Question

  help

Câu 2:

Cho \( A = \left( \frac{2x-1}{x+3} + \frac{x}{x-3} - \frac{3-10x}{x^2-9} \right) : \frac{x+2}{x-3} \)

a) Tìm điều kiện xác định của \( A \) và rút gọn \( A \)

b) Tìm \( x \) nguyên để \( A \) có giá trị nguyên

Answer 1

a: ĐKXĐ: \(x\notin\left\{3;-3;-2\right\}\)

\(A=\left(\dfrac{2x-1}{x+3}+\dfrac{x}{x-3}-\dfrac{3-10x}{x^2-9}\right):\dfrac{x+2}{x-3}\)

\(=\dfrac{\left(2x-1\right)\left(x-3\right)+x\left(x+3\right)+10x-3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x-3}{x+2}\)

\(=\dfrac{2x^2-7x+3+x^2+3x+10x-3}{\left(x+3\right)\left(x+2\right)}\)

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

b: Để A là số nguyên thì \(3x⋮x+3\)

=>\(3x+9-9⋮x+3\)

=>\(-9⋮x+3\)

=>\(x+3\in\left\{1;-1;3;-3;9;-9\right\}\)
=>\(x\in\left\{-2;-4;0;-6;6;-12\right\}\)

Kết hợp ĐKXĐ, ta được; \(x\in\left\{-4;0;-6;6;-12\right\}\)

Answer 2

a: ĐKXĐ: \(x\notin\left\{3;-3;-2\right\}\)

\(A=\left(\dfrac{2x-1}{x+3}+\dfrac{x}{x-3}-\dfrac{3-10x}{x^2-9}\right):\dfrac{x+2}{x-3}\)

\(=\dfrac{\left(2x-1\right)\left(x-3\right)+x\left(x+3\right)+10x-3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x-3}{x+2}\)

\(=\dfrac{2x^2-7x+3+x^2+3x+10x-3}{\left(x+3\right)\left(x+2\right)}\)

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

b: Để A là số nguyên thì \(3x⋮x+3\)

=>\(3x+9-9⋮x+3\)

=>\(-9⋮x+3\)

=>\(x+3\in\left\{1;-1;3;-3;9;-9\right\}\)
=>\(x\in\left\{-2;-4;0;-6;6;-12\right\}\)

Kết hợp ĐKXĐ, ta được; \(x\in\left\{-4;0;-6;6;-12\right\}\)