1/x-1+1/x-2=1/x+2+1/x+1
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27 tháng 2 2021
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: \(S=\varnothing\)
NM
Nguyễn Minh Quang
Giáo viên
5 tháng 3 2022
ta có :
\(\left|x+1\right|+\left|x-1\right|=1+\left|\left(x-1\right)\left(x+1\right)\right|\)
\(\Leftrightarrow\left|x-1\right|\left|x+1\right|-\left|x-1\right|-\left|x+1\right|+1=0\)
\(\Leftrightarrow\left(\left|x-1\right|-1\right)\left(\left|x+1\right|-1\right)=0\Leftrightarrow\orbr{\begin{cases}\left|x-1\right|=1\\\left|x+1\right|=1\end{cases}}\)
\(\Leftrightarrow x\in\left\{-2,0,2\right\}\)
7 tháng 3 2021
\(\frac{1-x}{1+x}+3=\frac{2x+3}{x+1}\left(ĐKXĐ:x\ne-1\right)\)
\(\Leftrightarrow\frac{1-x}{x+1}+\frac{3\left(x+1\right)}{x+1}=\frac{2x+3}{x+1}\)
\(\Leftrightarrow\frac{1-x+3\left(x+1\right)}{x+1}=\frac{2x+3}{x+1}\)
\(\Rightarrow1-x+3\left(x+1\right)=2x+3\)
\(\Leftrightarrow1-x+3x+3=2x+3\)
\(\Leftrightarrow2x+4=2x+3\)
\(\Leftrightarrow0x=-1\)(vô nghiệm)
Vậy phương trình vô nghiệm.
7 tháng 3 2021
\(\frac{\left(x+2\right)^2}{2x-3}-1=\frac{x^2-10}{2x-3}\left(ĐKXĐ:x\ne\frac{3}{2}\right)\)
\(\Leftrightarrow\frac{x^2+4x+4}{2x-3}-\frac{2x-3}{2x-3}=\frac{x^2-10}{2x-3}\)
\(\Leftrightarrow\frac{x^2+4x+4-2x+3}{2x-3}=\frac{x^2-10}{2x-3}\)
\(\Rightarrow x^2+4x+4-2x+3=x^2-10\)
\(\Leftrightarrow2x+7=-10\)
\(\Leftrightarrow2x=-17\)
\(\Leftrightarrow x=\frac{-17}{2}\)(thỏa mãn ĐKXĐ)
Vậy phương trình có nghiệm duy nhất : \(x=\frac{-17}{2}\)
7 tháng 12 2021
d: \(\dfrac{x}{x+3}=\dfrac{x^2-3x}{\left(x+3\right)\left(x-3\right)}\)
\(\dfrac{1}{3-x}=\dfrac{-1}{x-3}=\dfrac{-x-3}{\left(x-3\right)\left(x+3\right)}\)
\(\dfrac{1}{x^2-9}=\dfrac{1}{\left(x+3\right)\left(x-3\right)}\)
21 tháng 4 2019
b,\(\frac{2}{x-1}=\frac{6}{x+1}\)
\(2x+2=6x-6\)
\(4x=8\)
\(x=2\)
Dài quá, mk làm đến vậy thôi, cs j sai góp ý !
ĐKXĐ : \(x\ne\pm1;\pm2\)
\(\frac{1}{x-1}+\frac{1}{x-2}=\frac{1}{x+2}+\frac{1}{x+1}\)
\(\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x+2}-\frac{1}{x+1}=0\)
\(\frac{\left(x-2\right)\left(x+2\right)\left(x+1\right)}{\left(x-1\right)\left(x-2\right)\left(x+2\right)\left(x+1\right)}+\frac{\left(x-1\right)\left(x+2\right)\left(x+1\right)}{\left(x-1\right)\left(x-2\right)\left(x+2\right)\left(x+1\right)}-\frac{\left(x-1\right)\left(x-2\right)\left(x+1\right)}{\left(x-1\right)\left(x-2\right)\left(x+2\right)\left(x+1\right)}-\frac{\left(x-1\right)\left(x-2\right)\left(x+2\right)}{\left(x-1\right)\left(x-2\right)\left(x+2\right)\left(x+1\right)}=0\)
\(\left(x-2\right)\left(x+2\right)\left(x+1\right)+\left(x-1\right)\left(x+2\right)\left(x+1\right)-\left(x-1\right)\left(x-2\right)\left(x+1\right)-\left(x-1\right)\left(x-2\right)\left(x+2\right)=0\)
\(\frac{1}{x-1}+\frac{1}{x-2}=\frac{1}{x+2}+\frac{1}{x+1}\) \(\left(ĐK:x\ne\pm1;x\ne\pm2\right)\)
\(< =>\frac{\left(x-2\right)+\left(x-1\right)}{\left(x-1\right)\left(x-2\right)}=\frac{\left(x+1\right)+\left(x+2\right)}{\left(x+1\right)\left(x+2\right)}\)
\(< =>\frac{2x-3}{x^2-3x-2}=\frac{2x+3}{x^2+3x+2}\)
\(< =>\left(2x-3\right)\left(x^2+3x+2\right)=\left(2x+3\right)\left(x^2-3x-2\right)\)
\(< =>2x^3+6x^2+4x-3x^2-9x-6=2x^3-6x^2-4x+3x^2-9x-6\)
\(< =>2x^3+3x^2-5x-6=2x^3-3x^2-13x-6\)
\(< =>2x^3-2x^3+3x^2-3x^2-5x+13x-6+6=0\)
\(< =>8x=0< =>x=0\left(tmđk\right)\)
Vậy nghiệm của pt trên là : 0