Question

tìm x:

a) 3(x-\(\frac{1}{2}\))-5(x+\(\frac{3}{5}\))= -x+\(\frac{1}{5}\)

b)(2x-1) (x+\(\frac{2}{3}\))=0

mọi người giải giúp mình với nhé mình đang cần gấp. Cảm ơn nhé

Answer 1

\(b,\left(2x-1\right).\left(x+\frac{2}{3}\right)=0\)

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

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

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

Vậy : \(x\in\left\{\frac{1}{2},-\frac{2}{3}\right\}\)

Answer 2

a) \(3.\left(x-\frac{1}{2}\right)-5.\left(x+\frac{3}{5}\right)=-x+\frac{1}{5}\)

\(\Leftrightarrow3x-\frac{1}{6}-5x-\frac{3}{25}=-x+\frac{1}{5}\)

\(\Leftrightarrow3x-5x+x=\frac{1}{5}+\frac{1}{6}+\frac{3}{25}\)

\(\Leftrightarrow-x=\frac{73}{150}\)

\(\Leftrightarrow x=-\frac{73}{150}\)

Vậy : \(x=-\frac{73}{150}\)

Answer 3

b) \(\left(2x-1\right).\left(x+\frac{2}{3}\right)=0\)

=> \(\left[{}\begin{matrix}2x-1=0\\x+\frac{2}{3}=0\end{matrix}\right.\) => \(\left[{}\begin{matrix}2x=0+1=1\\x=0-\frac{2}{3}\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=1:2\\x=-\frac{2}{3}\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{2}{3}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{1}{2};-\frac{2}{3}\right\}.\)

Mình chỉ làm câu b) thôi nhé.

Chúc bạn học tốt!