a) ĐKXĐ: \(x\ne\pm2\)
\(A=\dfrac{1}{x-2}-\dfrac{2x}{4-x^2}+\dfrac{1}{2+x}\)
\(=\dfrac{1}{x-2}+\dfrac{2x}{x^2-4}+\dfrac{1}{2+x}\)
\(=\dfrac{1}{x-2}+\dfrac{2x}{\left(x-2\right)\left(x+2\right)}+\dfrac{1}{2+x}\)
\(=\dfrac{x+2+2x+x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{4x}{\left(x-2\right)\left(x+2\right)}\)\(=\dfrac{4x}{x^2-4}\)
Vậy \(A=\dfrac{4x}{x^2-4}\) với \(x\ne\pm2\)
b) Sửa đề: Tìm gtrị của \(x\) khi A = 1
ĐK: \(x\ne\pm2\)
\(A=1\Leftrightarrow\dfrac{4}{4-x^2}=1\Leftrightarrow\dfrac{4}{4-x^2}-1=0\)\(\Leftrightarrow\dfrac{4-4+x^2}{4-x^2}=0\)\(\Leftrightarrow\dfrac{x^2}{x^2-4}=0\Leftrightarrow x^2=0\Leftrightarrow x=0\)
Vậy \(x=0\) khi A = 1
a ) Đ/k : \(x\ne2;-2\)
\(A=\dfrac{1}{x-2}-\dfrac{2x}{4-x^2}+\dfrac{1}{2+x}\)
\(\Rightarrow A=\dfrac{-1}{2-x}-\dfrac{2x}{4-x^2}+\dfrac{1}{2+x}\)
\(\Rightarrow A=\dfrac{-1\left(x+2\right)}{\left(2-x\right)\left(x+2\right)}-\dfrac{2x}{\left(2-x\right)\left(2+x\right)}+\dfrac{2-x}{\left(2+x\right)\left(2-x\right)}\)
\(\Rightarrow A=\dfrac{-1\left(x+2\right)-2x+2-x}{\left(2-x\right)\left(2+x\right)}\)
\(\Rightarrow A=\dfrac{-x-2-2x+2-x}{4-x^2}\)
\(\Rightarrow A=\dfrac{-4x}{4-x^2}\)
b ) \(A=1\)
\(\Leftrightarrow-4x=4-x^2\)
\(\Leftrightarrow-4x+x^2=4\)
\(\Leftrightarrow x^2-2.x.2+2^2=8\)
\(\Leftrightarrow\left(x-2\right)^2=8\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=\sqrt{8}\\x-2=-\sqrt{8}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{8}+2\\x=-\sqrt{8}+2\end{matrix}\right.\)
sr bn nhé, câu b mk lộn :))
b) Tìm x khi A = 1
Đk: \(x\pm\ne2\)
A = 1 \(\Leftrightarrow\dfrac{4}{x^2-4}=1\Leftrightarrow\dfrac{4}{x^2-4}-1=0\)\(\Leftrightarrow\dfrac{4-x^2+4}{x^2-4}=0\) \(\Leftrightarrow8-x^2=0\Leftrightarrow x=\sqrt{8}=2\sqrt{2}\)
