a: \(C=\frac{x^4+2x^3+8x+16}{x^4-2x^3+8x^2-8x+16}\)
\(=\frac{x^3\left(x+2\right)+8\left(x+2\right)}{x^4-2x^3+4x^2+4x^2-8x+16}\)
\(=\frac{\left(x+2\right)\left(x^3+8\right)}{\left(x^2-2x+4\right)\left(x^2+4\right)}=\frac{\left(x+2\right)^2}{x^2+4}\ge0\forall x\)
Dấu '=' xảy ra khi x+2=0
=>x=-2
b:
ĐKXĐ: x<>1
\(D=\frac{x^2-3x+3}{x^2-2x+1}\)
\(=\frac{x^2-2x+1-x+2}{x^2-2x+1}=1+\frac{-x+2}{\left(x-1\right)^2}=1+\frac{-x+1+1}{\left(x-1\right)^2}\)
\(=1-\frac{1}{x-1}+\frac{1}{\left(x-1\right)^2}\)
\(=\frac{1}{\left(x-1\right)^2}-2\cdot\frac{1}{x-1}\cdot\frac12+\frac14+\frac34=\left(\frac{1}{x-1}-\frac12\right)^2+\frac34\ge0\forall x\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi \(\frac{1}{x-1}-\frac12=0\)
=>\(\frac{1}{x-1}=\frac12\)
=>x-1=2
=>x=3(nhận)
