b. `|x + 1| + |2x - 3| = |3x - 2|`

Ta có: \(\left|x+1\right|+\left|2x-3\right|\ge\left|x+1+2x-3\right|=\left|3x-2\right|\)

\(\Leftrightarrow\left|3x-2\right|=\left|3x-2\right|\) (luôn đúng với mọi x)

Vậy phương trình có vô số nghiệm.

\(a.ĐK:x\ne3;1\)

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

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

\(\Leftrightarrow x-1+2\left(3x-10\right)=7\left(x-1\right)\left(x-3\right)\)

\(\Leftrightarrow x-1+6x-20=7\left(x^2-4x+3\right)\)

\(\Leftrightarrow7x-21=7x^2-28x+21\)

\(\Leftrightarrow7x^2-35x+42=0\)

\(\Leftrightarrow7\left(x^2-5x+6\right)=0\)

\(\Leftrightarrow x^2-5x+6=0\)

\(\Leftrightarrow x^2-2x-3x+6=0\)

\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=3\left(ktm\right)\end{matrix}\right.\)

b.\(ĐK:x\ne2;4\)

\(\Rightarrow\dfrac{x-1}{x-2}-\dfrac{x+3}{4-x}=\dfrac{2}{\left(x-2\right)\left(4-x\right)}\)

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

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

\(\Leftrightarrow4x-x^2-4+x-x^2+2x-3x+6-2=0\)

\(\Leftrightarrow-2x^2+4x=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=2\left(ktm\right)\end{matrix}\right.\)

a: \(\Leftrightarrow\dfrac{1}{2\left(x-3\right)}+\dfrac{3x-10}{\left(x-1\right)\left(x-3\right)}=\dfrac{7}{2}\)

\(\Leftrightarrow x-1+2\left(3x-10\right)=7\left(x-1\right)\left(x-3\right)\)

\(\Leftrightarrow7\left(x^2-4x+3\right)=x-1+6x-20=7x-21\)

\(\Leftrightarrow\left(x-3\right)\left(7x-7\right)-7\left(x-3\right)=0\)

=>(x-3)(7x-14)=0

=>x=3(loại) hoặc x=2(nhận)

b: \(\Leftrightarrow\left(x-1\right)\left(x-4\right)+\left(x+3\right)\left(x-2\right)=-2\)

\(\Leftrightarrow x^2-5x+4+x^2+x-6=-2\)

\(\Leftrightarrow2x^2-4x=0\)

=>2x(x-2)=0

=>x=0(nhận) hoặc x=2(loại)

* Trả lời:

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

\(\Leftrightarrow-3+6x-4-12x=-5x+5\)

\(\Leftrightarrow6x-12x+5x=3+4+5\)

\(\Leftrightarrow x=12\)

\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)

\(\Leftrightarrow6x-15-6+24x=-3x+7\)

\(\Leftrightarrow6x+24x+3x=15+6+7\)

\(\Leftrightarrow33x=28\)

\(\Leftrightarrow x=\dfrac{28}{33}\)

\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)

\(\Leftrightarrow1-3x-6x+12=-4x-5\)

\(\Leftrightarrow-3x-6x+4x=-1-12-5\)

\(\Leftrightarrow-5x=-18\)

\(\Leftrightarrow x=\dfrac{18}{5}\)

\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)

\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)

\(\Leftrightarrow-x-5x=-7\)

\(\Leftrightarrow-6x=-7\)

\(\Leftrightarrow x=\dfrac{7}{6}\)

\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)

\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)

\(\Leftrightarrow-15x+3x=4\)

\(\Leftrightarrow-12x=4\)

\(\Leftrightarrow x=-\dfrac{1}{3}\)

b) \(\left||3x+1|+3\right|=2\)

Mà \(\left|3x+1\right|\ge0\)nên \(\left|3x+1\right|+3\ge3\)

Vậy biểu thức trong dấu GTTĐ luôn dương

\(\Rightarrow\left|3x+1\right|+3=2\)

