Question

Tìm x , y biết:

\(\left(2x-5\right)^{2016}+\left(3y+4\right)^{2020}\le0\)

Answer 1

x=5/2,y=-4/3

Answer 2

Vì \(\left(2x-5\right)^{2016}\ge0\forall x;\left(3y+4\right)^{2020}\ge0\forall y\)

\(\Rightarrow\left(2x-5\right)^{2016}+\left(3y+4\right)^{2020}\ge0\)

Mà đề lại cho \(\left(2x-5\right)^{2016}+\left(3y+4\right)^{2020}\le0\)

Nên \(\hept{\begin{cases}\left(2x-5\right)^{2016}=0\\\left(3y+4\right)^{2020}=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{5}{2}\\y=-\frac{4}{3}\end{cases}}}\)

Vậy ..........

Answer 3

vì (2x-5)²⁰¹⁶  và (3y+4)²⁰²⁰ >hoặc=0 với mọi x

=>2x-5=3y+4=0

=>x=2/5;y=-4/3

Answer 4

a^xyz=(a^x)^yz=(bc)^yz=(b^y)^z.(c^z)^y
               =(ca)^z.(ab)^y=c^z.a^z.a^y.b^y
               =a^z+y.a^2.(b.c)=a^z+y+z.a^x=a^x+y+z+z
           =>xyz=x+y+z+2 (đpcm)​