

Phương pháp: Các bước giải phương trình chứa ẩn ở mẫu:

• • Bước 1: Tìm điều kiện xác định.
  • Bước 2: Quy đồng mẫu hai vế và khử mẫu.
  • Bước 3: Giải phương trình vừa nhận được.
  • Bước 4: Xem xét các giá trị của ẩn vừa tìm được có thỏa mãn ĐKXĐ không và kết luận về nghiệm của phương trình.

2. Ví dụ

Ví dụ 1: Giải các phương trình sau:

a/ ${\displaystyle \frac{4}{x-1}}-{\displaystyle \frac{5}{x-2}}=-3$

b/ $3x-{\displaystyle \frac{1}{x-2}}={\displaystyle \frac{x-1}{2-x}}$

c/ ${\displaystyle \frac{x+4}{x^2-3x+2}}+{\displaystyle \frac{x+1}{x^2-4x+3}}={\displaystyle \frac{2x+5}{x^2-4x+3}}$

d/ ${\displaystyle \frac{2}{x^2-4}}-{\displaystyle \frac{1}{x(x-2)}}+{\displaystyle \frac{x-4}{x(x+2)}}=0$

Giải

a/ ${\displaystyle \frac{4}{x-1}}-{\displaystyle \frac{5}{x-2}}=-3$

ĐKXĐ:
$\{\begin{array}{l}x-1\ne 0\\ x-2\ne 0\end{array}\iff \{\begin{array}{l}x\ne 1\\ x\ne 2\end{array}$

${\displaystyle \frac{4}{x-1}}-{\displaystyle \frac{5}{x-2}}=-3$

$\iff {\displaystyle \frac{4(x-2)-5(x-1)}{(x-1)(x-2)}}={\displaystyle \frac{-3(x-1)(x-2)}{(x-1)(x-2)}}$

$\Rightarrow 4x-8-5x+5=-3(x^2-3x+2)$
$\iff -x-3+3(x^2-3x+2)=0$
$\iff 3x^2-10x+3=0$
$\iff (3x-1)(x-3)=0$
$\iff [\begin{array}{l}3x-1=0\\ x-3=0\end{array}\iff [\begin{array}{lc}x={\displaystyle \frac{1}{3}}& \text{(nhận)}\\ x=3& \text{(nhận)}\end{array}$
Vậy $S=\{{\displaystyle \frac{1}{3}};3\}$

 

b/ $3x-{\displaystyle \frac{1}{x-2}}={\displaystyle \frac{x-1}{2-x}}$

ĐKXĐ:
$\{\begin{array}{l}x-2\ne 0\\ 2-x\ne 0\end{array}\iff x\ne 0$

$3x-{\displaystyle \frac{1}{x-2}}={\displaystyle \frac{x-1}{2-x}}$

$\iff {\displaystyle \frac{3x(x-2)-1}{x-2}}=-{\displaystyle \frac{x-1}{x-2}}$

$\Rightarrow 3x^2-6x-1=-x+1$
$\iff 3x^2-5x-2=0$
$\iff (3x+1)(x-2)=0$
$\iff [\begin{array}{l}3x+1=0\\ x-2=0\end{array}\iff [\begin{array}{lc}x=–{\displaystyle \frac{1}{3}}& \text{(nhận)}\\ x=2& \text{(loại)}\end{array}$
Vậy $S=\{-{\displaystyle \frac{1}{3}};\}$

 

c/ ${\displaystyle \frac{x+4}{x^2-3x+2}}+{\displaystyle \frac{x+1}{x^2-4x+3}}={\displaystyle \frac{2x+5}{x^2-4x+3}}$

$\iff {\displaystyle \frac{x+4}{(x-2)(x-1)}}+{\displaystyle \frac{x+1}{(x-3)(x-1)}}={\displaystyle \frac{2x+5}{(x-3)(x-1)}}$

