Question

Giải phương trình sau:

a) \(\frac{1}{x-1}+\frac{2}{x^2+x+1}=\frac{3x^2}{x^3-1}\)

b) \(\frac{1}{2-x}+1=\frac{1}{x+2}-\frac{6-x}{3x^2-12}\)

c) \(\frac{x-2}{x+2}+\frac{3}{2-x}=\frac{2\left(x-11\right)}{x^2-4}\)

d) \(x^2-6x-2+\frac{14}{x^2-6x+7}=0\)

Answer 1

a) ĐKXĐ: $x\neq 1$

PT \(\Leftrightarrow \frac{x^2+x+1+2(x-1)}{(x-1)(x^2+x+1)}=\frac{3x^2}{x^3-1}\)

\(\Leftrightarrow \frac{x^2+3x-1}{x^3-1}=\frac{3x^2}{x^3-1}\)

\(\Rightarrow x^2+3x-1=3x^2\Leftrightarrow 2x^2-3x+1=0\)

\(\Leftrightarrow (x-1)(2x-1)=0\)

Mà $x\neq 1$ nên $2x-1=0\Rightarrow x=\frac{1}{2}$ là nghiệm

b) ĐK: $x\neq \pm 2$

PT \(\Leftrightarrow \frac{3-x}{2-x}=\frac{1}{x+2}-\frac{6-x}{3x^2-12}\)

\(\Leftrightarrow \frac{1}{x+2}-\frac{3-x}{2-x}=\frac{6-x}{3(x^2-4)}\)

\(\Leftrightarrow \frac{1}{x+2}+\frac{3-x}{x-2}=\frac{6-x}{3(x-2)(x+2)}\)

\(\Leftrightarrow \frac{-x^2+2x+4}{(x-2)(x+2)}=\frac{6-x}{3(x-2)(x+2)}\)

\(\Rightarrow 3(-x^2+2x+4)=6-x\)

\(\Leftrightarrow -3x^2+7x+6=0\)

\(\Leftrightarrow (x-3)(3x+2)=0\Rightarrow x=3\) hoặc $x=-\frac{2}{3}$

Vậy........

Answer 2

c) ĐK: $x\neq \pm 2$

PT \(\Leftrightarrow \frac{x-2}{x+2}-\frac{3}{x-2}=\frac{2(x-11)}{x^2-4}\)

\(\Leftrightarrow \frac{(x-2)^2-3(x+2)}{(x+2)(x-2)}=\frac{2(x-11)}{(x-2)(x+2)}\)

\(\Leftrightarrow \frac{x^2-7x-2}{(x-2)(x+2)}=\frac{2x-22}{(x-2)(x+2)}\)

\(\Rightarrow x^2-7x-2=2x-22\)

\(\Leftrightarrow x^2-9x+20=0\Leftrightarrow (x-4)(x-5)=0\Rightarrow x=4\) hoặc $x=5$

(đều thỏa mãn)

d) ĐK: \(x^2-6x+7\neq 0\)

PT \(\Leftrightarrow (x^2-6x+7)+\frac{14}{x^2-6x+7}-9=0\)

\(\Rightarrow (x^2-6x+7)^2-9(x^2-6x+7)+14=0\)

\(\Leftrightarrow (x^2-6x+7-2)(x^2-6x+7-7)=0\)

\(\Leftrightarrow (x^2-6x+5)(x^2-6x)=0\)

\(\Leftrightarrow (x-1)(x-5)x(x-6)=0\)

\(\Rightarrow x\in \left\{1;5;0;6\right\}\) (đều thỏa mãn)

Vậy.........