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

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

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

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

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

      \(x=8\)

ét ô ét

a.25/27                                                                                                                 b.0                          c.0

Lời giải:

a. 

$\frac{2}{3}x-\frac{7}{6}=\frac{12}{7}-\frac{1}{2}=\frac{17}{14}$

$\frac{2}{3}x=\frac{17}{14}+\frac{7}{6}=\frac{50}{21}$

$x=\frac{50}{21}: \frac{2}{3}=\frac{25}{7}$

b.

$(1\frac{1}{2}+\frac{5}{3}-\frac{1}{6}):x=\frac{3}{4}-\frac{1}{2}$

$3:x=\frac{1}{4}$

$x=3: \frac{1}{4}=12$

4, \(\Leftrightarrow4x+4+9\left(2x+1\right)=4x+6\left(x+1\right)+7+12x\)

\(\Leftrightarrow22x+13=22x+13\)vậy pt có vô số nghiệm 

5, \(\dfrac{2x}{3}+\dfrac{2x-1}{6}=4-\dfrac{x}{3}\Rightarrow4x+2x-1=24-2x\)

\(\Leftrightarrow8x=25\Leftrightarrow x=\dfrac{25}{8}\)

6, \(\dfrac{x-1}{2}+\dfrac{x-1}{4}=1-\dfrac{2\left(x-1\right)}{3}\Rightarrow6x-6+3x-3=12-8\left(x-1\right)\)

\(\Leftrightarrow9x-9=20-8x\Leftrightarrow17x=29\Leftrightarrow x=\dfrac{29}{17}\)

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

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

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

\(x+\dfrac{7}{8}=\dfrac{3}{4}\)

\(x=\dfrac{3}{4}-\dfrac{7}{8}\)

\(x=\dfrac{-1}{8}\)

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

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

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

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

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

Câu D ko bt

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 \(x=\dfrac{2}{3}\)

1: Ta có: \(\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}\)

\(\Leftrightarrow5x+20+12x-28=7x+2\)

\(\Leftrightarrow17x-7x=2+8=10\)

hay x=1

2: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)

\(\Leftrightarrow\dfrac{6x}{36}+\dfrac{4\left(1-3x\right)}{36}=\dfrac{3\left(-x+1\right)}{36}\)

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

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

hay \(x=\dfrac{1}{3}\)

3: Ta có: \(\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}\)

\(\Leftrightarrow4x-12-x-2=6x-3\)

\(\Leftrightarrow3x-14-6x+3=0\)

\(\Leftrightarrow-3x=11\)

hay \(x=-\dfrac{11}{3}\)

4: Ta có: \(\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)

\(\Leftrightarrow3x-6-8x-12=x+6\)

\(\Leftrightarrow-5x-x=6+18\)

hay x=-4

5: Ta có: \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)

\(\Leftrightarrow6x-3+2x-6=-1\)

\(\Leftrightarrow8x=8\)

hay x=1

giúp mình giải bài toán trên với. Mình cảm ơn rất nhiều

a: =>1/2x-3/4x=-5/6+7/3

=>-1/4x=14/6-5/6=3/2

=>x=-3/2*4=-6

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

=>-7/10x=17/10

=>x=-17/7

c: =>6/5x+6/20=6/5-1/3x

=>6/5x+1/3x=6/5-3/10=12/10-3/10=9/10

=>x=27/46

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

=>8x=1/2-3/2-4/5=-1-4/5=-9/5

=>x=-9/40

a) ^{_{\(\dfrac{2}{7}\times\dfrac{4}{3}+\dfrac{2}{6}\)}}

^{_{\(=\dfrac{8}{21}+\dfrac{2}{6}\)}}

^{_{\(=\dfrac{5}{7}\)}}

b) ^{_{\(\dfrac{4}{7}:2+\dfrac{4}{7}\)}}

^{_{\(=\dfrac{2}{7}+\dfrac{4}{7}\)}}

^{_{\(=\dfrac{6}{7}\)}}

c) ^{_{\(\dfrac{1}{3}+\dfrac{3}{4}-\dfrac{1}{5}:\dfrac{2}{3}\)}}

^{_{\(=\dfrac{13}{12}-\dfrac{3}{10}\)}}

^{_{\(=\dfrac{47}{60}\)}}

a \(\dfrac{8}{21}+\dfrac{2}{6}=\dfrac{16}{42}+\dfrac{14}{42}=\dfrac{30}{42}=\dfrac{5}{7}\)

\(b\) \(\dfrac{4}{7}\times\dfrac{1}{2}+\dfrac{4}{7}=\dfrac{2}{7}+\dfrac{4}{7}=\dfrac{6}{7}\)

\(c\) \(\dfrac{1}{3}+\dfrac{3}{4}-\dfrac{1}{5}\times\dfrac{3}{2}=\dfrac{1}{3}+\dfrac{3}{4}-\dfrac{3}{10}=\dfrac{20}{60}+\dfrac{45}{60}-\dfrac{18}{60}=\dfrac{47}{60}\)

a) \(\dfrac{2}{7}\times\dfrac{4}{3}+\dfrac{2}{6}=\dfrac{8}{21}+\dfrac{2}{6}=\dfrac{16}{42}+\dfrac{14}{42}=\dfrac{30}{42}=\dfrac{5}{7}\)

b) \(\dfrac{4}{7}:2+\dfrac{4}{7}=\dfrac{4}{7}\times\dfrac{1}{2}+\dfrac{4}{7}=\dfrac{2}{7}+\dfrac{4}{7}=\dfrac{6}{7}\)

c) \(\dfrac{1}{3}+\dfrac{3}{4}-\dfrac{1}{5}:\dfrac{2}{3}=\dfrac{1}{3}+\dfrac{3}{4}-\dfrac{1}{5}\times\dfrac{3}{2}=\dfrac{1}{3}+\dfrac{3}{4}-\dfrac{3}{10}=\dfrac{20}{60}+\dfrac{45}{60}-\dfrac{18}{60}=\dfrac{20+45-18}{60}=\dfrac{47}{60}\)