ĐKXĐ:
$\{\begin{array}{l}x-2\ne 0\\ x-1\ne 0\\ x-3\ne 0\end{array}\iff \{\begin{array}{l}x\ne 2\\ x\ne 1\\ x\ne 3\end{array}$

$\iff {\displaystyle \frac{x+4}{(x-2)(x-1)}}+{\displaystyle \frac{x+1}{(x-3)(x-1)}}={\displaystyle \frac{2x+5}{(x-3)(x-1)}}$

$\iff {\displaystyle \frac{(x+4)(x-3)+(x+1)(x-2)}{(x-1)(x-2)(x-3)}}={\displaystyle \frac{(2x+5)(x-2)}{(x-1)(x-2)(x-3)}}$

$\Rightarrow (x^2+x-12)+(x^2-x-2)=(2x^2+x-10)$
$\iff (x^2+x-12)+(x^2-x-2)–(2x^2+x-10)=0$
$\iff -x–4=0$
$\iff x=-4$ (nhận)
Vậy $S=\{-4\}$

 

d/ ${\displaystyle \frac{2}{x^2-4}}-{\displaystyle \frac{1}{x(x-2)}}+{\displaystyle \frac{x-4}{x(x+2)}}=0$

$\iff {\displaystyle \frac{2}{(x+2)(x-2)}}-{\displaystyle \frac{1}{x(x-2)}}+{\displaystyle \frac{x-4}{x(x+2)}}=0$

ĐKXĐ:
$\{\begin{array}{l}x-2\ne 0\\ x+2\ne 0\\ x\ne 0\end{array}\iff \{\begin{array}{l}x\ne 2\\ x\ne -2\\ x\ne 0\end{array}$

$\iff {\displaystyle \frac{2}{(x+2)(x-2)}}-{\displaystyle \frac{1}{x(x-2)}}+{\displaystyle \frac{x-4}{x(x+2)}}=0$

$\iff {\displaystyle \frac{2x-(x+2)+(x-4)(x-2)}{x(x-2)(x+2)}}=0$

$\Rightarrow 2x-x-2+x^2-4x-2x-8=0$
$\iff x^2-5x-8=0$
$\iff (x-2)(x-3)=0$
$\iff [\begin{array}{l}x-2=0\\ x-3=0\end{array}\iff [\begin{array}{lc}x=2& \text{(loại)}\\ x=3& \text{(nhận)}\end{array}$
Vậy $S=\left\{3\right\}$

Ví dụ 2: Giải các phương trình sau:

a/ ${\displaystyle \frac{x^2+1}{x+1}}+{\displaystyle \frac{x^2+2}{x-2}}=-2$

b/ ${\displaystyle \frac{1}{x-1}}+{\displaystyle \frac{2x^2-5}{x^3-1}}={\displaystyle \frac{4}{x^2+x+1}}$

c/ ${\displaystyle \frac{12x+1}{6x-2}}-{\displaystyle \frac{9x-5}{3x+1}}={\displaystyle \frac{108x-36x^2-9}{4(9x^2-1)}}$

d/ $x+{\displaystyle \frac{1}{x}}=x^2+{\displaystyle \frac{1}{x^2}}$

Giải

a/ ${\displaystyle \frac{x^2+1}{x+1}}+{\displaystyle \frac{x^2+2}{x-2}}=-2$

ĐKXĐ:
$\{\begin{array}{l}x+1\ne 0\\ x-2\ne 0\end{array}\iff \{\begin{array}{l}x\ne -1\\ x\ne 2\end{array}$

${\displaystyle \frac{x^2+1}{x+1}}+{\displaystyle \frac{x^2+2}{x-2}}=-2$

$\iff {\displaystyle \frac{(x^2+1)(x-2)+(x^2+2)(x+1)}{(x+1)(x-2)}}={\displaystyle \frac{-2(x+1)(x-2)}{(x+1)(x-2)}}$

