|x+1|+|3x-1|+|x-1|=3

=} vs cả trong dấu giá trị tuyệt đối >0 thì=}

x+1+3x-1+x-1=3{=}5x=4{=}x=4/5

=}vs cả trong giá trị tuyệt đối <0 thì=}

x+1+3x-1+x-1=-3{=}5x=-4{=}x=-4/5

`123-x=38`

`=> x= 123-38`

`=>x=85`

Vậy `x=85`

__

`(x+15) -7=33`

`=>x+15=33+7`

`=>x+16= 40`

`=>x=40-16`

`=>x=24`

Vậy `x=24`

__

`3^(x+3) -13=230`

`=> 3^(x+3) = 230+13`

`=>3^(x+3)=243`

`=> 3^(x+3)=3^5`

`=> x+3=5`

`=>x=5-3`

`=>x=2`

Vậy `x=2`

giup minh voi

b: \(\left(2x+1\right)^2=25\)

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

=>\(\left[{}\begin{matrix}2x=4\\2x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

c: \(\left(1-3x\right)^3=64\)

=>\(\left(1-3x\right)^3=4^3\)

=>1-3x=4

=>3x=1-4=-3

=>x=-3/3=-1

d: \(\left(4-x\right)^3=-27\)

=>\(\left(4-x\right)^3=\left(-3\right)^3\)

=>4-x=-3

=>x=4+3=7

e: \(x^2-5x=0\)

=>\(x\left(x-5\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

a: \(\Leftrightarrow x^2-x+4x-4+5⋮x-1\)

\(\Leftrightarrow x-1\in\left\{1;-1;5;-5\right\}\)

hay \(x\in\left\{2;0;6;-4\right\}\)

c: \(\Leftrightarrow4x-5=13k\left(k\in Z\right)\)

=>4k=13k+5

hay \(x=\dfrac{13k+5}{4}\)

Bài 1: 

a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)

\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)

\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)

\(\Leftrightarrow-12x^2+14x+13=0\)

\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)

b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)

\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)

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

ai giúp mik vs

Bài 3: 

a) Ta có: \(2x-3=x+\dfrac{1}{2}\)

\(\Leftrightarrow2x-x=\dfrac{1}{2}+3\)

\(\Leftrightarrow x=\dfrac{7}{2}\)

b) Ta có: \(4x-\left(x+\dfrac{1}{2}\right)=2x-\left(\dfrac{1}{2}-5\right)\)

\(\Leftrightarrow3x-\dfrac{1}{2}-2x+\dfrac{1}{2}-5=0\)

\(\Leftrightarrow x=5\)

Không rõ đề lắm bạn ơi

câu hỏi của bạn có sai ko ạ

a) \(x=\dfrac{1}{2}+\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{2}{4}+\dfrac{3}{4}\\ \Rightarrow x=\dfrac{5}{4}\)

b) \(\dfrac{x}{5}=\dfrac{1}{2}-\dfrac{2}{3}\)

\(\Rightarrow\dfrac{x}{5}=\dfrac{3}{6}-\dfrac{4}{6}\)

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

\(\Rightarrow x=-\dfrac{1}{6}.5\)

\(\Rightarrow x=-\dfrac{5}{6}\)

c) \(x=\dfrac{1}{2}-\dfrac{2}{3}\)

\(\Rightarrow x=\dfrac{3}{6}-\dfrac{4}{6}\)

\(\Rightarrow x=-\dfrac{1}{6}\)

d) \(\dfrac{x}{3}=\dfrac{2}{3}-\dfrac{1}{7}\)

\(\Rightarrow\dfrac{x}{3}=\dfrac{14}{21}-\dfrac{3}{21}\)

\(\Rightarrow\dfrac{x}{3}=\dfrac{11}{21}\)

\(\Rightarrow x=\dfrac{11}{21}.3\)

\(\Rightarrow x=\dfrac{33}{21}\)