Question

Tim x, y, z biết 

\(\left|x^2-4\right|+\left(2y+5\right)^{10}+\left(4z-3\right)^{20}\le0\)

Answer 1

`|x^2 -4| + (2y+5)^10 +(4z -3)^20 <= 0(1)`

Ta có `|x^2 -4|>=0 AAx`

        ` (2y+5)^10>=0 AAy`

         `(4z -3)^20 >=0 AA z`

 `=>|x^2 -4| + (2y+5)^10 +(4z -3)^20 >=0 AA x;y;z(2)` 

`(1)và(2)=> |x^2 -4| + (2y+5)^10 +(4z -3)^20 =0`

`=> {(x^2-4=0),(2y+5=0),(4z-3=0):}`

`=> {(x^2=4),(2y=-5),(4z=3):}`

`=>{(x=+-2),(y=-5/2),(z=3/4):}`

Answer 2

=>x^2-4=0 và 2y+5=0 và 4z-3=0

=>\(\left\{{}\begin{matrix}x\in\left\{2;-2\right\}\\y=-\dfrac{5}{2}\\z=\dfrac{3}{4}\end{matrix}\right.\)