$\Rightarrow x^3-2x^2+x-2+x^3+x^2+2x+2=-2(x^2-x-2)$
$\iff 2x^3+x^2+x-4=0$
$\iff 2x^3-2x^2+3x^2-3x+4x-4=0$
$\iff 2x^2(x-1)+3x(x-1)+4(x-1)=0$
$\iff (x-1)(2x^2+3x+4)=0$
$\iff [\begin{array}{l}x-1=0\\ 2x^2+3x+4=0\end{array}$

$\iff [\begin{array}{ll}x=1& \text{(nhận)}\\ 2{(x+{\displaystyle \frac{3}{4}})}^2+{\displaystyle \frac{23}{8}}=0& \Rightarrow \text{ Phương trình vô nghiệm vì }2{(x+{\displaystyle \frac{3}{4}})}^2+{\displaystyle \frac{23}{8}}⩾{\displaystyle \frac{23}{8}}\end{array}$

Vậy $S=\left\{1\right\}$

 

b/ ${\displaystyle \frac{1}{x-1}}+{\displaystyle \frac{2x^2-5}{x^3-1}}={\displaystyle \frac{4}{x^2+x+1}}$

$\iff {\displaystyle \frac{1}{x-1}}+{\displaystyle \frac{2x^2-5}{(x-1)(x^2+x+1)}}={\displaystyle \frac{4}{x^2+x+1}}$

ĐKXĐ:
$\{\begin{array}{l}x-1\ne 0\\ x^2+x+1\ne 0\end{array}\iff \{\begin{array}{l}x\ne 1\\ {(x+{\displaystyle \frac{1}{2}})}^2+{\displaystyle \frac{3}{4}}>0\end{array}$

$\iff {\displaystyle \frac{1}{x-1}}+{\displaystyle \frac{2x^2-5}{(x-1)(x^2+x+1)}}={\displaystyle \frac{4}{x^2+x+1}}$

$\iff {\displaystyle \frac{(x^2+x+1)+(2x^2-5)}{(x-1)(x^2+x+1)}}={\displaystyle \frac{4(x-1)}{(x-1)(x^2+x+1)}}$

$\Rightarrow (x^2+x+1)+(2x^2-5)=4x-4$
$\iff 3x^2-3x=0$
$\iff 3x(x-1)=0$
$\iff [\begin{array}{l}x=0\\ x-1=0\end{array}\iff [\begin{array}{lc}x=0& \text{(nhận)}\\ x=1& \text{(loại)}\end{array}$
Vậy $S=\left\{0\right\}$

 

c/ ${\displaystyle \frac{12x+1}{6x-2}}-{\displaystyle \frac{9x-5}{3x+1}}={\displaystyle \frac{108x-36x^2-9}{4(9x^2-1)}}$

$\iff {\displaystyle \frac{12x+1}{2(3x-1)}}-{\displaystyle \frac{9x-5}{3x+1}}={\displaystyle \frac{108x-36x^2-9}{4(3x+1)(3x-1)}}$

ĐKXĐ:
$\{\begin{array}{l}3x-1\ne 0\\ 3x+1\ne 0\end{array}\iff \{\begin{array}{l}x\ne {\displaystyle \frac{1}{3}}\\ x\ne -{\displaystyle \frac{1}{3}}\end{array}$

$\iff {\displaystyle \frac{12x+1}{2(3x-1)}}-{\displaystyle \frac{9x-5}{3x+1}}={\displaystyle \frac{108x-36x^2-9}{4(3x+1)(3x-1)}}$

$\iff {\displaystyle \frac{2(3x+1)(12x+1)-4(3x-1)(9x-5)}{4(3x+1)(3x-1_{)}}}={\displaystyle \frac{108x-36x^2-9}{4(3x+1)(3x-1)}}$

$\Rightarrow 2(36x^2+15x+1)-4(27x^2-24x+5)=108x-36x^2-9$
$\iff 18x=9$
$\iff x={\displaystyle \frac{1}{2}}$ (nhận)
Vậy $S=\left\{{\displaystyle \frac{1}{2}}\right\}$

