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1) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
Ta có: \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{x^2-1}\)
\(\Leftrightarrow\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)
Suy ra: \(x^2+2x+1-\left(x^2-2x+1\right)=4\)
\(\Leftrightarrow x^2+2x+1-x^2+2x-1=4\)
\(\Leftrightarrow4x=4\)
hay x=1(loại)
Vậy: \(S=\varnothing\)
2) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{x+2}{x-2}+\dfrac{x}{x+2}=2\)
\(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2\left(x^2-4\right)}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(x^2+4x+4+x^2-2x=2x^2-8\)
\(\Leftrightarrow2x^2+2x+4-2x^2-8=0\)
\(\Leftrightarrow2x-4=0\)
\(\Leftrightarrow2x=4\)
hay x=2(loại)
Vậy: \(S=\varnothing\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b,\(\frac{2}{x-1}=\frac{6}{x+1}\)
\(2x+2=6x-6\)
\(4x=8\)
\(x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có :
\(\left|x+1\right|+\left|x-1\right|=1+\left|\left(x-1\right)\left(x+1\right)\right|\)
\(\Leftrightarrow\left|x-1\right|\left|x+1\right|-\left|x-1\right|-\left|x+1\right|+1=0\)
\(\Leftrightarrow\left(\left|x-1\right|-1\right)\left(\left|x+1\right|-1\right)=0\Leftrightarrow\orbr{\begin{cases}\left|x-1\right|=1\\\left|x+1\right|=1\end{cases}}\)
\(\Leftrightarrow x\in\left\{-2,0,2\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a. ĐKXĐ: $x\neq 0;-1$
\(=\left(\frac{2x^2+3x}{(x+1)(x^2-x+1)}+\frac{x+1}{(x+1)(x^2-x+1)}\right).\frac{x^2-x+1}{x}\)
\(=\frac{2x^2+3x+x+1}{(x+1)(x^2-x+1)}.\frac{x^2-x+1}{x}=\frac{2x^2+4x+1}{x(x+1)}\)
b. ĐKXĐ: $x\neq 0; 1;2$
\(=\frac{x-(x-1)}{x(x-1)}:\frac{(x+1)(x-1)-(x-2)(x+2)}{(x-2)(x-1)}=\frac{1}{x(x-1)}:\frac{3}{(x-2)(x-1)}\)
\(=\frac{1}{x(x-1)}.\frac{(x-2)(x-1)}{3}=\frac{x-2}{3x}\)
c. ĐKXĐ: $x\neq 0; -1$
\(=\frac{x+1+x^2}{x(x+1)}.\frac{x(x+1)}{x}=\frac{x^2+x+1}{x}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1-x}{1+x}+3=\frac{2x+3}{x+1}\left(ĐKXĐ:x\ne-1\right)\)
\(\Leftrightarrow\frac{1-x}{x+1}+\frac{3\left(x+1\right)}{x+1}=\frac{2x+3}{x+1}\)
\(\Leftrightarrow\frac{1-x+3\left(x+1\right)}{x+1}=\frac{2x+3}{x+1}\)
\(\Rightarrow1-x+3\left(x+1\right)=2x+3\)
\(\Leftrightarrow1-x+3x+3=2x+3\)
\(\Leftrightarrow2x+4=2x+3\)
\(\Leftrightarrow0x=-1\)(vô nghiệm)
Vậy phương trình vô nghiệm.
\(\frac{\left(x+2\right)^2}{2x-3}-1=\frac{x^2-10}{2x-3}\left(ĐKXĐ:x\ne\frac{3}{2}\right)\)
\(\Leftrightarrow\frac{x^2+4x+4}{2x-3}-\frac{2x-3}{2x-3}=\frac{x^2-10}{2x-3}\)
\(\Leftrightarrow\frac{x^2+4x+4-2x+3}{2x-3}=\frac{x^2-10}{2x-3}\)
\(\Rightarrow x^2+4x+4-2x+3=x^2-10\)
\(\Leftrightarrow2x+7=-10\)
\(\Leftrightarrow2x=-17\)
\(\Leftrightarrow x=\frac{-17}{2}\)(thỏa mãn ĐKXĐ)
Vậy phương trình có nghiệm duy nhất : \(x=\frac{-17}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\frac{x+1}{x-2}-\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{2x^2+4}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow x^2+2x+x+2-\left(x^2-2x-x+2\right)=2x^2+4\)
\(\Leftrightarrow x^2+3x+2-x^2+2x+x-2=2x^2+4\)
\(\Leftrightarrow6x=2x^2+4\)
\(\Leftrightarrow2x^2+4-6x=0\)
\(\Leftrightarrow2x^2+4-6x=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=-3\end{cases}}\)
\(b,\frac{2x+1}{x-1}=\frac{5\left(x-1\right)}{x+1}\)
\(\Leftrightarrow\left(2x+1\right)\left(x+1\right)=5\left(x-1\right)\left(x-1\right)\)
\(\Leftrightarrow2x^2+2x+x+1=5\left(x^2-2x+1\right)\)
\(\Leftrightarrow2x^2+3x+1=5x^2-10x+5\)
\(\Leftrightarrow5x^2-2x^2-10x-3x+5-1=0\)
\(\Leftrightarrow3x^2-13x+4=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-\frac{1}{3}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=\frac{1}{3}\end{cases}}}\)
\(\left(x+1\right)\left(x^2-x+1\right)+\left(x+1\right)\left(x-1\right)\)
\(=x^3+1+x^2-1=x^3+x^2=x^2.\left(x+1\right)\)
\(\left(x+1\right)\left(x^2-x+1\right)+\left(x+1\right)\left(x-1\right)\)
\(=x^3+1+x^2-1\)
\(=x^3+x^2\)
\(=x^2.\left(x+1\right)\)