Tìm x; y biết:x 2 - y 2 + 2x -4y-10 =0 với x,y nguyên dương2)cho abc=2 rút gọn biểu thức:
$\frac{a}{ab+a+2}+\frac{b}{bc+b+1}+\frac{2c}{ac+2c+2}$+$\frac{a}{ab+a+2}+\frac{b}{bc+b+1}+\frac{2c}{ac+2c+2}$
ta có:$\frac{2a}{ab+a+2}+\frac{b}{bc+b+1}+\frac{2c}{ac+2c+2}$
=$\frac{a}{ab+a+2}+\frac{ab}{abc+ab+a}+\frac{4c}{2ac+4c+4}$
=$\frac{a}{ab+a+2}+\frac{ab}{ab+a+2}+\frac{a^{2}b^{2}c^{3}}{a^{2}bc^{2}+a^{2}b^{2}c^{3}+a^{2}b^{2}c^{2}}$
=$\frac{a}{ab+a+2}+\frac{ab}{ab+a+2}+\frac{2}{ab+a+2}$
=1