Question

[thanh dat nguyen]

giải chi tiết giúp mình ạ

Answer 1

[Nguyễn Ngọc Huy Toàn]

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

Answer 2

[Shinichi Kudo]

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)