Question

[Sakura]

Tìm x,y biết :

a) ( x - 3 )² + ( y + 2 )² = 0

b) 2x + 2^x+3 = 136

c) ( x - 12 + y )²⁰⁰ + ( x - 4 -y )²⁰⁰ = 0

d) ( 2x - 5)²⁰⁰⁰+(3y+4)²⁰⁰²< hoặc = 0

Answer 1

[Trần Minh Hưng]

a) Ta thấy:

\(\left(x-3\right)^2\ge0\)

\(\left(y+2\right)^2\ge0\)

\(\Rightarrow\left(x-3\right)^2+\left(y+2\right)^2\ge0\)

Để \(\left(x-3\right)^2+\left(y+2\right)^2=0\)

\(\Rightarrow\begin{cases}\left(x-3\right)^2=0\\\left(y+3\right)^2=0\end{cases}\)

\(\Rightarrow\begin{cases}x-3=0\\y+3=0\end{cases}\)

\(\Rightarrow\begin{cases}x=3\\y=-3\end{cases}\)

Vậy \(\begin{cases}x=3\\y=-3\end{cases}\)

Answer 2

[Trần Minh Hưng]

c) Ta thấy:

\(\left(x-12+y\right)^{200}\ge0\)

\(\left(x-4-y\right)^{200}\ge0\)

\(\Rightarrow\left(x-12+y\right)^{200}+\left(x-4-y\right)^{200}\ge0\)

Để \(\left(x-12+y\right)^{200}+\left(x-4-y\right)^{200}=0\)

\(\Rightarrow\begin{cases}\left(x-12+y\right)^{200}=0\\\left(x-4-y\right)^{200}=0\end{cases}\)

\(\Rightarrow\begin{cases}x-12+y=0\\x-4-y=0\end{cases}\)

\(\Rightarrow\begin{cases}x+y=12\\x-y=4\end{cases}\)

\(\Rightarrow\begin{cases}x=\left(12+4\right):2\\y=\left(12-4\right):2\end{cases}\)

\(\Rightarrow\begin{cases}x=8\\y=4\end{cases}\)

Vậy \(\begin{cases}x=8\\y=4\end{cases}\)