 

d/ $x+{\displaystyle \frac{1}{x}}=x^2+{\displaystyle \frac{1}{x^2}}$

ĐKXĐ: $x\ne 0$

$x+{\displaystyle \frac{1}{x}}=x^2+{\displaystyle \frac{1}{x^2}}$

$\iff {\displaystyle \frac{x^3+x}{x^2}}={\displaystyle \frac{x^4+1}{1}}{x}^2$

$\Rightarrow x^3+x=x^4+1$
$\iff -x^4+x^3+x-1=0$
$\iff -x^3(x-1)+(x-1)=0$
$\iff (x-1)(-x^3+1)=0$
$\iff (x-1)(1-x)(1+x+x^2)=0$
$\iff [\begin{array}{l}x-1=0\\ 1-x=0\\ 1+x+x^2=0\end{array}$

$\iff [\begin{array}{ll}x=1& \text{(nhận)}\\ {(x+{\displaystyle \frac{1}{2}})}^2+{\displaystyle \frac{3}{4}}=0& \Rightarrow \text{ Phương trình vô nghiệm vì }{(x+{\displaystyle \frac{1}{2}})}^2+{\displaystyle \frac{3}{4}}⩾{\displaystyle \frac{3}{4}}\end{array}$
Vậy $S=\left\{1\right\}$

 Ví dụ 3: Giải các phương trình sau:

a/ ${\displaystyle \frac{x}{x+1}}+{\displaystyle \frac{x+1}{x+2}}+{\displaystyle \frac{x+2}{x}}={\displaystyle \frac{25}{6}}$

b/ $x^2+{\displaystyle \frac{2x}{x-1}}=8$

c/ ${\displaystyle \frac{2}{x-14}}-{\displaystyle \frac{5}{x-13}}={\displaystyle \frac{2}{x-9}}-{\displaystyle \frac{5}{x-11}}$

d/ ${\displaystyle \frac{1}{x^2+5x+6}}+{\displaystyle \frac{1}{x^2+7x+12}}+{\displaystyle \frac{1}{x^2+9x+20}}+{\displaystyle \frac{1}{x^2+11x+30}}={\displaystyle \frac{1}{8}}$

Giải

a/ ${\displaystyle \frac{x}{x+1}}+{\displaystyle \frac{x+1}{x+2}}+{\displaystyle \frac{x+2}{x}}={\displaystyle \frac{25}{6}}$

ĐKXĐ:
$\{\begin{array}{l}x+1\ne 0\\ x+2\ne 0\\ x\ne 0\end{array}\iff \{\begin{array}{l}x\ne -1\\ x\ne -2\\ x\ne 0\end{array}$

${\displaystyle \frac{x}{x+1}}+{\displaystyle \frac{x+1}{x+2}}+{\displaystyle \frac{x+2}{x}}={\displaystyle \frac{25}{6}}$

$\iff {\displaystyle \frac{6x^2(x+2)+6x(x+1{)}^2+6(x+1)(x+2{)}^2}{6x(x+1)(x+2)}}={\displaystyle \frac{25x(x+1)(x+2)}{6x(x+1)(x+2)}}$

