a: Ta có: \(\left|\dfrac{2}{5}-x\right|+\dfrac{1}{2}=3.5\)

\(\Leftrightarrow\left|x-\dfrac{2}{5}\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{2}{5}=3\\x-\dfrac{2}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{5}\\x=-\dfrac{13}{5}\end{matrix}\right.\)

b: Ta có: \(\dfrac{21}{5}+3:\left|\dfrac{x}{4}-\dfrac{2}{3}\right|=6\)

\(\Leftrightarrow3:\left|\dfrac{1}{4}x-\dfrac{2}{3}\right|=6-\dfrac{21}{5}=\dfrac{9}{5}\)

\(\Leftrightarrow\left|\dfrac{1}{4}x-\dfrac{2}{3}\right|=\dfrac{5}{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{4}x-\dfrac{2}{3}=\dfrac{5}{3}\\\dfrac{1}{4}x-\dfrac{2}{3}=-\dfrac{5}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{4}x=\dfrac{7}{3}\\\dfrac{1}{4}x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{28}{3}\\x=-4\end{matrix}\right.\)

a) | 5/4x -7/2| - | 5/8x + 3/5| = 0

|5/4x - 7/2| = | 5/8x + 3/5|

TH1: 5/4x - 7/2 = 5/8x + 3/5

=> 5/4x - 5/8x = 3/5 +7/2

5/8x = 41/10

x = 41/10:5/8

x = 164/25

TH2: 5/4x - 7/2 = -5/8x - 3/5

=> 5/4x + 5/8x  = -3/5 +7/2

15/8x  = 29/10

x = 29/10 : 15/8

x = 116/75

KL: x = 164/25 hoặc x = 116/75

các bài cn lại b lm tương tự nha! h lm dài lắm!

\(\left(5x+8\right)-\left(2x-15\right)+21=2x-5\)

\(\Leftrightarrow5x+8-2x+15+21=2x-5\)

\(\Leftrightarrow\left(5x-2x\right)+\left(8+15+21\right)=2x-5\)

\(\Leftrightarrow3x+44=2x-5\)

\(\Leftrightarrow3x-2x=-44-5\)

\(\Leftrightarrow x=-49\)

~ rất vui vì giúp đc bn ~

a: =>|5/4x-7/2|=|5/8x+3/5|

=>5/4x-7/2=5/8x+3/5 hoặc 5/4x-7/2=-5/8x-3/5

=>5/8x=41/10 hoặc 15/8x=29/10

=>x=164/25 hoặc x=116/75

b: =>3:|x/4-2/3|=6-21/5=9/5

=>|1/4x-2/3|=5/3

=>1/4x-2/3=5/3 hoặc 1/4x-2/3=-5/3

=>1/4x=7/3 hoặc 1/4x=-1

=>x=28/3 hoặc x=-4

c: \(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\\left(2x-x-9\right)\left(2x+x+9\right)=0\end{matrix}\right.\Leftrightarrow x=9\)

e: =>|2x-7|=2x-7

=>2x-7>=0

=>x>=7/2

\(\dfrac{2x+19}{21}-\dfrac{2x+17}{23}=\dfrac{2x+7}{33}-\dfrac{2x+5}{35}\)

\(\Rightarrow\dfrac{2x+19}{21}-\dfrac{2x+17}{23}-\dfrac{2x+7}{33}+\dfrac{2x+5}{35}=0\)

\(\Rightarrow\left(\dfrac{2x+19}{21}+1\right)-\left(\dfrac{2x+17}{23}+1\right)-\left(\dfrac{2x+7}{33}+1\right)+\left(\dfrac{2x+5}{35}+1\right)=0\)

\(\Rightarrow\dfrac{2x+40}{21}-\dfrac{2x+40}{23}-\dfrac{2x+40}{33}+\dfrac{2x+40}{35}=0\)

\(\Rightarrow\left(2x+40\right)\left(\dfrac{1}{21}-\dfrac{1}{23}-\dfrac{1}{33}+\dfrac{1}{35}\right)=0\)

\(\Rightarrow2x+40=0\Rightarrow x=-20\)( do \(\dfrac{1}{21}-\dfrac{1}{23}-\dfrac{1}{33}+\dfrac{1}{35}>0\))

Ta có: \(\dfrac{2x+19}{21}-\dfrac{2x+17}{23}=\dfrac{2x+7}{33}-\dfrac{2x+5}{35}\)

\(\Leftrightarrow\left(2x+40\right)\left(\dfrac{1}{21}-\dfrac{1}{23}-\dfrac{1}{33}+\dfrac{1}{35}\right)=0\)

