\(\frac{1}{a+b-x}=\frac{1}{a}+\frac{1}{b}-\frac{1}{x}\) (ĐKXĐ: x \(\ne\) 0 và x \(\ne\) a + b)

<=> \(\frac{1}{a+b-x}+\frac{1}{x}-\frac{1}{a}-\frac{1}{b}=0\)

<=> \(\frac{x}{x\left(a+b-x\right)}+\frac{a+b-x}{x\left(a+b-x\right)}-\frac{b}{ab}-\frac{a}{ab}\)

<=> \(\frac{a+b}{x\left(a+b-x\right)}-\frac{a+b}{ab}=0\)

<=> \(\left(a+b\right)\left(\frac{1}{x\left(a+b-x\right)}-\frac{1}{ab}\right)=0\)

* Nếu a = - b thì tập nghiệm cuả pt là S = R

* Nếu a \(\ne\) b thì \(\frac{1}{x\left(a+b-x\right)}-\frac{1}{ab}=0\)

<=> \(\frac{ab}{abx\left(a+b-x\right)}-\frac{x\left(a+b-x\right)}{abx\left(a+b-x\right)}=0\)

<=> \(\frac{ab-\text{ax}-bx+x^2}{abx\left(a+b-x\right)}=0\)

<=> \(\frac{b\left(a-x\right)-x\left(a-x\right)}{abx\left(a+b-x\right)}=0\)

<=> \(\frac{\left(a-x\right)\left(b-x\right)}{abx\left(a+b-x\right)}=0\)

<=> \(\left[\begin{matrix}a-x=0\\b-x=0\end{matrix}\right.\)

<=> \(\left[\begin{matrix}x=a\\x=b\end{matrix}\right.\)

Vậy tập nghiệm của pt là S = {a ; b}

\(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{3}{x\left(x^4+x^2+1\right)}\) (ĐKXĐ: x \(\ne\) 0

<=> \(\frac{x\left(x+1\right)\left(x^2-x+1\right)}{x\left(x^2+x+1\right)\left(x^2-x+1\right)}-\frac{x\left(x-1\right)\left(x^2+x+1\right)}{x\left(x^2-x+1\right)\left(x^2+x+1\right)}=\frac{3}{x\left(x^2+x+1\right)\left(x^2-x+1\right)}\)

=> \(\left(x^4+x\right)-\left(x^4-x\right)=3\)

<=> \(2x-3=0\)

<=> \(x=\frac{3}{2}\) (nhận)

Vậy S = {1,5}

a) Qui đồng rồi khử mẫu ta được:

   3(3x+2)-(3x+1)=2x.6+5.2

<=> 9x+6-3x-1 = 12x+10

<=> 9x-3x-12x  = 10-6+1

<=> -6x            = 5

<=> x               = -5/6

Vậy ....

b) ĐKXĐ: \(x\ne\pm2\)

Qui đồng rồi khử mẫu ta được:

   (x+1)(x+2)+(x-1)(x-2) = 2(x²+2)

<=> x²+3x+2+x²-3x+2 = 2x²+4

<=> x²+x²-2x²+3x-3x = 4-2-2

<=> 0x             = 0

<=> x vô số nghiệm

Vậy x vô số nghiệm với x khác 2 và x khác -2

c) \(\left(2x+3\right)\left(\frac{3x+7}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\) (ĐKXĐ:x khắc 2/7)

\(\Leftrightarrow\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)-\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)=0\)

\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left[\left(2x+3\right)-\left(x-5\right)\right]=0\)

\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left(x+8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{3x+8}{2-7x}+1=0\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{3x+8}{2-7x}=-1\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}3x+8=-1\left(2-7x\right)\\x=0-8\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x+8=-2+7x\\x=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}-4x=-10\\x=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-8\end{cases}}}\) (nhận)

Vậy ...... 

d) (x+1)²-4(x²-2x+1) = 0

<=> x²+2x+1-4x²+8x-4 = 0

<=> -3x²+10x-3 = 0

giải phương trình

Bài 2:

a) ĐK: $x\geq \pm \frac{1}{2}; x\neq 0$

\(\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}=\frac{(2x+1)^2-(2x-1)^2}{(2x-1)(2x+1)}.\frac{10x-5}{4x}\)

\(\frac{4x^2+4x+1-(4x^2-4x+1)}{(2x-1)(2x+1)}.\frac{5(2x-1)}{4x}=\frac{8x}{(2x-1)(2x+1)}.\frac{5(2x-1)}{4x}\)

\(=\frac{10}{2x+1}\)

b) ĐK : $x\neq 0;-1$

\(\left(\frac{1}{x^2+x}-\frac{2-x}{x+1}\right):\left(\frac{1}{x}+x-2\right)=\left(\frac{1}{x(x+1)}-\frac{x(2-x)}{x(x+1)}\right):\frac{1+x^2-2x}{x}\)

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

Bài 3:
a) ĐKXĐ: \(x\neq \pm 1\)

b)

\(A=\left(\frac{x+1}{2x-2}-\frac{3}{1-x^2}-\frac{x+3}{2x+2}\right).\frac{4x^2-4}{5}\)

\(=\left[\frac{(x+1)^2}{2(x-1)(x+1)}+\frac{6}{2(x-1)(x+1)}-\frac{(x+3)(x-1)}{2(x+1)(x-1)}\right].\frac{4(x^2-1)}{5}\)

\(=\frac{(x+1)^2+6-(x^2+2x-3)}{2(x-1)(x+1)}.\frac{4(x-1)(x+1)}{5}\)

\(=\frac{10}{2(x-1)(x+1)}.\frac{4(x-1)(x+1)}{5}=4\)

a) ĐKXĐ: x∉{2;5}

Ta có: \(\frac{6x+1}{x^2-7x+10}+\frac{5}{x-2}=\frac{3}{x-5}\)

\(\Leftrightarrow\frac{6x+1}{\left(x-2\right)\left(x-5\right)}+\frac{5\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\frac{3\left(x-2\right)}{\left(x-5\right)\left(x-2\right)}=0\)

\(\Leftrightarrow6x+1+5x-25-3\left(x-2\right)=0\)

\(\Leftrightarrow11x-24-3x+6=0\)

\(\Leftrightarrow8x-18=0\)

\(\Leftrightarrow8x=18\)

hay \(x=\frac{9}{4}\)(tm)

Vậy: \(x=\frac{9}{4}\)

b) ĐKXĐ: x∉{0;2;-2}

Ta có: \(\frac{2}{x^2-4}-\frac{x-1}{x\left(x-2\right)}+\frac{x-4}{x\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{2x}{x\left(x-2\right)\left(x+2\right)}-\frac{\left(x-1\right)\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}+\frac{\left(x-4\right)\left(x-2\right)}{x\left(x+2\right)\left(x-2\right)}=0\)

\(\Leftrightarrow2x-\left(x^2+x-2\right)+x^2-6x+8=0\)

\(\Leftrightarrow2x-x^2-x+2+x^2-6x+8=0\)

\(\Leftrightarrow-5x+10=0\)

\(\Leftrightarrow-5x=-10\)

hay x=2(ktm)

Vậy: x∈∅