Question

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

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

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

Answer 3

Để biểu thức a = 0 thì x - 3 = 0 và y + 2 = 0

Xét x-3=0

=>x=3

Xét y+2=0

=>y=-2

Vậy x=0;y=-2

Answer 4

Ta thấy:

\(\left(2x-5\right)^{2000}\ge0\)

\(\left(3y+4\right)^{2002}\ge0\)

\(\Rightarrow\left(2x-5\right)^{2000}+\left(3y+4\right)^{2002}\ge0\)

Mà theo bài ra ta có: \(\left(2x-5\right)^{2000}+\left(3y+4\right)^{2002}\le0\)

\(\Rightarrow\left(2x-5\right)^{2000}+\left(3y+4\right)^{2002}=0\)

\(\Rightarrow\begin{cases}\left(2x-5\right)^{2000}=0\\\left(3y+4\right)^{2002}=0\end{cases}\)

\(\Rightarrow\begin{cases}2x-5=0\\3y+4=0\end{cases}\)

\(\Rightarrow\begin{cases}2x=5\\3y=-4\end{cases}\)

\(\Rightarrow\begin{cases}x=\frac{5}{2}\\y=-\frac{4}{3}\end{cases}\)

Vậy \(\begin{cases}x=\frac{5}{2}\\y=-\frac{4}{3}\end{cases}\)