Question

Tìm x, y :

\(|3x+1|+|2y-1|=0\)

Answer 1

Với mọi giá trị của \(x;y\in R\) ta có:

\(\left|3x+1\right|+\left|2y-1\right|\ge0\)

Để \(\left|3x+1\right|+\left|2y-1\right|=0\) thì

\(\left\{{}\begin{matrix}\left|3x+1\right|=0\\\left|2y-1\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{3}\\y=\dfrac{1}{2}\end{matrix}\right.\)

Vậy.........................

Answer 2

Giải.

a) \(\left|3x+1\right|+\left|2y-1\right|=0\)

Nhận xét : Do \(\left|3x+1\right|\ge0\) với \(\forall x\)

và \(\left|2y-1\right|\ge0\) với \(\forall y\)

Nên \(\left|3x+1\right|+\left|2y-1\right|=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left|3x+1\right|=0\\\left|2y-1\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+1=0\\2y-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-1}{3}\\y=\dfrac{1}{2}\end{matrix}\right.\)

Vậy \(x=\dfrac{-1}{3}\) và \(y=\dfrac{1}{2}\)

tik mik nha !!!

Answer 3

a, \(\left|3x+1\right|+\left|2y-1\right|=0\)

Nhận xét : Do \(\left|3x+1\right|\ge0\) với \(\forall x\) và \(\left|2y-1\right|\ge0\) với \(\forall y\)

Nên \(\left|3x+1\right|+\left|2y-1\right|=0\)

=> \(\left\{{}\begin{matrix}\left|3x+1\right|=0\\\left|2y-1\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+1=0\\2y-1=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-1}{3}\\y=\dfrac{1}{2}\end{matrix}\right.\)

Vậy \(x=\dfrac{-1}{3};y=\dfrac{1}{2}\)