a: =>2/3-1/3x+1/2-x-1/2=5

=>-4/3x+2/3=5

=>-4/3x=13/3

=>x=-13/4

b: \(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\x-\dfrac{3}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\)

c: =>1/3x+3/5x+3/5=0

=>14/15x=-3/5

=>x=-3/5:14/15=-3/5x15/14=-45/70=-9/14

d: =>x>8/2

e: =>x:1/45=1/2

=>x=1/90

g: =>1/2:x=-2/15

=>x=-1/2:2/15=-15/4

a,  \((\frac{3}{7}-\frac{2}{3})\)  .x  =\(\frac{10}{21}\)                                                                   

  \(\frac{-5}{21}\).x=\(\frac{10}{21}\)

x= -2

 Mk chỉ làm 1 phần  các phằn còn lại tương tự

Bài rút gọn biểu thức

a) M=|2x-3|+|x-1| với x > 1,5

b) N=|2-x|-3|x+1| với x < -1

c) P=|3x-5|+|x-2|

d) Q=|x-3|-2.|-5x|

a) \(\left|4-x\right|+2x=3\)

<=> \(\left|4-x\right|=3-2x\)

<=> \(\orbr{\begin{cases}4-x=3-2x\left(x\le4\right)\\x-4=3-2x\left(x>4\right)\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-1\left(tm\right)\\3x=7\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-1\\x=\frac{7}{3}\left(ktm\right)\end{cases}}\)

Vậy x = -1

b) \(\left|x-7\right|+2x+5=6\)

<=> \(\left|x-7\right|=1-2x\)

<=> \(\orbr{\begin{cases}x-7=1-2x\left(đk:x\ge7\right)\\x-7=2x-1\left(đk:x< 7\right)\end{cases}}\)

<=> \(\orbr{\begin{cases}3x=8\\x=-6\left(tm\right)\end{cases}}\)

<=> \(\orbr{\begin{cases}x=\frac{8}{3}\left(ktm\right)\\x=-6\left(tm\right)\end{cases}}\)

Vậy x = -6

c) \(3x-\left|2x+1\right|=2\)

<=> \(\left|2x+1\right|=3x-2\)

<=> \(\orbr{\begin{cases}2x+1=3x-2\left(đk:x\ge-\frac{1}{2}\right)\\2x+1=2-3x\left(đk:x< -\frac{1}{2}\right)\end{cases}}\)

<=> \(\orbr{\begin{cases}x=3\left(tm\right)\\5x=1\end{cases}}\)

<=> \(\orbr{\begin{cases}x=3\\x=\frac{1}{5}\left(ktm\right)\end{cases}}\)

Vậy x = 3

d) \(\left|x+2\right|-x=2\)

<=> \(\left|x+2\right|=x+2\)

<=> \(\orbr{\begin{cases}x+2=x+2\left(đk:x\ge-2\right)\\x+2=-x-2\left(x< -2\right)\end{cases}}\)

<=> \(\orbr{\begin{cases}0x=0\\2x=-4\end{cases}}\)

<=> 0x = 0 (luôn đúng) và x = -2 (ktm)

Vậy x \(\ge\)-2

e) \(\left|x-3\right|=21\)

<=> \(\orbr{\begin{cases}x-3=21\\3-x=21\end{cases}}\)

<=> \(\orbr{\begin{cases}x=24\\x=-18\end{cases}}\)

Vậy x = 24 hoặc x = -18

f) \(\left|2x+3\right|-\left|x-3\right|=0\)

<=> \(\left|2x+3\right|=\left|x-3\right|\)

<=> \(\orbr{\begin{cases}2x+3=x-3\\2x+3=3-x\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-6\\3x=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-6\\x=0\end{cases}}\)

Vậy x thuộc {-6; 0}

g) Ta có: \(\left|x+\frac{1}{8}\right|\ge0\forall x\)

          \(\left|x+\frac{2}{8}\right|\ge0\forall x\)

    \(\left|x+\frac{5}{8}\right|\ge0\forall x\)

=> VT = \(\left|x+\frac{1}{8}\right|+\left|x+\frac{2}{8}\right|+\left|x+\frac{5}{8}\right|\ge0\forall x\)

=> VP \(\ge0\) => \(4x\ge0\) => \(x\ge0\)

Do đó: \(x+\frac{1}{8}+x+\frac{2}{8}+x+\frac{5}{8}=4x\)

<=> \(3x+1=4x\) <=> \(x=1\left(tm\right)\)

Vậy x = 1

h) \(\left|x-2\right|-\left|2x+3\right|-x=-2\)

