Question

Tìm \(x\) biết:

a) \(\dfrac{1}{3}\).( \(\dfrac{1}{2}\) - \(6\)) + 5\(x\) = \(x\) - \(\dfrac{3}{5}\)

b) \(\dfrac{3}{2}\). \(x\) - \(1\dfrac{1}{2}\) = \(x\) - \(\dfrac{3}{4}\)

c)\(x\) - \(\dfrac{5}{4}\) = \(\dfrac{1}{3}-\dfrac{3}{4}\) . \(x\)

d) \(\dfrac{3}{2}.\left(x-\dfrac{5}{3}\right)\)- \(\dfrac{7}{5}\) = \(x+1\)

e)\(\dfrac{2}{3}.\left|4x+3\right|=\dfrac{6}{15}\)

giúp mk với, mai mk phải nộp rồi

cách làm nữa nhé

Answer 1

\(a.\)

\(\dfrac{1}{3}\left(\dfrac{1}{2}-6\right)+5x=x-\dfrac{3}{5}\)

\(\Rightarrow\dfrac{1}{6}-2+5x=x-\dfrac{3}{5}\)

\(\Rightarrow\dfrac{1}{6}-2+\dfrac{3}{5}=-5x+x\)

\(\Rightarrow-4x=-\dfrac{37}{30}\)

\(\Rightarrow4x=\dfrac{37}{30}\)

\(\Rightarrow x=\dfrac{37}{120}\)

\(b.\)

\(\dfrac{3}{2}x-1\dfrac{1}{2}=x-\dfrac{3}{4}\)

\(\Rightarrow\dfrac{3}{2}x-\dfrac{3}{2}=x-\dfrac{3}{4}\)

\(\Rightarrow\dfrac{3}{2}x-x=\dfrac{3}{2}-\dfrac{3}{4}\)

\(\Rightarrow\dfrac{1}{2}x=\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{3}{2}\)

\(c.\)

\(x-\dfrac{5}{4}=\dfrac{1}{3}-\dfrac{3}{4}x\)

\(\Rightarrow x+\dfrac{3}{4}x=\dfrac{1}{3}+\dfrac{5}{4}\)

\(\Rightarrow\dfrac{7}{4}x=\dfrac{19}{12}\)

\(\Rightarrow x=\dfrac{19}{21}\)

\(d.\)

\(\dfrac{3}{2}\left(x-\dfrac{5}{3}\right)-\dfrac{7}{5}=x+1\)

\(\Rightarrow\dfrac{3}{2}x-\dfrac{5}{2}-\dfrac{7}{5}=x+1\)

\(\Rightarrow\dfrac{3}{2}x-\dfrac{39}{10}=x+1\)

\(\Rightarrow\dfrac{3}{2}x-x=\dfrac{39}{10}+1\)

\(\Rightarrow\dfrac{1}{2}x=\dfrac{49}{10}\)

\(\Rightarrow x=\dfrac{49}{5}\)

\(e.\)

\(\dfrac{2}{3}\left|4x+3\right|=\dfrac{6}{15}\)

\(\Rightarrow\left|4x+3\right|=\dfrac{3}{5}\)

\(\Rightarrow\left[{}\begin{matrix}4x+3=\dfrac{3}{5}\\4x+3=-\dfrac{3}{5}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}4x=-\dfrac{12}{5}\\4x=-\dfrac{18}{5}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=-\dfrac{9}{10}\end{matrix}\right.\)

Answer 2

a) \(\dfrac{1}{3}.\left(\dfrac{1}{2}-6\right)+5x=x-\dfrac{3}{5}\Leftrightarrow\dfrac{1}{6}-2+5x=x-\dfrac{3}{5}\)

\(\Leftrightarrow5x-x=-\dfrac{3}{5}-\dfrac{1}{6}+2\Leftrightarrow4x=\dfrac{37}{30}\Leftrightarrow x=\dfrac{\dfrac{37}{30}}{4}=\dfrac{37}{120}\)

vậy \(x=\dfrac{37}{120}\)

b) \(\dfrac{3}{2}x-1\dfrac{1}{2}=x-\dfrac{3}{4}\Leftrightarrow\dfrac{3}{2}x-x=\dfrac{-3}{4}+1\dfrac{1}{2}\Leftrightarrow\dfrac{1}{2}x=\dfrac{-3}{4}+\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{1}{2}x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{4}.2=\dfrac{6}{4}=\dfrac{3}{2}\) vậy \(x=\dfrac{3}{2}\)

c) \(x-\dfrac{5}{4}=\dfrac{1}{3}-\dfrac{3}{4}x\Leftrightarrow x+\dfrac{3}{4}x=\dfrac{1}{3}+\dfrac{5}{4}\Leftrightarrow\dfrac{7}{4}x=\dfrac{19}{12}\)

