Gợi ý: Các biểu thức mũ chẵn đều không âm.
\(a^{2n}+b^{2n}\le0\Leftrightarrow a^{2n}+b^{2n}=0\Leftrightarrow a=b=0\)
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 \(x=\frac{2}{5}\)\(y=\frac{3}{7}\)
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 \(x=0,7\)\(y=-6,3\)
Mình xin lỗi, chỗ câu a thì \(y=-3,7\) nhé
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 \(x=5\)\(y=-2\)
cái này của lớp 5 ?!?!?!? ai mà giải được !