Question

Tìm x , y biết :

a) 2 . ( x² - 4 )⁴ + 5 . ( y³ + 8 )² = 0 .

b) 3 . |2x² - 8 | + 7 . ( 2y - 1 )² = 0 .

Answer 1

a) \(2\left(x^2-4\right)^4+5\left(y^3+8\right)^2=0\)

Có 2\(\left(x^2-4\right)^4\) và \(5\left(y^3+8\right)^2\ge0\)

Mà \(2\left(x^2-4\right)^4+5\left(y^3+8\right)^2=0\)

=> \(2\left(x^2-4\right)^4=0\) và \(5\left(y^3+8\right)=0\)

+) \(2\left(x^2-4\right)^4=0\) => \(x^2-4=0=>x^2=4=>x=2\)

b) \(3\left|2x^2-8\right|+7\left(2y-1\right)^2=0\)

Có \(3\left|2x^2-8\right|\ge0\) ; \(7\left(2y-1\right)^2\ge0\)

Mà ​​​​​​​​​\(3\left|2x^2-8\right|+7\left(2y-1\right)^2=0\)​

=> \(3\left|2x^2-8\right|=0\) ; \(7\left(2y-1\right)^2=0\)\

+) ​​\(3\left|2x^2-8\right|=0\) => \(2x^2-8=0=>2x^2=8=>x^2=4=>x=2\)

+) \(7\left(2y-1\right)^2=0\)

=> 2y-1=0

=> 2y = 1

=> y= \(\dfrac{1}{2}\)