Question

Cho biểu thức

\(\frac{x}{x-2}+\frac{2-x}{x+2}+\frac{12-10x}{x^2-4}\)

a Rút gọn A

b,Tìm x thuộc Z để A thuộc Z

Answer 1

a) Ta có: A= \(\frac{x}{x-2}+\frac{2-x}{x+2}+\frac{12-10x}{x^2-4}\)

A = \(\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(2-x\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{12-10x}{\left(x-2\right)\left(x+2\right)}\)

A = \(\frac{x^2+2x-x^2+4x-4+12-10x}{\left(x-2\right)\left(x+2\right)}\)

A = \(\frac{-4x+8}{\left(x-2\right)\left(x+2\right)}\)

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

b) ĐKXĐ: x \(\ne\) \(\pm\)2

Để A \(\in\)Z <=> \(-\frac{4}{x+2}\in Z\) <=> -4 \(⋮\)x + 2

<=> x + 2 \(\in\)Ư(-4) = {1; -1; 2; -2; 4; -4}

Lập bảng :

x + 2	1	-1	2	-2	4	-4
x	-1	-3	0	-4	2(ktm)	-6

Answer 2

a) Rút gọn:

\(A=\frac{x}{x-2}+\frac{2-x}{x+2}+\frac{12-10x}{x^2-4}\)

\(A=\frac{x}{x-2}+\frac{2-x}{x+2}+\frac{12-10x}{\left(x-2\right).\left(x+2\right)}\)

\(A=\frac{x.\left(x+2\right)}{\left(x-2\right).\left(x+2\right)}+\frac{\left(2-x\right).\left(x-2\right)}{\left(x-2\right).\left(x+2\right)}+\frac{12-10x}{\left(x-2\right).\left(x+2\right)}\)

\(A=\frac{x^2+2x}{\left(x-2\right).\left(x+2\right)}+\frac{2x-4-x^2+2x}{\left(x-2\right).\left(x+2\right)}+\frac{12-10x}{\left(x-2\right).\left(x+2\right)}\)

\(A=\frac{x^2+2x}{\left(x-2\right).\left(x+2\right)}+\frac{4x-4-x^2}{\left(x-2\right).\left(x+2\right)}+\frac{12-10x}{\left(x-2\right).\left(x+2\right)}\)

\(A=\frac{x^2+2x+4x-4-x^2+12-10x}{\left(x-2\right).\left(x+2\right)}\)

\(A=\frac{8-4x}{\left(x-2\right).\left(x+2\right)}\)

\(A=\frac{4.\left(2-x\right)}{\left(x-2\right).\left(x+2\right)}\)

\(A=\frac{4}{x+2}.\)

Chúc bạn học tốt!

Answer 3

a) \(\frac{x}{x-2}+\frac{2-x}{x+2}+\frac{12-10x}{x^2-4}\)

=\(\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(2-x\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{12-10x}{\left(x-2\right)\left(x+2\right)}\)

=\(\frac{x\left(x+2\right)+\left(2-x\right)\left(x-2\right)+12-10x}{\left(x-2\right)\left(x+2\right)}\)

=\(\frac{x^2+2x+2x-4-x^2+2x+12-10x}{\left(x-2\right)\left(x+2\right)}\)

=\(\frac{-4x+8}{\left(x-2\right)\left(x+2\right)}\)

=\(\frac{-4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)

=\(\frac{-4}{x+2}\)

b)(ĐKXĐ của A là x\(\ne\pm2\))

Với x\(\ne\pm2\) ta có:

A\(\in Z\)

\(\Leftrightarrow\frac{-4}{x+2}\in Z\)

\(\Rightarrow x+2\inƯ_{\left(-4\right)}=\left\{\pm1;\pm2;\pm4\right\}\)

Ta có bảng sau :

x+2	-4	-2	-1	1	2	4
x	-6	-4	-3	-1	0	2
NX	tm	tm	tm	tm	tm	loại

Vậy để \(A\in Z\) thì x = {-6,-4,-3,-1,0}