a: \(\Leftrightarrow x^2+4x+3=x^2+4x+0.5x+2\)

=>0,5x+2=3

=>0,5x=1

hay x=2

b: \(\Leftrightarrow\left|7x-9\right|=5x-3\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{3}{5}\\\left(7x-9-5x+3\right)\left(7x-9+5x-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{3}{5}\\\left(2x-6\right)\left(12x-12\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{3;1\right\}\)

c: =>10x+7>-37 và 10x+7<37

=>10x>-44 và 10x<30

=>-4,4<x<3

Lời giải:
PT $\Leftrightarrow \frac{11}{17}x+\frac{3}{17}=\frac{78}{17}x-\frac{22}{17}$

$\Leftrightarrow \frac{25}{17}=\frac{67}{17}x$

$x=\frac{25}{67}$

\(x-\dfrac{3}{17}\left(2x-1\right)=\dfrac{7}{34}\left(1-2x\right)+\dfrac{10x-3}{2}\)

<=>\(x-\dfrac{3}{17}\left(2x-1\right)-\dfrac{7}{34}\left(1-2x\right)=\dfrac{10x-3}{2}\)

<=>\(x-\dfrac{3}{17}\left(2x-1\right)+\dfrac{7}{34}\left(2x-1\right)=\dfrac{10x-3}{2}\)

<=>\(x+\left(-\dfrac{3}{17}+\dfrac{7}{34}\right)\left(2x-1\right)-\dfrac{10x-3}{2}=0\)

<=>\(x+\dfrac{1}{34}\left(2x-1\right)-\dfrac{10x-3}{2}=0\)

<=>\(x+\dfrac{1}{17}x-\dfrac{1}{34}-\dfrac{10x-3}{2}=0\)

<=>\(\dfrac{18}{17}x-\dfrac{85x-25.5}{17}=\dfrac{0.5}{17}\)

=> \(18x-85x+25.5=0.5\)
<=>\(-67x=0.5-25.5\)
<=>\(-67x=-25\)
<=> \(x=\dfrac{25}{67}\)
đây là cách làm khác (dù dài hơn) nếu bạn không hiểu cách bên dưới 
chúc bạn học tốt!

I don't now

or no I don't

..................

sorry

a)\(x+12x+x+12x+1=0,5x+2x+30,5x+2x+3\)

\(\Leftrightarrow26x+1=35x+3\)

\(\Leftrightarrow26x+1-\left(35x+3\right)=0\)

\(\Leftrightarrow26x+1-35x-3=0\)

\(\Leftrightarrow-9x+\left(-2\right)=0\)

\(\Leftrightarrow-9x=2\)

\(\Leftrightarrow x=-\frac{2}{9}\)

Bn Nguyễn Quốc Thịnh ơi 30,5x+2x ở đâu vậy

a: \(\Leftrightarrow x\cdot\dfrac{1}{4}+\dfrac{3}{4}=3-\dfrac{1}{2}x-\dfrac{1}{2}-\dfrac{1}{3}x-\dfrac{2}{3}\)

=>13/12x=13/12

hay x=1

b: \(\Leftrightarrow\dfrac{3x-11}{11}-\dfrac{x}{3}=\dfrac{3x-5}{7}-\dfrac{5x-3}{9}\)

\(\Leftrightarrow\dfrac{3}{11}x-1-\dfrac{1}{3}x=\dfrac{3}{7}x-\dfrac{5}{7}-\dfrac{5}{9}x+\dfrac{1}{3}\)

\(\Leftrightarrow x\cdot\dfrac{46}{693}=\dfrac{13}{21}\)

hay x=429/46

một đòn bẫy dài một mét .đặt ở đâu để có thể dùng 3600n có thể nâng tảng đá nặng 120kg?

\(a,\frac{-2}{3}x=8\)<=> \(x=-12\)

\(b,\frac{1}{4}:x=-3+\frac{3}{4}\)<=>\(\frac{1}{4}:x=\frac{-9}{4}\)<=>\(x=-9\)

\(c,\orbr{\begin{cases}2x-\frac{1}{3}=0\\0.5x+0.25=0\end{cases}}\)<=>\(\orbr{\begin{cases}x=\frac{1}{6}\\x=\frac{-3}{4}\end{cases}}\)

a) \(\frac{-2}{3}x+4=12\)

\(\Rightarrow\frac{-2}{3}x=12-4\)

\(\Rightarrow\frac{-2}{3}x=8\)

\(\Rightarrow x=8:\frac{-2}{3}\)

\(\Rightarrow x=-12\)

Vậy x = -12

b) \(\frac{-3}{4}+\frac{1}{4}:x=-3\)

\(\Rightarrow\frac{1}{4}:x=-3-\left(\frac{3}{4}\right)\)

\(\Rightarrow\frac{1}{4}:x=\frac{-9}{4}\)

\(\Rightarrow x=\frac{1}{4}:\frac{-9}{4}\)

\(\Rightarrow x=\frac{-1}{9}\)

Vậy  \(x=\frac{-1}{9}\)

c) \(\left(2x-\frac{1}{3}\right)\left(0,5x+0,25\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{3}=0\\0,5x+0,25=0\end{cases}}\Rightarrow\orbr{\begin{cases}2x=\frac{1}{3}\\0,5x=-0,25\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{6}\\x=-0,5\end{cases}}\)

Vậy  \(x=\frac{1}{6}\)hoặc \(x=-0,5\)

_Chúc bạn học tốt_

\(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}\Rightarrow\left(x+1\right)\left(x+3\right)=\left(0,5x+2\right)\left(2x+1\right)\)

\(\Rightarrow x^2+3x+x+3=x^2+0,5x+4x+2\)

\(\Rightarrow x^2+3x+x-x^2-0,5x-4x=2-3\)

\(\Rightarrow-\frac{1}{2}x=-1\)

\(\Rightarrow x=2\)

\(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}\)

\(\Rightarrow\left(x+1\right)\left(x+3\right)=\left(2x+1\right)\left(0,5x+2\right)\)

\(\Rightarrow x^2+3x+x+3=x^2+4x+0,5x+2\)

\(\Rightarrow x^2+3x+x+3-x^2-4x-0,5x-2=0\)

\(\Rightarrow1-0,5x=0\)

\(\Rightarrow\frac{1}{2}x=1\Leftrightarrow x=2\)

\(\Leftrightarrow\left(x+1\right)\left(x+3\right)=\left(2x+1\right)\left(x+3\right)\)

\(\Leftrightarrow x^2+4x+3=x^2+4.5x+2\)

\(\Leftrightarrow0.5x+2=3\Leftrightarrow0.5x=1\Leftrightarrow x=2\)

\(\frac{x+1}{2x+1}\)\(=\)\(\frac{0,5x+2}{x+3}\)

  => (x+1).(x+3) = (2x+1).(0,5x+2)

  => x²+4x+3 = x²+4,5x+2

  => x²+4x-x²-4,5x = 2-3

  => -0,5x = -1

 =>  x      = -1:-0,5

 =>  x      = 2 

\(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}\)

\(\Leftrightarrow\left(x+1\right)\left(x+3\right)=\left(0,5x+2\right)\left(2x+1\right)\)

\(\Leftrightarrow x^2+3x+x+3=x^2+0,5x+4x+2\)

\(\Leftrightarrow x^2+4x+3=x^2+4,5x+2\)

\(\Leftrightarrow4x+3=4,5x+2\)

\(\Leftrightarrow0,5x=-1\)

\(\Leftrightarrow x=-2\)

Tíck cho mìk vs nha Vân Nguyễn!