Question

Giai phương trình

các bạn giúp mink với mai mink nộp rui\(\frac{x+2}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)

Answer 1

Ta có: \(\frac{x+2}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)

ĐKXĐ: \(x\ne\pm2\)

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

\(\Rightarrow\left(x+2\right)^2+\left(x-1\right)\left(x-2\right)=2\left(x^2+2\right)\)

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

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

\(\Leftrightarrow x+2=0\)

\(\Leftrightarrow x=-2\left(ktmđk\right)\)

Vậy: \(x=\varnothing\)

Answer 2

\(\Rightarrow S=\left\{\varnothing\right\}\)

Answer 3

Giải

\(\frac{x+2}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\\ĐKXĐ:x\ne2;-2\\ MTC:\left(x-2\right)\left(x+2\right)\\ \frac{\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{2\left(x^2+2\right)}{\left(x-2\right)\left(x+2\right)}\\ \left(x+2\right)\left(x-2\right)+\left(x-1\right)\left(x-2\right)=2\left(x^2+2\right)\\ x^2-4+x^2-3x+2=2x^2+4\\ x^2-4+x^2-3x+2-2x^2=4\\-3x=6\\ x=-2 \)