Question

[Wendy ~]

Tìm x,y,z biết

\(\left(2x-1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y-z\right|=0\)

Answer 1

[Suri Anh]

ta có: \(\left(\text{2x − 1}\right)^{2018}\) ≥ 0

\(\left(y-\frac{2}{5}\right)^{2018}\) ≥ 0

\(\left|x+y-z\right|\) ≥ 0

⇒ \(\left(\text{2x − 1 }\right)^{2018}\)+ \(\left(y-\frac{2}{5}\right)^{2018}\) +\(\left|\text{ x + y − z }\right|\) ≥ 0

để \(\left(\text{2x − 1}\right)^{2018}\) + \(\left(y-\frac{2}{5}\right)^{2018}\) + \(\left|\text{x + y − z}\right|\) = 0

⇒ \(\left(\text{2x − 1}\right)^{2018}\) = 0 ⇒ 2x − 1 = 0 ⇒ x = \(\frac{1}{2}\)

\(\left(y-\frac{2}{5}\right)^{2018}\) = 0 ⇒ y − \(\frac{2}{5}\) = 0⇒ \(\frac{2}{5}\)

\(\left|\text{x + y − z}\right|\) = 0 ⇒ x + y − z = 0 ⇒ z = x + y ⇒z = \(\frac{1}{2}\) + \(\frac{2}{5}\) = \(\frac{9}{10}\)

KL: x = \(\frac{1}{2}\); y = \(\frac{2}{5}\); z = \(\frac{9}{10}\)

( mình nghĩ nó còn có nhiều đáp số lắm, nhưng mình ko biết cách làm)

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