Question

\(Fin\text{d}\)  \(x,y,x\)

I \(5.x-3\) I + \(\left(\frac{1}{2}y-27\right)^{108}+\left(37.x+2\right)^{10}<\text{or}=0\)

Answer 1

Vì: | 5x - 3 | \(\ge0\)

\(\left(\frac{1}{2}y-27\right)^{108}\ge0\)

( 37x + 2 )¹⁰ \(\ge0\)

Để \(\left|5x-3\right|+\left(\frac{1}{2}y-27\right)^{108}+\left(37x+2\right)^{10}=0\)

<=> | 5x - 3 | = 0

5x = 3

x = 3/5

<=> \(\left(\frac{1}{2}y-27\right)^{108}=0\) khi \(\frac{1}{2}y-27=0\)

1/2y = 27

y = 54

<=> ( 37x + 2 )¹⁰ = 0

37x + 2 = 0

37x = -2

x = -2/37

KL: x,y,x = ........................................

Answer 2

đù, lm mà ko khí lik-e @@