Tìm x, y biết x^2 + (y-10)^4=0

1/

a/ \(x^2+\left(y-10\right)^2=0\)

vì: \(\left\{{}\begin{matrix}x^2\ge0\forall x\\\left(y-10\right)^4\ge0\forall y\end{matrix}\right.\)

=> Dấu ''='' xảy ra khi \(\left\{{}\begin{matrix}x=0\\y-10=0\Rightarrow y=10\end{matrix}\right.\)

vậy......

b/ \(\left(0,5x-5\right)^{20}+\left(y^2-0,25\right)^{10}\le0\)

vì: \(\left\{{}\begin{matrix}\left(0,5x-5\right)^{20}\ge0\forall x\\\left(y^2-0,25\right)^2\ge0\forall y\end{matrix}\right.\)=> \(\left(0,5x-5\right)^{20}+\left(y^2-0,25\right)^{10}\ge0\)

=> Dấu ''='' xảy ra khi :

\(\left\{{}\begin{matrix}0,5x-5=0\\y^2-0,25=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{0,5}=10\\y^2=0,25\Rightarrow\left[{}\begin{matrix}y=0,5\\y=-0,5\end{matrix}\right.\end{matrix}\right.\)

Vậy........

2/ Ta có: \(2011\equiv1\left(mod10\right)\)

\(2011^{201}\equiv1^{201}\equiv1\left(mod10\right)\);

Có: \(1997^3\equiv3\left(mod10\right)\)

\(\left(1997^3\right)^4\equiv3^4\equiv1\left(mod10\right)\)

\(\left(1997^{12}\right)^{14}\equiv1^{14}\equiv1\left(mod10\right)\) hay \(1997^{168}\equiv1\left(mod10\right)\)

=> \(2011^{201}-1997^{168}\equiv1-1\equiv0\left(mod10\right)\)

hay \(2011^{201}-1997^{168}\) chia hết cho 10

=> Đpcm