a: \(x^2-2x+2=\left(x-1\right)^2+1>0\)
\(x^2+1>0\)
Do đó: \(\dfrac{x^2-2x+2}{x^2+1}>0\forall x\)
b: \(=\dfrac{x^3\left(x+1\right)-\left(x+1\right)}{x^2-1}=\dfrac{\left(x+1\right)\left(x-1\right)\left(x^2+x+1\right)}{x^2-1}=x^2+x+1\)
\(=x^2+2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
\(\dfrac{x^2-2x+2}{x^2+1}=\dfrac{x^2-2x+1+1}{x^2+1}=\dfrac{\left(x-1\right)^2+1}{x^2+1}\)
Có : \(\left(x-1\right)^2\ge0\) với mọi x
1>0
\(x^2\ge0\) với mọi x
\(\Rightarrow\dfrac{\left(x-1\right)^2+1}{x^2+1}>0\) với mọi x hay \(\dfrac{x^2-2x+2}{x^2+1}\) luôn dương với mọi giá trị x
b)\(\dfrac{x^4+x^3-x-1}{x^2-1}=\dfrac{x^3\left(x+1\right)-\left(x+1\right)}{x^2-1}=\dfrac{\left(x^3-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=x^2+x+1\)
Có \(x^2+x+1=x^2+x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
\(\left(x+\dfrac{1}{2}\right)^2\ge0\) với mọi x
\(\dfrac{3}{4}>0\)
=>...
