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[Quoc Tuan Qbdh]

Giải hệ phương trình 

$1)$ $\left\{\begin{matrix}\sqrt{xy}-2\sqrt{yz}+3\sqrt{zx}=1 \\ x^{2}-3zx+yz=\frac{1}{3} \\ 2xyz-y^{3}+y^{2}z=2 \end{matrix}\right.$

$2)$ $\left\{\begin{matrix}\sqrt[4]{a-1}-\sqrt{b-1}=1 \\ a^{2}+2b=5 \end{matrix}\right.$

Edited by Quoc Tuan Qbdh, 31-07-2015 - 03:56.

[vutienhoang]

2) pt 1 <=> $\sqrt[4]{a-1}=\sqrt{b-1}+1\geq 1$ => a$\geq$ 2 = > $a^{2}\geq 4$ có  $b \geq 1$ => $a^{2}+2b\geq 6$