Question

[Nguyen Sinh Phuc]

c) (x² - 1)10 _{+ (y² - 36)^{12 ≤ 0}}

_{^{help me mai phai nop roi T_T}}

Answer 1

[Vũ Minh Tuấn]

c) \(\left(x^2-1\right)^{10}+\left(y^2-36\right)^{12}\le0\)

Ta có:

\(\left\{{}\begin{matrix}\left(x^2-1\right)^{10}\ge0\\\left(y^2-36\right)^{12}\ge0\end{matrix}\right.\forall x,y.\)

\(\Rightarrow\left(x^2-1\right)^{10}+\left(y^2-36\right)^{12}\ge0\) \(\forall x,y.\)

Mà \(\left(x^2-1\right)^{10}+\left(y^2-36\right)^{12}\le0\)

\(\Rightarrow\left(x^2-1\right)^{10}+\left(y^2-36\right)^{12}=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x^2-1\right)^{10}=0\\\left(y^2-36\right)^{12}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x^2-1=0\\y^2-36=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x^2=1\\y^2=36\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\\\left[{}\begin{matrix}y=6\\y=-6\end{matrix}\right.\end{matrix}\right.\)

Vậy \(\left(x;y\right)\in\left\{1;6\right\},\left\{-1;-6\right\}.\)

Chúc bạn học tốt!

Answer 2

[Nguyen Sinh Phuc]

cảm ơn bạn nhiều nhiều nha