a) ĐK:
x=0;x=1
E=(x−1)2x(x+1):x(x−1)x2−1+x+2−x2=(x−1)2x(x+1).x+1x(x−1)=x−1x2
b)
E>1⇔x−1x2>1⇔x−1x2−1>0⇔x−1x2−x+1>0⇔x−1>0⇔x>1
c) $|2x+1|=5\Leftrightarrow x=-3 \ or \ x=2\\
* \ x=-3\Rightarrow E=\dfrac{(-3)^2}{-3-1}=\dfrac{9}{-4}=\dfrac{-9}{4}\\
* \ x=2\Rightarrow E=\dfrac{2^2}{2-1}=\dfrac{4}{1}=4$
d)
E=x−1x2=x−1x2−1+1=x+1+x−11=x−1+x−11+2≥2+2=4
Dấu '=' xảy ra khi
x=2