Giải hệ phương trình : $
    \begin{cases}x^{3}-7y^{3}+3xy(x+y)-24y^{2}+3x-27y=14\\ \sqrt{3-x} +\sqrt{y+4}=x^{3}+y^{2}-5\end{cases}     $   $(x,y   \in  R)$
PT(1) $\Leftrightarrow (x-y-2)(x^2+(4y+2)x+7y^2+10y+7)=0$
$\Leftrightarrow (x-y-2)((x+2y+1)^2+3y^2+6y+6)=0\Leftrightarrow x=y-2$
Thay PT(2), nhân 3 cả 2 vế:
$3x^3+3x^2-12x-12+(5-x-3\sqrt{3-x})+(x+4-3\sqrt{x+2})=0$
$\Leftrightarrow (x-2)(x+1)(...)=0$
P/s: đề này nặng về tính toán quá :(
câu hỏi và trl- tỉ lệ 1:5! -_- –  Confusion 11-04-16 01:07 PM
chênh lệch j vậy chị? –  111 11-04-16 08:56 AM
chênh lệch giữa câu hỏi –  Confusion 10-04-16 10:20 PM
đúng click V và voteup giùm em với ạ –  111 10-04-16 08:29 AM

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