<=> \(\left|x-2\right|-\left|2x+3\right|=x-2\)(*)

Lập bảng xét dấu: 

x                     -3/2              2

x - 2        2 - x    |        2 - x    0        x - 2

2x + 3  -2x - 3   0      2x + 3  |          2x + 3

Xét x < -3/2 => pt (*) trở thành: 2 - x + 2x + 3 = x - 2

<=> x + 5 = x - 2 <=> 0x = -7 (vô lí)

Xét -3/2 \(\le\) x < 2 => pt (*) trở thành: 2 - x - 2x - 3 = x - 2

<=> 4x = 1 <=> x = 1/4 ((tm)

Xét x \(\ge\) 2 => pt (*) trở thành x - 2 - 2x - 3 = x - 2

<=> 2x = -3 <=>  x = -3/2 (ktm)

Vậy x = 1/4

i) |2x - 3| - x = |2 - x|

<=> |2x - 3| - |2 - x| = x (*)

Lập bảng xét dấu

x                    3/2               2

2x - 3   3 - 2x   0     2x - 3   |  2x - 3

2 - x     2 - x     |       2 - x    0   x - 2

Xét x < 3/2 => pt (*) trở thành: 3 - 2x - 2 + x =  x

<=> 2x = 1 <=> x = 1//2 ((tm)
Xét \(\frac{3}{2}\le x< 2\)=> pt (*) trở thành: 2x - 3 - 2 + x = x

<=> 2x = 5 <=> x = 5/2 (ktm)

Xét x \(\ge\)2 ==> pt (*) trở thành: 2x - 3 - x + 2 = x

<=> 0x = -5 (vô lí)

Vậy x = 1/2

k) 2|x - 3| - |4x - 1| = 0

<=> 2|x - 3| = |4x - 1|

<=> \(\orbr{\begin{cases}2\left(x-3\right)=4x-1\\2\left(x-3\right)=1-4x\end{cases}}\)

<=> \(\orbr{\begin{cases}2x-6=4x-1\\2x-6=1-4x\end{cases}}\)

<=> \(\orbr{\begin{cases}2x=-5\\6x=7\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-\frac{5}{2}\\x=\frac{7}{6}\end{cases}}\) Vậy ...

\(a,\left(\frac{3}{8}+-\frac{3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)

   =  \(\left(-\frac{3}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)

    = \(\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)

     = \(\frac{1}{4}+\frac{1}{2}\)

      =  \(\frac{3}{4}\)

b)\(-\frac{7}{3}.\frac{5}{9}+\frac{4}{9}.\left(-\frac{3}{7}\right)+\frac{17}{7}\)

    =\(-\frac{35}{27}+\left(-\frac{4}{21}\right)+\frac{17}{7}\)

   = \(-\frac{35}{27}+\frac{47}{21}\)

   =        \(\frac{178}{189}\)

c) \(\frac{117}{13}-\left(\frac{2}{5}+\frac{57}{13}\right)\)

  = \(\frac{117}{13}-\frac{311}{65}\)

 =       \(\frac{274}{65}\)

d) \(\frac{2}{3}-0,25:\frac{3}{4}+\frac{5}{8}.4\)

= \(\frac{2}{3}-\frac{1}{4}:\frac{3}{4}+\frac{5}{8}.4\)

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

=     \(\frac{1}{3}+\frac{5}{2}\)

=         \(\frac{17}{6}\)

\(a,\frac{1}{2}x+\frac{5}{2}=\frac{7}{2}x-\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{2}x+\frac{5}{2}-\frac{7}{2}x=-\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{2}x-\frac{7}{2}x+\frac{5}{2}=-\frac{3}{4}\)

\(\Leftrightarrow-3x+\frac{5}{2}=-\frac{3}{4}\)

\(\Leftrightarrow-3x=-\frac{13}{4}\)

\(\Leftrightarrow x=-\frac{13}{4}:(-3)=-\frac{13}{4}:\frac{-3}{1}=-\frac{13}{4}\cdot\frac{-1}{3}=\frac{13}{12}\)

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

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

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

\(\Leftrightarrow\frac{1}{6}x-\frac{2}{5}=-\frac{1}{3}\)

\(\Leftrightarrow\frac{1}{6}x=\frac{1}{15}\)

\(\Leftrightarrow x=\frac{1}{15}:\frac{1}{6}=\frac{1}{15}\cdot6=\frac{6}{15}=\frac{2}{5}\)

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

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

\(\Leftrightarrow\frac{11}{15}x=-\frac{2}{5}\)

\(\Leftrightarrow x=-\frac{6}{11}\)

d,e,f Tương tự