$\Rightarrow 6x^3+12x^2+6x(x^2+2x+1)+6(x+1)(x^2+4x+4)=25x(x^2+3x+2)$

$\iff 6x^3+12x^2+6x^3+12x^2+6x+6x^3+24x^2+24x+6x^2+24x+24=25x^3+75x^2+50x$
$\iff 7x^3+21x^2-4x-24=0$
$\iff 7x^3-7x^2+28x^2-28x+24x-24=0$
$\iff 7x^2(x-1)+28x(x-1)+24(x-1)=0$
$\iff 7(x-1)(x^2+4x+{\displaystyle \frac{24}{7}})=0$
$\iff 7(x-1)[(x+2{)}^2-{\displaystyle \frac{4}{7}}]=0$
$\iff 7(x-1)(x+2+{\displaystyle \frac{2}{\sqrt{7}}})(x+2-{\displaystyle \frac{2}{\sqrt{7}}})=0$
$\iff [\begin{array}{l}x-1=0\\ x+2+{\displaystyle \frac{2}{\sqrt{7}}}=0\\ x+2-{\displaystyle \frac{2}{\sqrt{7}}}=0\end{array}\iff [\begin{array}{ll}x=1& \text{(nhận)}\\ x=-2-{\displaystyle \frac{2}{\sqrt{7}}}& \text{(nhận)}\\ x=-2+{\displaystyle \frac{2}{\sqrt{7}}}& \text{(nhận)}\end{array}$
Vậy $S=\{1;-2-{\displaystyle \frac{2}{\sqrt{7}}};-2+{\displaystyle \frac{2}{\sqrt{7}}}\}$

 

b/ $x^2+{\displaystyle \frac{2x}{x-1}}=8$

ĐKXĐ: $x-1\ne 0\iff x\ne 1$

$x^2+{\displaystyle \frac{2x}{x-1}}=8$

$\iff {\displaystyle \frac{x^2(x-1)+2x}{x-1}}={\displaystyle \frac{8(x-1)}{x-1}}$

$\Rightarrow x^3-x^2+2x=8x-8$
$\iff x^3-x^2-6x+8=0$
$\iff x^3-2x^2+x^2-2x-4x+8=0$
$\iff x^2(x-2)+x(x-2)-4(x-2)=0$
$\iff (x-2)(x^2+x-4)=0$
$\iff (x-2)[{(x+{\displaystyle \frac{1}{2}})}^2-{\displaystyle \frac{17}{4}}]=0$
$\iff (x-2)(x+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{\sqrt{17}}{2}})(x+{\displaystyle \frac{1}{2}}-{\displaystyle \frac{\sqrt{17}}{2}})=0$
$\iff [\begin{array}{l}x-2=0\\ x+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{\sqrt{17}}{2}}=0\\ x+{\displaystyle \frac{1}{2}}-{\displaystyle \frac{\sqrt{17}}{2}}=0\end{array}\iff [\begin{array}{ll}x=2& \text{(nhận)}\\ x=-{\displaystyle \frac{1}{2}}-{\displaystyle \frac{\sqrt{17}}{2}}& \text{(nhận)}\\ x=-{\displaystyle \frac{1}{2}}+{\displaystyle \frac{\sqrt{17}}{2}}& \text{(nhận)}\end{array}$

Vậy $S=\{2;-{\displaystyle \frac{1}{2}}-{\displaystyle \frac{\sqrt{17}}{2}};-{\displaystyle \frac{1}{2}}+{\displaystyle \frac{\sqrt{17}}{2}}\}$

 

c/ ${\displaystyle \frac{2}{x-14}}-{\displaystyle \frac{5}{x-13}}={\displaystyle \frac{2}{x-9}}-{\displaystyle \frac{5}{x-11}}$

ĐKXĐ:
$\{\begin{array}{l}x-14\ne 0\\ x-13\ne 0\\ x-9\ne 0\\ x-11\ne 0\end{array}\iff \{\begin{array}{l}x\ne 14\\ x\ne 13\\ x\ne 9\\ x\ne 11\end{array}$

${\displaystyle \frac{2}{x-14}}-{\displaystyle \frac{5}{x-13}}={\displaystyle \frac{2}{x-9}}-{\displaystyle \frac{5}{x-11}}$

$\iff 2({\displaystyle \frac{1}{x-14}}-{\displaystyle \frac{1}{x-9}})=5({\displaystyle \frac{1}{x-13}}-{\displaystyle \frac{1}{x-11}})$

$\iff 2{\displaystyle \frac{(x-9)-(x-14)}{(x-14)(x-9)}}=5{\displaystyle \frac{(x-11)-(x-13)}{(x-11)(x-13)}}$

$\iff {\displaystyle \frac{10}{(x-14)(x-9)}}={\displaystyle \frac{10}{(x-13)(x-11)}}$

