Question

Tìm x, y, z biết

/3x-5/+(2y+5)^20+(4z-3)^206≤0

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Answer 1

Xét \(\left|3x-5\right|\ge0\)

     \(\left(2y+5\right)^{20}\ge0\)

     \(\left(4z-3\right)^{206}\ge0\)

\(\Rightarrow\left|3x-5\right|+\left(2y+5\right)^{20}+\left(4z-3\right)^{206}\ge0\)(1)

Mà:  \(\left|3x-5\right|+\left(2y+5\right)^{20}+\left(4z-3\right)^{206}\le0\)(2)

(1)(2) suy ra: \(\left|3x-5\right|+\left(2y+5\right)^{20}+\left(4z-3\right)^{206}=0\)

\(\hept{\begin{cases}3x-5=0\Rightarrow3x=5\Rightarrow x=\frac{5}{3}\\\left(2y+5\right)^{20}=0\Rightarrow2y+5=0\Rightarrow2y=-5\Rightarrow y=-\frac{5}{2}\\\left(4z-3\right)^{206}=0\Rightarrow4z-3=0\Rightarrow4z=3\Rightarrow z=\frac{3}{4}\end{cases}}\)

Vậy............