\(4\left(x-3\right)+5=2x+1\)

\(4x-12+5=2x+1\)

\(4x-2x=1-5+12\)

\(2x=8\)

\(x=4\)

vay \(x=4\)

4.(x-3)+5=2x+1

\(\Rightarrow\)4x-12+5=2x+1

\(\Rightarrow\)4x-12=2x+1-5

\(\Rightarrow\)4x-12=2x+(-4)

\(\Rightarrow\)4x=2x+(-4)+12

\(\Rightarrow\)4x=2x+8

\(\Rightarrow\)4x-2x=8

\(\Rightarrow\)2x=8

\(\Rightarrow\) x=4

4.(x-3)+5=2x+1

⇒4x-12+5=2x+1

⇒4x-12=2x+1-5

⇒4x-12=2x+(-4)

⇒4x=2x+(-4)+12

⇒4x=2x+8

⇒4x-2x=8

⇒2x=8

⇒ x=4

Mình không chép lại đề nha

a)119-(x-6)=102

            x-6 =17

            x    =23

b)(6x-39):7=3

    6x-39     =21

    6x          =60

      x          =10

c, d bạn tự làm nha

\(A=\dfrac{2x+1+4}{2x+1}=1+\dfrac{4}{2x+1}\)

A min khi 2x+1=-1

=>x=-1

Bài làm:

a) \(\left|\frac{1}{2}x-\frac{5}{2}\right|-1=-\frac{1}{2}\)

\(\Leftrightarrow\left|\frac{1}{2}x-\frac{5}{2}\right|=\frac{1}{2}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x-\frac{5}{2}=\frac{1}{2}\\\frac{1}{2}x-\frac{5}{2}=-\frac{1}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x=3\\\frac{1}{2}x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=4\end{cases}}\)

+ Nếu x = 6

\(\left|12-\frac{1}{3}y\right|=\frac{5}{6}\)

\(\Leftrightarrow\orbr{\begin{cases}12-\frac{1}{3}y=\frac{5}{6}\\12-\frac{1}{3}y=-\frac{5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}y=\frac{67}{6}\\\frac{1}{3}y=\frac{77}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}y=\frac{67}{2}\\y=\frac{77}{2}\end{cases}}\)

+ Nếu x = 4

\(\left|8-\frac{1}{3}y\right|=\frac{5}{6}\)

\(\Leftrightarrow\orbr{\begin{cases}8-\frac{1}{3}y=\frac{5}{6}\\8-\frac{1}{3}y=-\frac{5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}y=\frac{43}{6}\\\frac{1}{3}y=\frac{53}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}y=\frac{43}{2}\\y=\frac{53}{2}\end{cases}}\)

Vậy ta có 4 cặp số (x;y) thỏa mãn: \(\left(6;\frac{67}{2}\right);\left(6;\frac{77}{2}\right);\left(4;\frac{43}{2}\right);\left(4;\frac{53}{2}\right)\)

b) \(\frac{3}{2}x-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{5}{3}\)

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

\(\Leftrightarrow x=\frac{4}{3}\)

Thay vào ta được:

\(\frac{2.\frac{4}{3}+y}{\frac{4}{3}-2y}=\frac{5}{4}\)

\(\Leftrightarrow\frac{32}{3}+4y=\frac{20}{3}-10y\)

\(\Leftrightarrow14y=-4\)

\(\Rightarrow y=-\frac{2}{7}\)

Vậy ta có 1 cặp số (x;y) thỏa mãn: \(\left(\frac{4}{3};-\frac{2}{7}\right)\)

=>(2x-1)^2=24^2

=>2x-1=24 hoặc 2x-1=-24

=>x=-23/2 hoặc x=25/2

\(\dfrac{2x-1}{-12}=\dfrac{48}{1-2x}\) (ĐK: \(x\ne\dfrac{1}{2}\))

\(\Leftrightarrow\dfrac{2x-1}{12}=\dfrac{48}{2x-1}\)

\(\Leftrightarrow-12\cdot48=\left(2x-1\right)\left(2x-1\right)\)

\(\Leftrightarrow567=\left(2x-1\right)^2\)

\(\Leftrightarrow24^2=\left(2x-1\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=-24\\2x-1=24\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=-23\\2x=25\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{23}{2}\left(tm\right)\\x=\dfrac{25}{2}\left(tm\right)\end{matrix}\right.\)

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

\(\frac{3}{4}x-\frac{1}{4}=2x-6+\frac{1}{4}x\)

\(\frac{3}{4}x-2x-\frac{1}{4}x=\frac{1}{4}-6\)

\(x\left(\frac{3}{4}-2-\frac{1}{4}\right)=-\frac{23}{4}\)

\(-\frac{3}{2}x=-\frac{23}{4}\)

\(x=-\frac{23}{4}\div\left(-\frac{3}{2}\right)\)

\(x=\frac{23}{6}\)

b) \(30\%-x+\frac{5}{6}=\frac{1}{3}\)

\(\frac{3}{10}-x+\frac{5}{6}=\frac{1}{3}\)

\(\frac{3}{10}-x=\frac{1}{3}-\frac{5}{6}\)

\(\frac{3}{10}-x=-\frac{1}{2}\)

\(x=\frac{3}{10}-\left(-\frac{1}{2}\right)\)

\(x=\frac{4}{5}\)

c) \(\left(2x-1\right).\left(3x+12\right)=0\)

\(\hept{\begin{cases}2x-1=0\\3x+12=0\end{cases}\Rightarrow\hept{\begin{cases}2x=1\\3x=-12\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{1}{2}\\x=-4\end{cases}}}\)

Vậy \(x\in\left\{\frac{1}{2};-4\right\}\)

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tớ chịu.

hi hi.