Question

a. \(\frac{2}{3}x+\frac{5}{7}=\frac{3}{10}\)

b.\(-\frac{21}{13}x+\frac{1}{3}=-\frac{2}{3}\)

c. |x-1,5|=2

d.\(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)

Answer 1

a) \(\frac{2}{3}x+\frac{5}{7}=\frac{3}{10}\)

\(\Leftrightarrow\frac{2}{3}x=\frac{3}{10}-\frac{5}{7}\)

\(\Leftrightarrow\frac{2}{3}x=-\frac{29}{70}\)

\(\Leftrightarrow x=-\frac{29}{70}:\frac{2}{3}\)

\(\Leftrightarrow x=-\frac{87}{140}\)

b) \(-\frac{21}{13}x+\frac{1}{3}=-\frac{2}{3}\)

\(\Leftrightarrow-\frac{21}{13}x=-\frac{2}{3}-\frac{1}{3}\)

\(\Leftrightarrow-\frac{21}{13}x=-1\)

\(\Leftrightarrow x=-1:\left(-\frac{21}{13}\right)\)

\(\Leftrightarrow x=\frac{13}{21}\)

c) \(\left|x-1,5\right|=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1,5=2\\x-1,5=-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=3,5\\x=-0,5\end{matrix}\right.\)

d) \(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)

\(\Leftrightarrow\left|x+\frac{3}{4}\right|=\frac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{4}\\x=\frac{5}{4}\end{matrix}\right.\)

Answer 2

a) \(\frac{2}{3}x+\frac{5}{7}=\frac{3}{10}\)

=> \(\frac{2}{3}x=\frac{3}{10}-\frac{5}{7}\)

=> \(\frac{2}{3}x=-\frac{29}{70}\)

=> \(x=-\frac{29}{70}:\frac{2}{3}\)

=> \(x=-\frac{29}{70}.\frac{3}{2}\)

=> \(x=-\frac{87}{140}\)

b) \(-\frac{21}{13}x+\frac{1}{3}=-\frac{2}{3}\)

=> \(-\frac{21}{13}x=-\frac{2}{3}-\frac{1}{3}\)

=> \(-\frac{21}{13}x=-\frac{3}{3}\)

=> \(-\frac{21}{13}x=1\)

=> \(x=1:\left(-\frac{21}{13}\right)\)

=> \(x=-\frac{13}{21}\)

c) \(\left|x-1,5\right|=2\)

=> \(\left[{}\begin{matrix}x-1,5=2\\x-1,5=-2\end{matrix}\right.=>\left[{}\begin{matrix}x=2+1,5\\x=-2+1,5\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=3,5\\x=-0,5\end{matrix}\right.=>\left[{}\begin{matrix}x=\frac{7}{2}\\x=-\frac{1}{2}\end{matrix}\right.\)(T/M)

d) \(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)

=> \(\left|x+\frac{3}{4}\right|=\frac{1}{2}\)

=> \(=>\left[{}\begin{matrix}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{matrix}\right.=>\left[{}\begin{matrix}x=\frac{1}{2}-\frac{3}{4}\\x=-\frac{1}{2}-\frac{3}{4}\end{matrix}\right.=>\left[{}\begin{matrix}x=-\frac{1}{4}\\x=-\frac{5}{4}\end{matrix}\right.\)(T/M)

HỌC TỐT

Answer 3

chj lm câu b) , d) sai kìa

Answer 4

hình như câu b em sai chj đúng còn câu d em đúng chj sai

do cả hai mắc chứng thiếu dấu