a, ĐKXĐ x \(\ne\) \(\pm\)2
b, \(\dfrac{x^{2^{ }}-4x+4}{x^2-4}=\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x-2}{x+2}\)
c, \(|x|=3\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
* Với x=3 thì \(\dfrac{x-2}{x+2}=\dfrac{3-2}{3+2}=\dfrac{1}{5}\)
* Với x= -3 thì \(\dfrac{x-2}{x+2}=\dfrac{\left(-3\right)-2}{\left(-3\right)+2}=\dfrac{-5}{-1}=5\)
d, \(\dfrac{x-2}{x+2}\)= 2 \(\Leftrightarrow x-2=2\left(x+2\right)\)
\(\Leftrightarrow x-2=2x+4\)
\(\Leftrightarrow x-2x=2+4\)
\(\Leftrightarrow-x=6\)
\(\Leftrightarrow x=6\)
Vậy ...
a) ĐKXĐ: x\(\ne\pm2\)
b) Rút gọn
\(\dfrac{x^2-4x+4}{x^2-4}=\dfrac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}=\dfrac{\left(x-2\right)}{\left(x+2\right)}\) (1)
c) Ta thay x = 3 vào biểu thức 1 ta đc :
\(\dfrac{x-2}{x+2}=\dfrac{3-2}{3+2}=\dfrac{1}{5}\)
Vậy tại x=3 giá trị của biểu thức (1) là 1/5
d) Ta có
\(\dfrac{x-2}{x+2}=2\Leftrightarrow\dfrac{x-2}{x+2}=\dfrac{2\left(x+2\right)}{x+2}\)
<=> x-2 = 2x+4
<=> x-2x = 4+2
<=> -x = 6
=> x= -6
a, ĐKXĐ:\(x\ne2,x\ne-2\)
b,\(\dfrac{x^2-4x+4}{x^2-4}=\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x-2}{x+2}\)
Tại x=|3|
x=3\(\Rightarrow\dfrac{x-2}{x+2}=\dfrac{3-2}{3+2}=\dfrac{1}{5}\)
x=-3\(\Rightarrow\dfrac{x-2}{x+2}=\dfrac{-3-2}{-3+2}=5\)
để \(\dfrac{x-2}{x+2}=2\Leftrightarrow x-2=2\left(x+2\right)\Leftrightarrow x-2=2x+4\)\(\Leftrightarrow x-2x=4+2\Leftrightarrow-x=6\Leftrightarrow x=-6\)
