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Question

Cho $A= \left ( \frac{2+x}{2-x} - \frac{2-x }{2+x} -\frac{4x^{2}}{x^2-4}\right )$ $\div$ $\frac{x-3}{2x -x^{2}}$

a) Rút gọn $A$

b) Tính $A$ khi x thỏa mãn $| x | = 1$

c) Tìm $X$ để $A = \frac{-4}{5}$

D ) Tìm $A$ để $A<1$

Kaito kid · Answer 1 · 6/18/2016 8:35:57 PM

|x|=1

=>x=1 or x= -1

X=1 =>A=-2

X= -1 =>A=-1

C, A= -4/5 => vô nghiệm

Trả lời 18-06-16 08:35 PM

Kaito kid
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http://toan.hoctainha.vn/Hỏi-Đáp/Câu-Hỏi/136273/đây-nữa-ak – Yêu Tatoo 18-06-16 08:37 PM

bh a lm hộ e bài bên kia đi – Yêu Tatoo 18-06-16 08:36 PM

score 4 · Answer 2

a. $A= \left ( \frac{2+x}{2-x} - \frac{2-x }{2+x} -\frac{4x^{2}}{x^2-4}\right )$ $\div$ $\frac{x-3}{2x -x^{2}}$

$A= \left ( \frac{-(2+x)^2+(2-x)^2-4x^{2}}{x^2-4}\right )$ $\div$ $\frac{x-3}{2x -x^{2}}$

$A= \frac{-4x^2-8x}{x^2-4}: \frac{x-3}{2x -x^{2}}$

$A= \frac{-4x(x+2)}{x^2-4}: \frac{x-3}{2x -x^{2}}$

$A= \frac{-4x}{x-2}* \frac{-x(x-2)}{x-3}=-\frac{4x^2}{x-3}$

b. $\left| {x} \right|=1\Rightarrow \left[ \begin{array}{l} x=1\Rightarrow A=2\\ x=-1\Rightarrow A=1\end{array} \right.$

c $A=-\frac{4}{5}\Leftrightarrow -\frac{4x^2}{x-3}=-\frac{4}{5}$

$\Leftrightarrow 5x^2=x-3$ (vô nghiệm)

2 Đáp án