Question

[Tạ Kim Thoa]

Tìm x, y biết:

(2×x- 5)²⁰⁰⁶ + (3×y+4)²⁰⁰⁸ = 0

Answer 1

[Nguyễn Ngọc Linh]

\(\left(2x-5\right)^{2006}+\left(3y+4\right)^{2008}=0\)

Vì \(\left\{{}\begin{matrix}\left(2x-5\right)^{2006}\ge0\forall x\\\left(3y+4\right)^{2008}\ge0\forall y\end{matrix}\right.\)\(\Rightarrow\left(2x-5\right)^{2006}+\left(3y+4\right)^{2008}\ge0\forall x,y\)

Dấu = xảy ra khi: \(\left\{{}\begin{matrix}\left(2x-5\right)^{2006}=0\\\left(3y+4\right)^{2008}=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-5=0\\3y+4=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\y=-\frac{4}{3}\end{matrix}\right.\)

Vậy \(x=\frac{5}{2},y=-\frac{4}{3}\)

Answer 2

[Vũ Minh Tuấn]

\(\left(2x-5\right)^{2006}+\left(3y+4\right)^{2008}=0\)

Ta có:

\(\left\{{}\begin{matrix}\left(2x-5\right)^{2006}\ge0\\\left(3y+4\right)^{2008}\ge0\end{matrix}\right.\forall x,y\)

\(\Rightarrow\left(2x-5\right)^{2006}+\left(3y+4\right)^{2008}\ge0\forall x,y.\)

\(\Rightarrow\left(2x-5\right)^{2006}+\left(3y+4\right)^{2008}=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(2x-5\right)^{2006}=0\\\left(3y+4\right)^{2008}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x-5=0\\3y+4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=5\\3y=-4\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=5:2\\y=\left(-4\right):3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\y=-\frac{4}{3}\end{matrix}\right.\)

Vậy \(\left(x;y\right)\in\left\{\frac{5}{2};-\frac{4}{3}\right\}.\)

Chúc bạn học tốt!