giúp với

Question

Giải hệ:

$\begin{cases}y(\sqrt{x}+y)=2x\sqrt{x}+2xy \\ y(\sqrt{x^2+1}-1)=\sqrt{3x^2+3} \end{cases}$

❄⊰๖ۣۜNgốc๖ۣۜ ⊱ ❄ · Answer 1 · 7/10/2017 6:24:25 AM

ĐK:

$x\geq 0$

$PT(1)\Leftrightarrow \sqrt{x}(2x-y)+y(2x-y)=0$

$\Leftrightarrow (2x-y)(\sqrt{x}+y)=0$

$\Leftrightarrow y=2x\veebar y=-\sqrt{x}$

+)Với

$y=2x$ pt (2)

$\Leftrightarrow 2x(\sqrt{x^2+1}-1)=\sqrt{3x^2+3}$

$\Leftrightarrow 2x(\sqrt{x^2+1}-2)+2x-\sqrt{3x^2+3}=0$

$\Leftrightarrow \frac{2x(x^2-3)}{\sqrt{x^2+1}+2}+\frac{x^2-3}{2x+\sqrt{3x^2+3}}=0$

$\Leftrightarrow (x^2-3)(\frac{2x}{\sqrt{x^2+1}+2}+\frac{1}{2x+\sqrt{3x^2+3}})=0$

$\Leftrightarrow x^2-3=0$

$\Leftrightarrow x=\sqrt{3}$

+) Với

$y=-\sqrt{x}$ pt (2)

$\Leftrightarrow -\sqrt{x}(\sqrt{x^2+1}-1)=\sqrt{3x^3+3}$ ( vô nghiệm)

Vậy ..........

~~~~

$Amen$ ~~~~

Trả lời 10-07-17 06:24 AM

❄⊰๖ۣۜNgốc๖ۣۜ ⊱ ❄
1

258K 575K

^^ kcj :v – ❄⊰๖ۣۜNgốc๖ۣۜ ⊱ ❄ 10-07-17 06:45 AM

thank b................ – TQT 10-07-17 06:26 AM

1 Đáp án