$\Rightarrow (x-14)(x-9)=(x-13)(x-11)$
$\iff x^2-23x+126=x^2-24x+143$
$\iff x=17$ (nhận)
Vậy $S=\left\{17\right\}$

 

d/ ${\displaystyle \frac{1}{x^2+5x+6}}+{\displaystyle \frac{1}{x^2+7x+12}}+{\displaystyle \frac{1}{x^2+9x+20}}+{\displaystyle \frac{1}{x^2+11x+30}}={\displaystyle \frac{1}{8}}$

${\displaystyle \frac{1}{(x+2)(x+3)}}+{\displaystyle \frac{1}{(x+3)(x+4)}}+{\displaystyle \frac{1}{(x+4)(x+5)}}+{\displaystyle \frac{1}{(x+5)(x+6)}}={\displaystyle \frac{1}{8}}$

ĐKXĐ:
$\{\begin{array}{l}x+2\ne 0\\ x+3\ne 0\\ x+4\ne 0\\ x+5\ne 0\\ x+6\ne 0\end{array}\iff \{\begin{array}{l}x\ne -2\\ x\ne -3\\ x\ne -4\\ x\ne -5\\ x\ne -6\end{array}$

${\displaystyle \frac{1}{(x+2)(x+3)}}+{\displaystyle \frac{1}{(x+3)(x+4)}}+{\displaystyle \frac{1}{(x+4)(x+5)}}+{\displaystyle \frac{1}{(x+5)(x+6)}}={\displaystyle \frac{1}{8}}$

$\iff {\displaystyle \frac{(x+4)+(x+2)}{(x+2)(x+3)(x+4)}}+{\displaystyle \frac{(x+6)+(x+4)}{(x+4)(x+5)(x+6)}}={\displaystyle \frac{1}{8}}$

$\iff {\displaystyle \frac{2(x+3)}{(x+2)(x+3)(x+4)}}+{\displaystyle \frac{2(x+5)}{(x+4)(x+5)(x+6)}}={\displaystyle \frac{1}{8}}$

$\iff {\displaystyle \frac{2}{(x+2)(x+4)}}+{\displaystyle \frac{2}{(x+4)(x+6)}}={\displaystyle \frac{1}{8}}$

$\iff {\displaystyle \frac{2(x+6)+2(x+2)}{(x+2)(x+4)(x+6)}}={\displaystyle \frac{1}{8}}$

$\iff {\displaystyle \frac{4(x+4)}{(x+2)(x+4)(x+6)}}={\displaystyle \frac{1}{8}}$

$\leftrightarrow {\displaystyle \frac{4}{(x+2)(x+6)}}={\displaystyle \frac{1}{8}}$

$\Rightarrow (x+2)(x+6)=32$
$\iff x^2+8x-20=0$
$\iff (x+10)(x-2)=0$
$\iff [\begin{array}{l}x+10=0\\ x-2=0\end{array}\iff [\begin{array}{lc}x=-10& \text{(nhận)}\\ x=2& \text{(nhận)}\end{array}$
Vậy $S=\{-10;2\}$

 

3. Bài tập tự luyện

Bài 1: Giải các phương trình sau:

a/ ${\displaystyle \frac{3x^2+7x-10}{x}}=0$

b/ ${\displaystyle \frac{4x-17}{2x^2+1}}=0$

c/ ${\displaystyle \frac{x-6}{x-4}}={\displaystyle \frac{x}{x-2}}$

d/ $1+{\displaystyle \frac{2x-5}{x-2}}-{\displaystyle \frac{3x-5}{x-1}}=0$

e/ ${\displaystyle \frac{x-3}{x-2}}-{\displaystyle \frac{x-2}{x-4}}=3{\displaystyle \frac{1}{5}}$

f/ ${\displaystyle \frac{x-3}{x-2}}+{\displaystyle \frac{x-2}{x-4}}=-1$

g/ ${\displaystyle \frac{3x-2}{x+7}}={\displaystyle \frac{6x+1}{2x-3}}$

h/ ${\displaystyle \frac{x+1}{x-2}}-{\displaystyle \frac{x-1}{x+2}}={\displaystyle \frac{2(x^2+2)}{x^2-4}}$