\(\Rightarrow\left|3x+1\right|=-1\)(vô lí)

Vậy pt vô nghiệm

a) \(\left|2x-1\right|-4=5\)

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

\(\Leftrightarrow\left|2x-1\right|=9\)

\(\Leftrightarrow2x-1=\pm9\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=9\\2x-1=-9\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-4\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=5\\x=-4\end{cases}}\)

c) \(\left|3x-2\right|=4-2x\)

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

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

d) \(\left|1-3x\right|=1+2x\)

\(\Leftrightarrow\orbr{\begin{cases}1-3x=1+2x\\-\left(1-3x\right)=1+2x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

         Bài 1:

- \(\dfrac{11}{2}x\) + 1 = \(\dfrac{1}{3}x-\dfrac{1}{4}\)

- \(\dfrac{11}{2}\)\(x\) - \(\dfrac{1}{3}\)\(x\) = - \(\dfrac{1}{4}\) - 1

-(\(\dfrac{33}{6}\) + \(\dfrac{2}{6}\))\(x\) = - \(\dfrac{5}{4}\)

- \(\dfrac{35}{6}\)\(x\) = - \(\dfrac{5}{4}\)

  \(x=-\dfrac{5}{4}\) : (- \(\dfrac{35}{6}\))

 \(x\) = \(\dfrac{3}{14}\)

Vậy \(x=\dfrac{3}{14}\)

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

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

         - 5\(x\)    = \(\dfrac{9}{6}\) - \(\dfrac{6}{6}\) + \(\dfrac{4}{6}\) 

        - 5\(x\)    = \(\dfrac{7}{6}\)

           \(x\)    = \(\dfrac{7}{6}\) : (- 5) 

          \(x\)    = - \(\dfrac{7}{30}\)

Vậy \(x=-\dfrac{7}{30}\)

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

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

\(\frac{-3}{2}-2x=\frac{-7}{4}\)

\(2x=\frac{-7}{4}+\frac{-3}{2}\)

\(2x=\frac{-13}{4}\)

\(x=\frac{-13}{4}:2\)

\(x=\frac{-13}{4}.\frac{1}{2}\)

\(x=\frac{-13}{8}\)

a) \(||2x-3|-4x|=5\)

TH1: \(|2x-3|-4x=5\)

\(\Leftrightarrow|2x-3|=5+4x\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3=5+4x\\2x-3=-5-4x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-4x=5+3\\2x+4x=-5+3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}-2x=8\\6x=-2\end{cases}}\)

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

TH2: \(|2x-3|-4x=-5\)

\(\Leftrightarrow|2x-3|=-5-4x\)<0 ( loại )

Vậy \(x\in\left\{-4;\frac{-1}{3}\right\}\)

phần a tui làm sairooif để làm lại

a)

\(\Leftrightarrow3x^2-3x+2x-2=0\)

\(\Leftrightarrow3x\left(x-1\right)+2\left(x-1\right)=0\)

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

Tới đây cho mỗi cái = 0 rồi tìm x

b)

\(\Leftrightarrow2x^2+4x=6x^2+12x-2x-4\)

\(\Leftrightarrow2x^2+4x-6x^2-12x+2x+4=0\)

\(\Leftrightarrow-4x^2-6x+4=0\)

\(\Leftrightarrow-4x^2+2x-8x+4=0\)

\(\Leftrightarrow-2x\left(2x-1\right)-4\left(2x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(-2x-4\right)=0\)

Tới đây cũng cho mỗi cái = 0 và tìm x

a, 3x ( x - 1 ) + 2 ( x - 1 ) = 0

<=> ( x - 1 ) ( 3x + 2 ) = 0 

\(\Rightarrow\orbr{\begin{cases}x-1=0\\3x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0+1=1\\3x=-2\Rightarrow x=\frac{-2}{3}\end{cases}}}\)

Vậy ...