b) Ta có: \(3+\left(x-5\right)=2\left(3x-2\right)\)

\(\Leftrightarrow3+x-5=6x-4\)

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

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

\(\Leftrightarrow-5x=-2\)

\(\Leftrightarrow x=\dfrac{2}{5}\)

Vậy: \(S=\left\{\dfrac{2}{5}\right\}\)

c) Ta có: \(2\left(x-0.5\right)+3=0.25\left(4x-1\right)\)

\(\Leftrightarrow2x-1+3=x-\dfrac{1}{4}\)

\(\Leftrightarrow2x+2-x+\dfrac{1}{4}=0\)

\(\Leftrightarrow x+\dfrac{9}{4}=0\)

\(\Leftrightarrow x=-\dfrac{9}{4}\)

Vậy: \(S=\left\{-\dfrac{9}{4}\right\}\)

d) Ta có: \(2\left(x-\dfrac{1}{4}\right)-4=-6\left(-\dfrac{1}{3}x+0.5\right)+2\)

\(\Leftrightarrow2x-\dfrac{1}{2}-4=2x-3+2\)

\(\Leftrightarrow2x-\dfrac{9}{2}=2x-1\)

\(\Leftrightarrow2x-2x=-1+\dfrac{9}{2}\)

\(\Leftrightarrow0x=\dfrac{7}{2}\)(vô lý)

Vậy: \(S=\varnothing\)

b. 

→ -2 + x = 6x - 4

→ -2 + 4 = 6x - x

→ 2 = 5x

→ x = \(\dfrac{2}{5}\) 

Vậy, phương trình có tập nghiệm S = {\(\dfrac{2}{5}\)}

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

\(\Leftrightarrow3-2x=-1\)

\(\Leftrightarrow-2x=-1-3\)

\(\Leftrightarrow-2x=-4\)

\(\Leftrightarrow x=2\)

Vậy \(x=2\)

b) \(1\dfrac{2}{3}x+0,2=x-\dfrac{7}{8}\)

\(\Leftrightarrow\dfrac{5}{3}x+\dfrac{1}{5}=x-\dfrac{7}{8}\)

\(\Leftrightarrow200x+24=120x-105\)

\(\Leftrightarrow80x=-129\)

\(\Leftrightarrow x=-\dfrac{129}{80}\)

Vậy \(x=-\dfrac{129}{80}\)

c) \(\dfrac{3}{4}-\left|x+0,5\right|=\dfrac{1}{5}\)

\(\Leftrightarrow-\left|x+0,5\right|=\dfrac{1}{5}-\dfrac{3}{4}\)

\(\Leftrightarrow-\left|x+0,5\right|=-\dfrac{11}{20}\)

\(\Leftrightarrow\left|x+0,5\right|=\dfrac{11}{20}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+0,5=\dfrac{11}{20}\\x+0,5=-\dfrac{11}{20}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{20}\\x=-\dfrac{21}{20}\end{matrix}\right.\)

Vậy \(x_1=-\dfrac{21}{20};x_2=\dfrac{1}{20}\)

d) \(\left(x+0,2\right)^2+0,75=1\)

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

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

\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{1}{4}\)

\(\Leftrightarrow x+\dfrac{1}{5}=\pm\dfrac{1}{2}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{10}\\x=-\dfrac{7}{10}\end{matrix}\right.\)

Vậy \(x_1=-\dfrac{7}{10};x_2=\dfrac{3}{10}\)

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

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

=>\(-\dfrac{1}{2}x=-1\)

=>\(x=-1:(-\dfrac{1}{2})\)

=>\(x=2\)

vậy ...........

b,\(1\dfrac{2}{3}x+0,2=x-\dfrac{7}{8}\)

=>\(\dfrac{5}{3}x+0,2=x-\dfrac{7}{8}\)

=>\(\dfrac{5}{3}x-x=-\dfrac{7}{8}-0,2\)

=>\(\dfrac{2}{3}x=-\dfrac{43}{40}\)

=>\(x=-\dfrac{43}{40}:\dfrac{2}{3}\)

=>\(x=-\dfrac{192}{80}\)

vậy...................

c,\(\dfrac{3}{4}-\left|x+0,5\right|=\dfrac{1}{5}\)

=>\(-\left|x+0,5\right|=\dfrac{1}{5}-\dfrac{3}{4}\)

=>\(-\left|x+0,5\right|=-\dfrac{11}{20}\)

=>\(\left|x+0,5\right|=\dfrac{11}{20}\)

=>\(\left[{}\begin{matrix}x+0.5=\dfrac{11}{20}\\x+0,5=-\dfrac{11}{20}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{1}{20}\\x=-\dfrac{21}{20}\end{matrix}\right.\)

vậy ....... hoặc.....

d,\((x+0,2)^2+0,75=1\)

=>\(\left(x+0,2\right)^2=1-0,75\)

=>\(\left(x+0,2\right)^2=0,25\)

=>\(\left[{}\begin{matrix}x+0,2=0,5\\x+0,2=-0,5\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=0,3\\x=-0,7\end{matrix}\right.\)

vậy..........................