Bài 2: Giải các phương trình sau:

a/ ${\displaystyle \frac{2x+1}{x-1}}={\displaystyle \frac{5(x-1)}{x+1}}$

b/ ${\displaystyle \frac{x-1}{x+2}}-{\displaystyle \frac{x}{x-2}}={\displaystyle \frac{5x-2}{4-x^2}}$

c/ ${\displaystyle \frac{x-2}{2+x}}-{\displaystyle \frac{3}{x-2}}={\displaystyle \frac{2(x-11)}{x^2-4}}$

d/ $\frac{x-1}{x+1}-{\displaystyle \frac{x^2+x-2}{x+1}}={\displaystyle \frac{x+1}{x-1}}-x-2$

e/ ${\displaystyle \frac{x+1}{x-1}}-{\displaystyle \frac{x-1}{x+1}}={\displaystyle \frac{4}{x^2-1}}$

f/ ${\displaystyle \frac{3}{4(x-5)}}+{\displaystyle \frac{15}{50-2x^2}}=-{\displaystyle \frac{7}{6(x+5)}}$

g/ ${\displaystyle \frac{8x^2}{3(1-4x^2)}}={\displaystyle \frac{2x}{6x-3}}-{\displaystyle \frac{1+8x}{4+8x}}$

h/ ${\displaystyle \frac{13}{(x-3)(2x+7)}}+{\displaystyle \frac{1}{2x+7}}={\displaystyle \frac{6}{x^2-9}}$

Bài 3: Giải các phương trình sau:

a/ ${\displaystyle \frac{x+1}{x-1}}-{\displaystyle \frac{x-1}{x+1}}={\displaystyle \frac{16}{x^2-1}}$

b/ ${\displaystyle \frac{12}{x^2-4}}-{\displaystyle \frac{x+1}{x-2}}+{\displaystyle \frac{x+7}{x+2}}=0$

c/ ${\displaystyle \frac{12}{8+x^3}}=1+{\displaystyle \frac{1}{x+2}}$

d/ ${\displaystyle \frac{x+25}{2x^2-50}}-{\displaystyle \frac{x+5}{x^2-5x}}={\displaystyle \frac{5-x}{2x^2+10x}}$

e/ ${\displaystyle \frac{4}{x^2+2x-3}}={\displaystyle \frac{2x-5}{x+3}}-{\displaystyle \frac{2x}{x-1}}$

f/ ${\displaystyle \frac{3}{x^2+x-2}}-{\displaystyle \frac{1}{x-1}}={\displaystyle \frac{-7}{x+2}}$

Bài 4: Giải các phương trình sau:

a/ ${\displaystyle \frac{2}{-x^2+6x-8}}-{\displaystyle \frac{x-1}{x-2}}=\frac{x+3}{x-4}$

b/ ${\displaystyle \frac{2}{x^3-x^2-x+1}}={\displaystyle \frac{3}{1-x^2}}-{\displaystyle \frac{1}{x+1}}$

c/ ${\displaystyle \frac{x+2}{x-2}}-{\displaystyle \frac{2}{x^2-2x}}={\displaystyle \frac{1}{x}}$

d/ ${\displaystyle \frac{5}{-x^2+5x-6}}+{\displaystyle \frac{x+3}{2-x}}=0$

e/ ${\displaystyle \frac{x}{2x+2}}-{\displaystyle \frac{2x}{x^2-2x-3}}={\displaystyle \frac{x}{6-2x}}$

f/ ${\displaystyle \frac{1}{x-1}}-{\displaystyle \frac{3x^2}{x^3-1}}={\displaystyle \frac{2x}{x^2+x+1}}$