Question

a) rút gọn biểu thức A=x/x^2-4+3/(x+2)^2

b) tìm x để A=0

Answer 1

ĐK:....

a) \(A=\frac{x}{x^2-4}+\frac{3}{\left(x+2\right)^2}\)

\(A=\frac{x\left(x+2\right)}{\left(x+2\right)^2\left(x-2\right)}+\frac{3\left(x-2\right)}{\left(x+2\right)^2\left(x-2\right)}\)

\(A=\frac{x^2+2x+3x-6}{\left(x+2\right)^2\left(x-2\right)}\)

\(A=\frac{x^2+5x-6}{\left(x+2\right)^2\left(x-2\right)}\)

b) Để \(A=0\Leftrightarrow x^2+5x-6=0\)

\(\Leftrightarrow x^2+6x-x-6=0\)

\(\Leftrightarrow x\left(x+6\right)-\left(x+6\right)=0\)

\(\Leftrightarrow\left(x+6\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=1\end{matrix}\right.\)( thỏa ĐKXĐ )