giải hpt

Question

$\begin{cases}(x+6y+3)\sqrt{xy+3y}=y(8y+3x+9) \\ \sqrt{-x^{2}+8x-24y+417}=(y+3)\sqrt{y-1} +3y+17\end{cases}$

Bloody's Rose · Answer 1 · 5/10/2016 5:42:27 PM

ĐK:

$x\geq -3$ ;

$y\geq 1$ ;...

pt(1)

$\Leftrightarrow$

$\left[ {(x+3)+6y} \right]\sqrt{y(x+3)}=\left[ {8y+3(x+3)} \right]y$ (1')

Đặt

$\begin{cases}\sqrt{x+3}=a \\ \sqrt{y} =b\end{cases}$ ;

$a\geq 0;b\geq 1$

(1') tt:

$(a^{2}+6b^{2})ab=(8b^{2}+3a^{2})b^{2}$

$\Leftrightarrow$

$b(a-2b)(a^{3}+5ab+4b^{2})$ =0

$\Rightarrow$

$a=2b$

$\Rightarrow$

$x=4y-3$

(2)tt

$\sqrt{-16y^{2}+32y+384}=(y+3)\sqrt{y-1}+3y+17$ (3)

Xét thấy:

$\sqrt{-16y^{2}+32y+384}=\sqrt{-16(y-1)^{2}+400}\leq20$ (4)

$(y+3)\sqrt{y-1}+3y+17\geq20( y\geq1)$ (5)

(3)

$\Leftrightarrow$ (4)&(5) xra dấu''=''

$\Leftrightarrow$ y=1

$\Rightarrow$ x=1(t/m đk)

KL:...

score 5 · Answer 2

http://i.imgur.com/BKGyMXJ.png

2 Đáp án