Question

tim x,y,zbiet

(3x-5)²⁰¹⁶+(y²-1)²⁰¹⁸+(x-z)²¹⁰⁰=0

Answer 1

Ta thấy : VT >= 0

Dấu "=" xảy ra <=> 3x-5=0 ; y^2-1=0 ; x-z=0

<=> x=z=5/3 ; y=-1 hoặc x=z=5/3 ; y=1

Vậy .........

Tk mk nha

Answer 2

\(\left(3x-5\right)^{2016}\ge0\)

\(\left(y^2-1\right)^{2018}\ge0\)

\(\left(x-z\right)^{2100}\ge0\)

suy ra \(\left(3x-5\right)^{2016}+\left(y^2-1\right)^{2018}+\left(x-z\right)^{2100}\ge0\)

Dấu bằng xảy ra khi và chỉ khi

\(\hept{\begin{cases}\left(3x-5\right)^{2016}=0\\\left(y^2-1\right)^{2018}=0\\\left(x-z\right)^{2100}=0\end{cases}}\)

\(\hept{\begin{cases}3x-5=0\\y^2-1=0\\x-z=0\end{cases}}\)

\(\hept{\begin{cases}3x=5\\y^2=1\\x=z\end{cases}}\)

\(\hept{\begin{cases}x=\frac{5}{3}\\y=\pm1\\z=\frac{5}{3}\end{cases}}\)

T I C K nha