\(\Leftrightarrow2x+40=0\)

hay x=-20

a: =>-3x=1/21*84/5=4/5

=>x=-4/5:3=-4/15

b: =>-1/6x=-3/4-1/5=-19/20

=>x=19/20:1/6=19/20*6=57/10

#\(N\)

`a,`\(2x-5x=-\dfrac{1}{21}\div-\dfrac{5}{84}\)

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

\(x.\left(2-5\right)=\dfrac{4}{5}\)

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

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

\(x=-\dfrac{4}{15}\)

`b,` \(\dfrac{1}{2}x-\dfrac{2}{3}x+\dfrac{1}{5}=-\dfrac{3}{4}\)

\(\dfrac{1}{2}x-\dfrac{2}{3}x=-\dfrac{3}{4}-\dfrac{1}{5}\)

\(\dfrac{1}{2}x-\dfrac{2}{3}x=-\dfrac{19}{20}\)

\(x.\left(\dfrac{1}{2}-\dfrac{2}{3}\right)=-\dfrac{19}{20}\)

\(x.-\dfrac{1}{6}=-\dfrac{19}{20}\)

\(x=-\dfrac{19}{20}\div-\dfrac{1}{6}\)

\(x=\dfrac{57}{10}\)

x=-18/17

f) \(\frac{2x-1}{21}=\frac{3}{2x+1}\)( ĐKXĐ : \(x\ne-\frac{1}{2}\))

\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)=21\cdot3\)

\(\Leftrightarrow4x^2-1=63\)

\(\Leftrightarrow4x^2=64\)

\(\Leftrightarrow x^2=16\)

\(\Leftrightarrow x^2=\left(\pm4\right)^2\)

\(\Leftrightarrow x=\pm4\)(tmđk)

h) \(\frac{10x+5}{6}=\frac{5}{x+1}\)( ĐKXĐ : \(x\ne-1\))

\(\Leftrightarrow\left(10x+5\right)\left(x+1\right)=6\cdot5\)

\(\Leftrightarrow10x^2+15x+5=30\)

\(\Leftrightarrow10x^2+15x+5-30=0\)

\(\Leftrightarrow10x^2+15x-25=0\)

\(\Leftrightarrow5\left(2x^2+3x-5\right)=0\)

\(\Leftrightarrow2x^2+3x-5=0\)

\(\Leftrightarrow2x^2-2x+5x-5=0\)

\(\Leftrightarrow2x\left(x-1\right)+5\left(x-1\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)(tmđk)

f) \(\frac{2x-1}{21}=\frac{3}{2x+1}\)

\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)=21.3\)

\(\Leftrightarrow4x^2-1=63\)

\(\Leftrightarrow4x^2=64\)

\(\Leftrightarrow x^2=16\)\(\Leftrightarrow x^2=4^2\)\(\Leftrightarrow x=4\)

Vậy \(x=4\)

h) \(\frac{10x+5}{6}=\frac{5}{x+1}\)

\(\Leftrightarrow\left(10x+5\right)\left(x+1\right)=5.6\)

\(\Leftrightarrow5\left(2x+1\right)\left(x+1\right)=30\)

\(\Leftrightarrow\left(2x+1\right)\left(x+1\right)=6\)

\(\Leftrightarrow2x^2+3x+1=6\)

\(\Leftrightarrow2x^2+3x-5=0\)

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

\(\Leftrightarrow2x\left(x-1\right)+5\left(x-1\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\2x=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{-5}{2}\end{cases}}\)

Vậy \(x\in\left\{\frac{-5}{2};1\right\}\)

\(\frac{2x-1}{21}=\frac{3}{2x+1}\)

=> (2x - 1)(2x + 1) = 63 (1)

Đặt 2x = t

Khi đó (1) <=> (t - 1)(t + 1) = 63

=> t² + t - t - 1 = 63

=> t² - 1 = 63

=> t² = 64

=> t = \(\pm\)8

Khi t = 8

=> 2x = 8

=> x = 4

Khi t = -8

=> 2x = -8

=> x = -4

Vậy \(x\in\left\{4;-4\right\}\)

h) \(\frac{10x+5}{6}=\frac{5}{x+1}\)

=> (10x  + 5)(x + 1) = 6.5

=> 5(2x + 1)(x + 1) = 30

=> (2x + 1)(x + 1) = 6

=> 2x² + 2x + x + 1 = 6

=> 2x² + 3x + 1 = 6

=>2x² + 3x - 5 = 0

=> 2x² - 2x + 5x - 5 = 0

=> 2x(x - 1) + 5(x - 1) = 0

=> (2x + 5)(x - 1) = 0

=> \(\orbr{\begin{cases}2x+5=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-2,5\\x=1\end{cases}}\)

Vậy \(x\in\left\{-2,5;1\right\}\)