HỌC TỐT NHA !!!!!

k,\(\dfrac{x}{3}-\dfrac{2x+1}{2}=\dfrac{x}{6}-x\)

giúp mk câu k nhé đề bài như trên

b: \(\Leftrightarrow4x+8-9=4x-4\)

=>-1=-4(loại)

d: \(\Leftrightarrow3\left(x-2\right)+2\left(x+1\right)=8x\)

=>8x=3x-6+2x+2=5x-4

=>3x=-4

=>x=-4/3

f: \(\Leftrightarrow3\left(x+2\right)+4\left(2x-3\right)=2\left(x-12\right)\)

=>3x+6+8x-12=2x-24

=>11x-6=2x-24

=>9x=-18

=>x=-2

c: Ta có: \(\dfrac{1}{3}-\dfrac{7}{8}x=\dfrac{1}{4}\)

\(\Leftrightarrow x\cdot\dfrac{7}{8}=\dfrac{1}{12}\)

\(\Leftrightarrow x=\dfrac{1}{12}\cdot\dfrac{8}{7}=\dfrac{2}{21}\)

d: Ta có: \(\dfrac{3}{2}x+\dfrac{1}{7}=\dfrac{7}{8}\cdot\dfrac{64}{49}\)

\(\Leftrightarrow x\cdot\dfrac{3}{2}=1\)

hay \(x=\dfrac{2}{3}\)

ét ô ét

a.25/27                                                                                                                 b.0                          c.0

tách đi bạn

a) (2x - 3)(6 - 2x) = 0

=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)

b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)

c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)

d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)

e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)

f)\(\dfrac{x-12}{4}=\dfrac{1}{2}=\dfrac{x-12}{4}=\dfrac{2}{4}\)

⇒\(x-12=2\)

   \(x=2+12\)

  x = 14

g)2\(\dfrac{1}{4}.\left(x-7\dfrac{1}{3}\right)=1,5\)

\(\dfrac{9}{4}.\left(x-\dfrac{22}{3}\right)=1,5\) 

      \(\left(x-\dfrac{22}{3}\right)=\dfrac{3}{2}:\dfrac{9}{4}\)

       \(x-\dfrac{22}{3}=\dfrac{2}{3}\)

       \(x=\dfrac{2}{3}+\dfrac{22}{3}\) 

      \(x=8\)

giúp mình với ạ câu nào cũng được

\(a,3-x=x+1,8\)

\(\Rightarrow-x-x=1,8-3\)

\(\Rightarrow-2x=-1,2\)

\(\Rightarrow x=0,6\)

\(b,2x-5=7x+35\)

\(\Rightarrow2x-7x=35+5\)

\(\Rightarrow-5x=40\)

\(\Rightarrow x=-8\)

\(c,2\left(x+10\right)=3\left(x-6\right)\)

\(\Rightarrow2x+20=3x-18\)

\(\Rightarrow2x-3x=-18-20\)

\(\Rightarrow-x=-38\)

\(\Rightarrow x=38\)

\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)

\(\Rightarrow8x-3+1=1+6x+x\)

\(\Rightarrow8x-3=7x\)

\(\Rightarrow8x-7x=3\)

\(\Rightarrow x=3\)

\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)

\(\Rightarrow-3x+x=\dfrac{4}{3}-\dfrac{2}{9}\)

\(\Rightarrow-2x=\dfrac{10}{9}\)

\(\Rightarrow x=-\dfrac{5}{9}\)

\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)

\(\Rightarrow\dfrac{1}{2}x-\dfrac{3}{4}x=-\dfrac{1}{2}-\dfrac{5}{6}\)

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

\(\Rightarrow x=\dfrac{16}{3}\)

\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)

\(\Rightarrow x-4=5-x\)

\(\Rightarrow x+x=5+4\)

\(\Rightarrow2x=9\)

\(\Rightarrow x=\dfrac{9}{2}\)

\(k,7x^2-11=6x^2-2\)

\(\Rightarrow7x^2-6x^2=-2+11\)

\(\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

\(m,5\left(x+3\cdot2^3\right)=10^2\)

\(\Rightarrow5\left(x+24\right)=100\)

\(\Rightarrow x+24=20\)

\(\Rightarrow x=-4\)

\(n,\dfrac{4}{9}-\left(\dfrac{1}{6^2}\right)=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)

\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{4}{9}-\dfrac{1}{36}\)

\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{5}{12}\)

\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2=0\)

\(\Rightarrow x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}\)

#\(Urushi\text{☕}\)

a: 3-x=x+1,8

=>-2x=-1,2

=>x=0,6

b: 2x-5=7x+35

=>-5x=40

=>x=-8

c: 2(x+10)=3(x-6)

=>3x-18=2x+20

=>x=38

d; 8(x-3/8)+1=6(1/6+x)+x

=>8x-3+1=1+6x+x

=>8x-2=7x+1

=>x=3

e: =>-3x+x=4/3-2/9

=>-2x=12/9-2/9=10/9

=>x=-5/9

g: =>3/4x-1/2x=5/6+1/2

=>1/4x=5/6+3/6=8/6=4/3

=>x=4/3*4=16/3

h: =>x-4=-x+5

=>2x=9

=>x=9/2

cac ban làm nhanh nhé

các bạn ơi làm hộ mình nhanh lên mình đang gấp quas