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

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

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

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

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

      \(x=8\)

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

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

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

ét ô ét

a.25/27                                                                                                                 b.0                          c.0

Bài 4: 

a) \(\dfrac{4}{3}+\left(1,25-x\right)=2,25\)

\(1,25-x=2,25-\dfrac{4}{3}=\dfrac{9}{4}-\dfrac{4}{3}\)

\(1,25-x=\dfrac{11}{12}\)

\(x=1,25-\dfrac{11}{12}=\dfrac{5}{4}-\dfrac{11}{12}\)

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

b) \(\dfrac{17}{6}-\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)

\(x-\dfrac{7}{6}=\dfrac{17}{6}-\dfrac{7}{4}=\dfrac{34}{12}-\dfrac{21}{12}\)

\(x-\dfrac{7}{6}=\dfrac{13}{12}\)

\(x=\dfrac{13}{12}+\dfrac{7}{6}=\dfrac{13}{12}+\dfrac{14}{12}\)

\(x=\dfrac{27}{12}=\dfrac{9}{4}\)

c) \(4-\left(2x+1\right)=3-\dfrac{1}{3}=\dfrac{9}{3}-\dfrac{1}{3}\)

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

\(2x+1=\dfrac{8}{3}+4=\dfrac{8}{3}+\dfrac{12}{3}\)

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

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

\(2x=\dfrac{17}{3}\)

\(x=\dfrac{17}{3}.\dfrac{1}{2}=\dfrac{17}{6}\)

Bài 15:

a) \(\left(\dfrac{-2}{3}\right)^9:x=\dfrac{-2}{3}\)

\(x=\left(\dfrac{-2}{3}\right)^9:\dfrac{-2}{3}=\left(\dfrac{-2}{3}\right)^{9-1}\)

\(=>x=\left(\dfrac{-2}{3}\right)^8\)

b) \(x:\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^4\)

\(x=\left(\dfrac{4}{9}\right)^4.\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^{4+5}\)

\(=>x=\left(\dfrac{4}{9}\right)^9\)

c) \(\left(x+4\right)^3=-125\)

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

\(=>x+4=-5\)

\(x=-5-4\)

\(=>x=-9\)

d) \(\left(10-5x\right)^3=64\)

\(\left(10-5x\right)^3=4^3\)

\(=>10-5x=4\)

\(5x=10-4\)

\(5x=6\)

\(=>x=\dfrac{6}{5}\)

e) \(\left(4x+5\right)^2=81\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(4x+5\right)^2=\left(-9\right)^2\\\left(4x+5\right)^2=9^2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+5=-9\\4x+5=9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=-14\\4x=4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-14}{4}\\x=1\end{matrix}\right.\)

Bài 16:

a) \(4-1\dfrac{2}{5}-\dfrac{8}{3}\)

\(=4-\dfrac{7}{5}-\dfrac{8}{3}\)

\(=\dfrac{60-21-40}{15}=\dfrac{-1}{15}\)

b) \(-0,6-\dfrac{-4}{9}-\dfrac{16}{15}\)

\(=\dfrac{-3}{5}+\dfrac{4}{9}-\dfrac{16}{15}\)

\(=\dfrac{\left(-27\right)+20-48}{45}=\dfrac{-55}{45}=\dfrac{-11}{9}\)

c) \(-\dfrac{15}{4}.\left(\dfrac{-7}{15}\right).\left(-2\dfrac{2}{5}\right)\)

\(=\dfrac{7}{4}.\dfrac{-12}{5}\)

\(=\dfrac{-21}{5}\)

\(#Wendy.Dang\)

Uh, chừa sau k dám học muộn nx

Đăng từng bài `1` thôi cậu ơii

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

1: \(\Leftrightarrow\left(\dfrac{x+1}{85}+1\right)+\left(\dfrac{x+3}{83}+1\right)=\left(\dfrac{x+5}{81}+1\right)+\left(\dfrac{x+7}{79}+1\right)\)

=>x+86=0

=>x=-86

2: \(\Leftrightarrow\left(\dfrac{x-1}{2015}+1\right)-\left(\dfrac{x+3}{2011}+1\right)=\left(\dfrac{x+7}{2007}+1\right)-\left(\dfrac{x+11}{2003}+1\right)\)

=>x+2014=0

=>x=-2014

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

=>4x=3x+12-2x+6

=>4x=x+18

=>3x=18

=>x=6

4: \(\Leftrightarrow15x-5\left(x+1\right)=3\left(2x+1\right)\)

=>15x-5x-5=6x+3

=>10x-5=6x+3

=>4x=8

=>x=2

5: \(\Leftrightarrow2\left(2x-7\right)+5\left(x+11\right)=-40\)

=>4x-14+5x+55=-40

=>9x+41=-40

=>x=-9

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

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

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

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

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

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

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

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

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

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

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

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 \(x=\dfrac{2}{3}\)

a: x=4/27-2/3=4/27-18/27=-14/27

b: =>3/4x-1/4x=1/6+7/3

=>1/2x=1/6+14/6=5/2

hay x=5

c: =>13/10x=7/2+5/2=6

=>x=13/10:6=13/60

d: (3x+2)(-2/5x-7)=0

=>3x+2=0 hoặc 2/5x+7=0

=>x=-2/3 hoặc x=-35/2

a) x = 4/27 - 2/3

    x = -14/27

a: x=4/27-2/3=4/27-18/27=-14/27

b: =>3/4x-1/4x=1/6+7/3

=>1/2x=1/6+14/6=5/2

hay x=5

c: =>13/10x=7/2+5/2=6

=>x=13/10:6=13/60

d: (3x+2)(-2/5x-7)=0

=>3x+2=0 hoặc 2/5x+7=0

=>x=-2/3 hoặc x=-35/2