Question

Answer 1

Lời giải:
a.

$x^2+x-2=(x-1)(x+2)$ nên để $f(x)=x^3+ax+b$ chia hết cho $x^2+x-2$ thì $f(x)\vdots x-1$ và $f(x)\vdots x+2$

$\Leftrightarrow f(1)=f(-2)=0$ (theo định lý Bê-du)

$\Leftrightarrow 1^3+a+b=(-2)^3-2a+b=0$

$\Leftrightarrow a+b=-1; -2a+b=8$

$\Rightarrow a=-3; b=2$

b.

\(A=\frac{xz}{xyz+xz+z}+\frac{yxz}{yz.xz+yxz+xz}+\frac{z}{xz+z+1}\)

\(=\frac{xz}{1+xz+z}+\frac{1}{z+1+xz}+\frac{z}{xz+z+1}=\frac{xz+z+1}{xz+z+1}=1\)