Question

Thực hiện phép tính:

a) \(A=\frac{x^2-yz}{1+\frac{y+z}{x}}+\frac{y^2-zx}{1+\frac{z+x}{y}}+\frac{z^2-xy}{1+\frac{x+y}{z}}\)

b) \(B=\frac{2}{3}.\left[\frac{1}{1+\frac{\left(2x+1\right)^2}{3}}+\frac{1}{1+\frac{\left(2x-1\right)^2}{3}}\right]\)

Answer 1

a) \(A=\frac{x\left(x^2-yz\right)}{x+y+z}+\frac{y\left(y^2-zx\right)}{x+y+z}+\frac{z\left(z^2-xy\right)}{x+y+z}\)

\(=\frac{x^3+y^3+z^3-3xyz}{x+y+z}\)

\(=\frac{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)}{x+y+z}\)

\(=x^2+y^2+z^2-xy-yz-xz\)

b) \(B=\frac{2}{3}.\left[\frac{3}{4x^2+4x+4}+\frac{3}{4x^2-4x+4}\right]\)

\(=\frac{2}{3}.\frac{3}{4}.\left(\frac{1}{x^2+x+1}+\frac{1}{x^2-x+1}\right)\)

\(=\frac{1}{2}.\frac{x^2-x+1+x^2+x+1}{\left(x^2+x+1\right)\left(x^2-x+1\right)}\)

\(=\frac{1}{2}.\frac{2\left(x^2+1\right)}{\left(x^2+x+1\right)\left(x^2-x+1\right)}\)

\(=\frac{x^2+1}{\left(x^2+x+1\right)\left(x^2-x+1\right)}\)

(vì \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)

và \(x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\))