# Violympic toán 9

Tìm x,y,z biết:

$2x+3y+4z+77=2\left(3\sqrt{2x-1}+5\sqrt{3y-2}+7\sqrt{4z-3}\right)$

Tân binh -
18 tháng 11 2020 lúc 22:47

ĐK: x ≥ $\frac{3}{4}$

2x + 3y + 4z + 77 = 2( $3\sqrt{2x-1}+5\sqrt{3y-2}+7\sqrt{4x-3}$)

⇔2x +3y +4z +77= $6\sqrt{2x-1}+10\sqrt{3y-2}+14\sqrt{4x-3}$

⇔2x - $6\sqrt{2x-1}$ + 3y- $10\sqrt{3y-2}$+ 4z-$14\sqrt{4x-3}$ +77=0

⇔2x-1-$6\sqrt{2x-1}$ +9 +3y-2-$10\sqrt{3y-2}$+25 +4z-3-$14\sqrt{4x-3}$ +49=0

$\left(\sqrt{2x-1}-3\right)^2+\left(\sqrt{3y-2}-5\right)^2+\left(\sqrt{4y-3}-7\right)^2$ =0

Có: $\left(\sqrt{2x-1}-3\right)^2$ ≥ 0 , ∀ x ;

$\left(\sqrt{3y-2}-5\right)^2$ ≥ 0, ∀ y ;

$\left(\sqrt{4z-3}-7\right)^2$ ≥ 0, ∀ z

$\left(\sqrt{2x-1}-3\right)^2+\left(\sqrt{3y-2}-5\right)^2+\left(\sqrt{4y-3}-7\right)^2$ ≥0, ∀ x, y, z.

$\left\{{}\begin{matrix}\left(\sqrt{2x-1}-3\right)^2=0\\\left(\sqrt{3y-2}-5\right)^2=0\\\left(\sqrt{4z-3}-7\right)^2=0\end{matrix}\right.$$\left\{{}\begin{matrix}\sqrt{2x-1}-3=0\\\sqrt{3y-2}-5=0\\\sqrt{4x-3}-7=0\end{matrix}\right.$

$\left\{{}\begin{matrix}\sqrt{2x-1}=3\\\sqrt{3y-2}=5\\\sqrt{4z-3}=7\end{matrix}\right.$$\left\{{}\begin{matrix}2x-1=9\\3y-2=25\\4z-3=49\end{matrix}\right.$$\left\{{}\begin{matrix}x=5\\y=9\\z=13\end{matrix}\right.$

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