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1) ĐKXĐ: \(x\notin\left\{1;-2\right\}\)

Ta có: \(\dfrac{2x}{x-1}-\dfrac{1}{x+2}=2\)

\(\Leftrightarrow\dfrac{2x\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}-\dfrac{x-1}{\left(x-1\right)\left(x+2\right)}=\dfrac{2\left(x-1\right)\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}\)

Suy ra: \(2x^2+4x-x+1=2\left(x^2+x-2\right)\)

\(\Leftrightarrow2x^2+3x+1=2x^2+2x-4\)

\(\Leftrightarrow2x^2+3x+1-2x^2-2x+4=0\)

\(\Leftrightarrow x+5=0\)

hay x=-5(thỏa ĐK)

Vậy: S={-5}

2) ĐKXĐ: \(x\notin\left\{5;-5\right\}\)

Ta có: \(\dfrac{x}{x^2-25}-\dfrac{1-x}{x-5}=\dfrac{1}{x+5}\)

\(\Leftrightarrow\dfrac{x}{\left(x-5\right)\left(x+5\right)}+\dfrac{\left(x-1\right)\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}\)

Suy ra: \(x+x^2+5x-x-5=x-5\)

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

\(\Leftrightarrow x^2+4x=0\)

\(\Leftrightarrow x\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)

Vậy: S={0;-4}

1 tháng 3 2021

a/ ĐKXĐ : \(x\ne1;-2\)

\(\dfrac{2x}{x-1}-\dfrac{1}{x+2}=2\)

\(\Leftrightarrow\dfrac{2x\left(x+2\right)-\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}=2\)

\(\Leftrightarrow2x^2+3x-x+1=2x^2+4x-2x-4\)

\(\Leftrightarrow2x+1=2x-4\)

\(\Leftrightarrow1=-4\left(loại\right)\)

Vậy...

b/ĐKXĐ :  \(x\ne\pm5\)

\(\dfrac{x}{x^2-25}-\dfrac{1-x}{x-5}=\dfrac{1}{x+5}\)

\(\Leftrightarrow\dfrac{x}{\left(x-5\right)\left(x+5\right)}+\dfrac{\left(x-1\right)\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}=\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}\)

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

\(\Leftrightarrow x^2+4x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)

Vậy...

6 tháng 3 2022

\(a,3x-2\left(x-3\right)=0\\ \Leftrightarrow3x-2x+6=0\\ \Leftrightarrow x=-6\\ b,\left(x+1\right)\left(2x-3\right)=\left(2x-1\right)\left(x+5\right)\\ \Leftrightarrow2x^2+2x-3x-3=2x^2-x+10x-5\\ \Leftrightarrow2x^2-x-3=2x^2+9x-5\\ \Leftrightarrow10x-2=0\\ \Leftrightarrow x=\dfrac{1}{5}\\ c,ĐKXĐ:x\ne\pm1\\ \dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\\ \Leftrightarrow\dfrac{2x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x-x^2+1}{\left(x+1\right)\left(x-1\right)}=0\)

\(\Rightarrow3x+1=0\\ \Leftrightarrow x=-\dfrac{1}{3}\left(tm\right)\)

\(d,\left(2x+3\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3x-5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{5}{3}\end{matrix}\right.\\ e,ĐKXĐ:x\ne\pm2\\ \dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\\ \Leftrightarrow\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-22}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\dfrac{x^2-4x+4-3x-6-2x+22}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow x^2-9x+20=0\\ \Leftrightarrow\left(x^2-5x\right)-\left(4x-20\right)=0\\ \Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)\\ \Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)

 

2 tháng 3 2021

Bài dài quá, lần sau chia nhỏ câu hỏi nhé!!!!!

12 tháng 9 2021

đúng vậy

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: \(S=\varnothing\)

a: ĐKXĐ: x<>0; x<>-1

PT =>x+1-2x=3

=>1-x=3

=>x=-2(nhận)

b: Sửa đề: \(\dfrac{1}{2x-3}-\dfrac{3}{x\left(2x-3\right)}=\dfrac{5}{x}\)

=>x-3=5(2x-3)

=>10x-15=x-3

=>9x=12

=>x=4/3(nhận)

c: ĐKXĐ: x<>0; x<>2

PT =>x(x+2)-x+2=2

=>x^2+2x-x=0

=>x(x+1)=0

=>x=-1