a)A=\(y-5x\sqrt{y}+6x^2\)
=\(y-2x\sqrt{y}-3x\sqrt{y}+6x^2\)
=\(\sqrt{y}\left(\sqrt{y}-2x\right)-3x\left(\sqrt{y}-2x\right)\)
=(\(\sqrt{y}-3x\))\(\left(\sqrt{y}-2x\right)\)
b)Có \(y=\frac{18}{4+\sqrt{7}}=\frac{18\left(4-\sqrt{7}\right)}{16-7}=2\left(4-\sqrt{7}\right)=8-2\sqrt{7}=\left(\sqrt{7}-1\right)^2\)
=> \(\sqrt{y}=\sqrt{\left(\sqrt{7}-1\right)^2}=\left|\sqrt{7}-1\right|=\sqrt{7}-1\)
Thay \(x=-\frac{2}{3}\),\(\sqrt{y}=\sqrt{7}-1\) vào A đã rút gọn có:
A=\(\left(\sqrt{7}-1-3.\frac{-2}{3}\right)\left(\sqrt{7}-1-2.\frac{-2}{3}\right)=\left(\sqrt{7}-1+2\right)\left(\sqrt{7}-1+\frac{4}{3}\right)=\left(\sqrt{7}+1\right)\left(\sqrt{7}+\frac{1}{3}\right)=7+\frac{\sqrt{7}}{3}+\sqrt{7}+\frac{1}{3}=\frac{22+4\sqrt{7}}{3}\)
Vậy A=\(\frac{22+4\sqrt{7}}{3}\)
