câu 3:
1, xét x=3/2 => là nghiệm
với x khác 3/2 có:
[tex]pt <=> 2\sqrt{2x-3}=\sqrt{x^2+19x-28}-\sqrt{x^2+3x-4}\\\\ <=> \frac{2.(2x-3)}{\sqrt{2x-3}}=\frac{8.(2x-3)}{\sqrt{x^2+19x-28}+\sqrt{x^2+3x-4}}\\\\ <=> \sqrt{x^2+19x-28}+\sqrt{x^2+3x-4}=4\sqrt{2x-3}\\\\ +, 2\sqrt{2x-3}=\sqrt{x^2+19x-28}-\sqrt{x^2+3x-4}\\\\ => \sqrt{x^2+19x-28}=3\sqrt{2x-3}\\\\ <=> ...[/tex]
2, xét y=0 => x=0 thay vô xem là nghiệm ko
xét y khác 0 => [tex]\sqrt{x+y}\neq \sqrt{x-2y}\\\\ => (1) <=> \frac{x+y-x+2y}{\sqrt{x+y}-\sqrt{x-2y}}=3y\\\\ <=> \sqrt{x+y}-\sqrt{x-2y}=1\\\\ +, \sqrt{x+y}+\sqrt{x-2y}=3y\\\\ => 2\sqrt{x+y}=3y+1\\\\ => 4(x+y)=(3y+1)^2\\\\ => 4x=9y^2+2y+1\\\\ (2) <=> (3x-1)\sqrt{4x+1}+4x\sqrt{3x-2}=15x+1\\\\ <=> (3x-1)(\sqrt{4x+1}-3)+4x(\sqrt{3x-2}-2)+2(x-2)=0\\\\ <=> (x-2).[\frac{4(3x-1)}{\sqrt{4x+1}+3}+\frac{12x}{\sqrt{3x-2}+2}+2]=0\\\\\ <=> x=2 => y=...[/tex]
câu 5:
GT <=> x+y=xy
[tex]L=\frac{1}{x^2+2y}+\frac{1}{y^2+2x}+\sqrt{(1+x^2).(1+y^2)}\\\\\ \geq \frac{4}{x^2+y^2+2(x+y)}+\sqrt{(1+xy)^2}\\\\\ => L\geq \frac{4}{(x+y)^2}+\frac{x+y}{16}+\frac{x+y}{16}+\frac{7.(x+y)}{16}+1[/tex]
[tex]x+y=xy\leq \frac{1}{4}(x+y)^2\\\\ =>x+y\geq 4[/tex]
suy ra: [tex]L\geq ...[/tex]
dấu "=" <=> x=y=2
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
