Question

[Cao Thị Thu Uyên]

\(A=\left(\frac{x}{x+3}+\frac{x-1}{x-3}+\frac{2x^2+x-3}{9-x^2}\right):\frac{-2}{x-3}\)

a) rút gọn A

b) Tính giá trị của A khi \(x^2-2x-3=0\)

c) tìm số nguyên x để A đạt giá trị nguyên

GIÚP MÌNH VOI TI NUA MINH DI HOC ROI

Answer 1

[ST]

a, DKXD: \(x\ne\pm3\)

\(A=\left(\frac{x}{x+3}+\frac{x-1}{x-3}+\frac{2x^2+x-3}{9-x^2}\right):\frac{-2}{x-3}\)

\(=\left(\frac{x\left(x+3\right)+\left(x-1\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{-2x^2-x+3}{x^2-9}\right):\frac{-2}{x-3}\)

\(=\left(\frac{2x^2+5x-3}{x^2-9}+\frac{-2x^2-x+3}{x^2-9}\right):\frac{-2}{x-3}\)

\(=\frac{4x}{x^2-9}:\frac{-2}{x-3}=\frac{4x}{\left(x-3\right)\left(x+3\right)}\cdot\frac{x-3}{-2}=\frac{4x}{-2\left(x+3\right)}=\frac{-2x}{x+3}\)

b, \(x^2-2x-3=0\Leftrightarrow x^2-3x+x-3=0\Leftrightarrow x\left(x-3\right)+\left(x-3\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}}\)

Thay x=-1 =>\(A=\frac{-2.\left(-1\right)}{-1+3}=1\)

thay x=3 =>\(A=\frac{-2.3}{3+3}=-1\)

c, De \(A\in Z\Leftrightarrow x+3\in U\left(-2\right)=\left\{1;-1;2;-2\right\}\)

<=>x thuoc {-2;-4;-1;-5}

Answer 2

[Pham Van Hung]

ĐK: \(x\ne\pm3\)

\(A=\left(\frac{x}{x+3}+\frac{x-1}{x-3}+\frac{2x^2+x-3}{9-x^2}\right):\frac{-2}{x-3}\)

\(=\left(\frac{x\left(x-3\right)+\left(x+3\right)\left(x-1\right)}{\left(x+3\right)\left(x-3\right)}+\frac{-2x^2-x+3}{x^2-9}\right).\frac{x-3}{-2}\)

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

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

b, \(x^2-2x-3=0\Rightarrow x\left(x-3\right)+\left(x-3\right)=0\Rightarrow\left(x-3\right)\left(x+1\right)=0\Rightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)

x = 3 không thỏa mãn ĐKXĐ

Với x = -1 (thỏa mãn ĐKXĐ) thì \(A=\frac{x}{x+3}=\frac{-1}{-1+3}=-\frac{1}{2}\)

c, \(A\in Z\Rightarrow\frac{x}{x+3}\in Z\Rightarrow x⋮\left(x+3\right)\)

\(\Rightarrow\left(x+3\right)-3⋮\left(x+3\right)\Rightarrow-3⋮\left(x+3\right)\Rightarrow x+3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\)

\(\Rightarrow x\in\left\{-6;-4;-2;0\right\}\) (thỏa mãn điều kiện)