Question

[nguyễn hữu kim]

cứu mìn b3c với với 4c

Answer 1

[HT.Phong (9A5)]

Bài 3:

a) \(DKXD:x\ne2;x\ne-2\)

b) \(A=\dfrac{5x-2}{x^2-4}-\dfrac{3}{x+2}+\dfrac{x}{x-2}\)

\(A=\dfrac{5x-2}{\left(x+2\right)\left(x-2\right)}-\dfrac{3\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{x\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}\)

\(A=\dfrac{5x-2-3x+6+x^2+2x}{\left(x+2\right)\left(x-2\right)}\)

\(A=\dfrac{x^2+4x+4}{\left(x+2\right)\left(x-2\right)}\)

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

\(A=\dfrac{x+2}{x-2}\)

c) Ta có: \(\left|x-3\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=5\left(x\ge3\right)\\x-3=-5\left(x< 3\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\left(tm\right)\\x=-2\left(tm\right)\end{matrix}\right.\)

Với \(x=8\Rightarrow A=\dfrac{8+2}{8-2}=\dfrac{10}{6}=\dfrac{5}{3}\)

Với \(x=-2\Rightarrow A=\dfrac{-2+2}{-2-2}=\dfrac{0}{-4}=0\)