Question

Tìm x,y biết :

a) \(\left(x-\frac{2}{5}\right)^{2010}\)+\(\left(y+\frac{3}{7}\right)^{468}\)\(\le0\)

B) \(\left(x+0,7\right)^{84}\)+ \(\left(y-6,3\right)^{262}\)\(\le0\)

c) \(\left(x-5\right)^{88}\)+\(\left(x+y+3\right)^{468}\)\(\le0\)

Answer 1

Gợi ý: Các biểu thức mũ chẵn đều không âm.

\(a^{2n}+b^{2n}\le0\Leftrightarrow a^{2n}+b^{2n}=0\Leftrightarrow a=b=0\)

Answer 2

a,\(\left(x-\frac{2}{5}\right)^{2010}+\left(y+\frac{3}{7}\right)^{468}\)< \(0\)

Vì \(\left(x-\frac{2}{5}\right)^{2010}\);\(\left(y+\frac{3}{7}\right)^{468}\)đều > \(0\)

=> \(\left(x-\frac{2}{5}\right)^{2010}=0\)

     \(\left(y+\frac{3}{7}\right)^{468}=0\)

=> \(\left(x-\frac{2}{5}\right)^{2010}=0^{2010}\)

     \(\left(y+\frac{3}{7}\right)^{468}=0^{468}\)

=> \(x-\frac{2}{5}=0\)

      \(y-\frac{3}{7}=0\)

=> \(x=\frac{2}{5}\)

      \(y=\frac{3}{7}\)

Vậy \(x=\frac{2}{5}\)\(y=\frac{3}{7}\)

Answer 3

b,\(\left(x+0,7\right)^{84}+\left(y-6,3\right)^{262}\)< \(0\)

Vì \(\left(x+0,7\right)^{84}\);\(\left(y-6,3\right)^{262}\)đều > \(0\)

=>\(\left(x+0,7\right)^{84}\)  = \(0\)

    \(\left(y-6,3\right)^{262}\)  = \(0\)

=> \(x+0,7=0\)

     \(y-6,3=0\)

=> \(x=0,7\)

     \(y=-6,3\)

Vậy \(x=0,7\)\(y=-6,3\)

Answer 4

Mình xin lỗi, chỗ câu a thì \(y=-3,7\)  nhé

Answer 5

quên câu a thì \(y=-37\)

Answer 6

c,\(\left(x-5\right)^{88}+\left(x+y+3\right)^{468}\) < \(0\)

Vì \(\left(x-5\right)^{88}\)và \(\left(x+y+3\right)^{468}\)đều > \(0\)

=> \(\left(x-5\right)^{88}\)\(=0\)

     \(\left(x+y+3\right)^{468}\)\(=0\)

=> \(x-5=0\)

     \(x+y+3=0\)

=> \(x=5\)

     \(x+y=3\)

=>\(x=5\)

     \(y=-2\)

Vậy \(x=5\)\(y=-2\)

Answer 7

cái này của lớp 5 ?!?!?!? ai mà giải được !