\(\Leftrightarrow x=\dfrac{\dfrac{19}{12}}{\dfrac{7}{4}}=\dfrac{19}{21}\) vậy \(x=\dfrac{19}{21}\)

d) \(\dfrac{3}{2}\left(x-\dfrac{5}{3}\right)-\dfrac{7}{5}=x+1\Leftrightarrow\dfrac{3}{2}x-\dfrac{5}{2}-\dfrac{7}{5}=x+1\)

\(\Leftrightarrow\dfrac{3}{2}x-x=1+\dfrac{5}{2}+\dfrac{7}{5}\Leftrightarrow\dfrac{1}{2}x=\dfrac{49}{10}\Leftrightarrow x=\dfrac{49}{10}.2=\dfrac{49}{5}\)

vậy \(x=\dfrac{49}{5}\)

e) \(\dfrac{2}{3}\left|4x+3\right|=\dfrac{6}{15}\Leftrightarrow\left|4x+3\right|=\dfrac{\dfrac{6}{15}}{\dfrac{2}{3}}=\dfrac{3}{5}\)

th1 : \(4x+3\ge0\Leftrightarrow4x\ge-3\Leftrightarrow x\ge\dfrac{-3}{4}\)

\(\Rightarrow\left|4x+3\right|=\dfrac{3}{5}\Leftrightarrow4x+3=\dfrac{3}{5}\Leftrightarrow4x=\dfrac{3}{5}-3=\dfrac{-12}{5}\)

\(\Leftrightarrow x=\dfrac{\dfrac{-12}{5}}{4}=\dfrac{-3}{5}\left(tmđk\right)\)

th2: \(4x+3< 0\Leftrightarrow4x< -3\Leftrightarrow x< \dfrac{-3}{4}\)

\(\Rightarrow\left|4x+3\right|=\dfrac{3}{5}\Leftrightarrow-\left(4x+3\right)=\dfrac{3}{5}\Leftrightarrow-4x-3=\dfrac{3}{5}\)

\(\Leftrightarrow4x=-3-\dfrac{3}{5}=\dfrac{-18}{5}\Leftrightarrow x=\dfrac{\dfrac{-18}{5}}{4}=\dfrac{-9}{10}\left(tmđk\right)\)

vậy \(x=\dfrac{-3}{5};x=\dfrac{-9}{10}\)

Answer 3

a, \(\dfrac{1}{3}\left(\dfrac{1}{2}-6\right)+5x=x-\dfrac{3}{5}\)

<=> \(\dfrac{1}{3}.\left(-\dfrac{11}{2}\right)+5x=x-\dfrac{3}{5}\)

<=> 5x - x = -\(\dfrac{3}{5}+\dfrac{11}{6}\)

<=> 4x = \(\dfrac{37}{30}\)

<=> x = \(\dfrac{37}{120}\)

@Nguyễn Trà My

Answer 4

b, \(\dfrac{3}{2}.x-1\dfrac{1}{2}=x-\dfrac{3}{4}\)

<=> \(\dfrac{3}{2}x-\dfrac{3}{2}=x-\dfrac{3}{4}\)

<=> \(\dfrac{3}{2}x-x=-\dfrac{3}{4}+\dfrac{3}{2}\)

<=> \(\dfrac{1}{2}x=\dfrac{3}{4}\)

<=> x = \(\dfrac{3}{2}\)

@Nguyễn Trà My

Answer 5

c, x - \(\dfrac{5}{4}=\dfrac{1}{3}-\dfrac{3}{4}x\)

<=> x + \(\dfrac{3}{4}x=\dfrac{1}{3}+\dfrac{5}{4}\)

<=> \(\dfrac{7}{4}x=\dfrac{19}{12}\)

<=> x = \(\dfrac{19}{21}\)

@Nguyễn Trà My

Answer 6

d, \(\dfrac{3}{2}\left(x-\dfrac{5}{3}\right)-\dfrac{7}{5}=x+1\)

<=> \(\dfrac{3}{2}x-\dfrac{3}{2}.\dfrac{5}{3}-\dfrac{7}{5}=x+1\)

<=> \(\dfrac{3}{2}x-\dfrac{5}{2}-\dfrac{7}{5}=x+1\)

<=> \(\dfrac{3}{2}x-\dfrac{39}{10}=x+1\)

<=> \(\dfrac{3}{2}x-x=1+\dfrac{39}{10}\)
<=> \(\dfrac{1}{2}x=\dfrac{49}{10}\)

<=> x = \(\dfrac{49}{5}\)

@Nguyễn Trà My

Answer 7

e, \(\dfrac{2}{3}.\left|4x+3\right|=\dfrac{6}{15}\)

<=> |4x + 3| = \(\dfrac{3}{5}\)
<=> \(\left[{}\begin{matrix}4x+3=\dfrac{3}{5}\\4x+3=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-\dfrac{12}{5}\\4x=-\dfrac{18}{5}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=-\dfrac{9}{10}\end{matrix}\right.\)

@Nguyễn Trà My