Question

[Lê Thu Oanh]

làm hộ tôi câu 1 b và câu 2

Answer 1

[Xyz OLM]

2) a) ĐKXĐ : \(x\ne\pm2\)

\(A=\frac{x^2-12}{x^2-4}-\frac{2}{2-x}-\frac{x+1}{x+2}\)

\(=\frac{x^2-12+2\left(x+2\right)-\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)

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

b) Ta có x² = 2x

<=> x(x - 2) = 0

<=> \(\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\left(\text{loại}\right)\end{cases}}\)

Khi x = 0 => A = 1,5

Vậy A = 1,5 khi x² = 2x

c) B = Ax = \(\frac{3x}{x+2}=\frac{3x+6-6}{x+2}=3-\frac{6}{x+2}\)

\(B\inℤ\Leftrightarrow x+2\inƯ\left(6\right)\)

<=> \(x+2\in\left\{1;2;3;6;-1;-2;-3;-6\right\}\)

<=> \(x\in\left\{-1;0;1;4;-3;-4;-5;-8\right\}\)