PPGNADPP

cách khác cho bài 3b

$x^{2}-y^{2}=xy+8 <=> 4x^{2}-4y^{2}=4xy+32 <=> \left ( 4x^{2}-4xy+1 \right )-4y^{2}=33 <=>\left ( 2x-1 \right )^{2}-4y^{2}=33 <=>\left ( 2x-2y-1 \right )\left ( 2x+2y-1 \right )=33$

vì $2x+2y-1\geq 2x-2y-1$ nên từ đây xét 4 trường hợp nữa là xong

4b

ĐK x,y\geq 0

$\sqrt{x}+\sqrt{y}=\sqrt{21}$

<=> $x+y+2\sqrt{xy}=21$

Không mất tính tổng quát giả sử $x\geq y$

<=> $x^{2}\geq y^{2}$

Có $y+y+2\sqrt{y.y}\leq x+y+2\sqrt{xy}=21$

<=> $4y\leq 21$

<=> $y=\left \{ 0;1;2;3;4;5 \right \}$

từ đó tìm ra x

Làm ra đi!!!!!!

$\inline M=x^{2}+y^{2}+\frac{3}{x+y+1}=\frac{3x^{2}}{4}+\frac{3y^{2}}{4}+\left ( \frac{x^{2}}{4}+\frac{1}{4} \right )+\left ( \frac{y^{2}}{4}+\frac{1}{4} \right )-\frac{1}{2}+\frac{3}{x+y+1}\geq \frac{3x^{2}}{4}+\frac{3y^{2}}{4}+\frac{x}{2}+\frac{y}{2}-\frac{1}{2}+\frac{3}{x+y+1}=\frac{x^{2}}{2}+\frac{y^{2}}{2}+\left ( \frac{x^{2}}{4}+\frac{y^{2}}{4}+\frac{1}{4}+\frac{xy}{2}+\frac{x}{2}+\frac{y}{2} \right )-\frac{1}{4}-\frac{1}{2}-\frac{xy}{2}+\frac{3}{x+y+1}=\frac{x^{2}}{2}+\frac{y^{2}}{2}+\left ( \frac{x}{2}+\frac{y}{2}+\frac{1}{2} \right )^{2}-\frac{5}{4}+\frac{3}{x+y+1}=\left ( \frac{x^{2}}{2}+\frac{y^{2}}{2} \right )+\left [ \frac{1}{4}\left ( x+y+1 \right )^{2}+\frac{\frac{27}{4}}{x+y+1}+\frac{\frac{27}{4}}{x+y+1} \right ]-\frac{\frac{21}{2}}{x+y+1}-\frac{5}{4}\geq xy+\frac{27}{4}-\frac{\frac{21}{2}}{2\sqrt{xy}+1}-\frac{5}{4}=1+\frac{27}{4}-\frac{\frac{21}{2}}{2.1+1}-\frac{5}{4}=3$

tách ra dùng AM-GM