1.\(ĐK:x\ne\pm2\)
\(\Leftrightarrow\dfrac{x+2}{x-2}-\dfrac{x-2}{x+2}=\dfrac{4}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x+2\right)-\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{4}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow\left(x+2\right)^2-\left(x-2\right)^2=4\)
\(\Leftrightarrow x^2+4x+4-x^2+4x-4=4\)
\(\Leftrightarrow8x=4\)
\(\Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)
Vậy S = \(\dfrac{1}{2}\)
2.\(ĐK:x\ne1;-3\)
\(\Leftrightarrow\dfrac{x+1}{x-1}-\dfrac{x+2}{x+3}=-\dfrac{4}{\left(x-1\right)\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+3\right)-\left(x+2\right)\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}=-\dfrac{4}{\left(x-1\right)\left(x+3\right)}\)
\(\Rightarrow x^2+3x+x+3-x^2+x-2x+2=-4\)
\(\Leftrightarrow3x=-9\)
\(\Leftrightarrow x=-3\left(ktm\right)\)
Vậy S vô nghiệm
1) ĐKXĐ: \(x\ne\pm2\)
\(\dfrac{x+2}{x-2}-\dfrac{x-2}{x+2}=\dfrac{4}{x^2-4}\)
\(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{4}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow x^2+4x+4-x^2+4x-4=4\)
\(\Leftrightarrow8x=4\)
\(\Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)
Vậy ....
2) ĐKXĐ:\(x\ne1,-3\)
\(\dfrac{x+1}{x-1}-\dfrac{x+2}{x+3}+\dfrac{4}{x^2+2x-3}=0\)
\(\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\dfrac{\left(x+2\right)\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}+\dfrac{4}{x^2-x+3x-3}=0\)
\(\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\dfrac{\left(x+2\right)\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}+\dfrac{4}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Rightarrow x^2+4x+3-x^2-x+2+4=0\)
\(\Leftrightarrow3x=-9\)
\(\Leftrightarrow x=-3\)(ktm)
\(1,\dfrac{x+2}{x-2}-\dfrac{x-2}{x+2}=\dfrac{4}{x^2-4}\) \(\left(đk:x\ne\pm2\right)\)
\(\Leftrightarrow\left(x+2\right)^2-\left(x-2\right)^2=4\)
\(\Leftrightarrow x^2+4x+4-x^2+4x-4=4\)
\(\Leftrightarrow8x=4\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
1: \(\Leftrightarrow x^2+4x+4-x^2+4x-4=4\)
=>8x=4
hay x=1/2(nhận)
2: \(\Leftrightarrow x^2+4x+4-\left(x+2\right)\left(x-1\right)+4=0\)
\(\Leftrightarrow x^2+4x+8-x^2+x-2x+2=0\)
=>3x+10=0
hay x=-10/3(nhận)
1, đk : x khác -2 ; 2
\(\Rightarrow x^2+4x+4-x^2+4x-4=4\Leftrightarrow8x=4\Leftrightarrow x=\dfrac{1}{2}\)(tm)
2, đk : x khác -3 ; 1
\(\Rightarrow x^2+4x+3-\left(x^2+x-2\right)+4=0\Leftrightarrow3x+9=0\Leftrightarrow x=-3\left(ktm\right)\)
Vậy pt vô nghiệm
