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\(\text{Đ}K\text{X}\text{Đ}:x\ne\pm2\)
Ta có: \(A=\left(\frac{2}{x+2}-\frac{4}{x^2+4x+4}\right)\div\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)
\(=\left(\frac{2x+2-4}{\left(x+2\right)^2}\right):\left(\frac{2-x-2}{\left(x+2\right)\left(x-2\right)}\right)=\frac{2x-2}{\left(x+2\right)^2}\cdot\frac{\left(x+2\right)\left(x-2\right)}{-x}\)
\(=\frac{2\left(x-1\right)\left(x-2\right)}{-x\left(x+2\right)}\)
\(\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2-\frac{x^2-10}{x+2}\right)\left(ĐK:x\ne\pm2\right)\)
\(=\frac{x-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}:\frac{\left(x-2\right)\left(x+2\right)-\left(x^2-10\right)}{x+2}\)
\(=\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\frac{x+2}{x^2-4-x^2+10}\)
\(=\frac{-6\left(x+2\right)}{6\left(x-2\right)\left(x+2\right)}=-\frac{1}{x-2}=\frac{1}{2-x}\)
\(A=\left(\frac{x-1}{x-2}+\frac{x+3}{x^2-4}\right):\left(\frac{x+2}{x-2}+\frac{1}{2-x}\right)\)
\(A=\frac{\left(x-1\right)\left(x+2\right)+x+3}{\left(x+2\right)\left(x-2\right)}:\left(\frac{x+2}{x-2}-\frac{1}{x-2}\right)\)
\(A=\frac{x^2+2x-x-2+x+3}{\left(x+2\right)\left(x-2\right)}:\frac{x+2-1}{x-2}\)
\(A=\frac{x^2+2x+1}{\left(x-2\right)\left(x+2\right)}.\frac{x-2}{x+1}\)
\(A=\frac{\left(x+1\right)^2}{x+2}.\frac{1}{x+1}\)
\(A=\frac{x+1}{x+2}\)
Bạn chú ý cách viết phương trình.
Phương trình chỉ có dạng f(x)=g(x) thôi, không có dạng A=f(x)=g(x) như bạn viết.
\(VT=\left[8\left(x+\frac{1}{x}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2\right]+4\left(x^2+\frac{1}{x^2}\right)^2\)
\(=4\left(x+\frac{1}{x}\right)^2\left(2-x^2-\frac{1}{x^2}\right)+4\left(x^2+\frac{1}{x^2}\right)^2\)
\(=-4\left(x+\frac{1}{x}\right)^2\left(x-\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2\)
\(=-4\left(x^2-\frac{1}{x^2}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2\)
\(=-4x^4+8-\frac{4}{x^4}+4x^4+8+\frac{4}{x^4}\)
\(=16\)
Phương trình đã cho trở thành
\(\left(x+4\right)^2=16\\ \Leftrightarrow\orbr{\begin{cases}x+4=-4\\x+4=4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-8\\x=0\end{cases}}\)