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27 tháng 7 2021

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

\(\Rightarrow x=-24\)

27 tháng 7 2021

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

\(x.\dfrac{-1}{6}=-4\)

\(x=-4:\dfrac{-1}{6}\)

\(x=-24\)

2 tháng 4 2017

a) \(0,5x-\dfrac{2}{3}x=\dfrac{7}{12}\)

\(\dfrac{1}{2}x-\dfrac{2}{3}x=\dfrac{7}{12}\)

\(\left(\dfrac{1}{2}-\dfrac{2}{3}\right)x=\dfrac{7}{12}\)

\(\dfrac{-1}{6}x=\dfrac{7}{12}\)

\(x=\dfrac{7}{12}:\dfrac{-1}{6}\)

\(x=\dfrac{-7}{2}\)

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

\(x:\dfrac{13}{3}=\dfrac{-5}{2}\)

\(x=\dfrac{13}{3}.\dfrac{-5}{2}\)

\(x=\dfrac{-65}{6}\)

c) \(5,5x=\dfrac{13}{15}\)

\(\dfrac{11}{2}x=\dfrac{13}{15}\)

\(x=\dfrac{13}{15}:\dfrac{11}{2}\)

\(x=\dfrac{26}{165}\)

d) \(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{28}\)

\(\left(\dfrac{3x}{7}+1\right)=\dfrac{-1}{28}.\left(-4\right)\)

\(\dfrac{3x}{7}+1=\dfrac{1}{7}\)

\(\dfrac{3x}{7}=\dfrac{1}{7}-1\)

\(\dfrac{3x}{7}=\dfrac{-6}{7}\)

\(\Rightarrow3x=-6\\ x=\left(-6\right):3\\ x=-2\)

2 tháng 4 2017

câu d mình nhầm bạn nha

d)\(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{28}\)

\(\dfrac{3x}{7}+1=\dfrac{-1}{28}.\left(-4\right)\)

\(\left(\dfrac{3x}{7}+1\right)=\dfrac{1}{7}\)

\(\dfrac{3x}{7}=\dfrac{1}{7}-1\)

\(\dfrac{3x}{7}=\dfrac{-6}{7}\)

Ta có : 3x =-6

x = -6:3

x=-2

2 tháng 5 2017

Bài 1:

a)

\(\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Leftrightarrow\dfrac{x-1}{9}=\dfrac{24}{9}\\ \Leftrightarrow x-1=24\\ x=24+1\\ x=25\)

b)

\(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{8}\\ \dfrac{3x}{7}+1=\dfrac{-1}{8}\cdot\left(-4\right)\\ \dfrac{3x}{7}+1=\dfrac{1}{2}\\ \dfrac{3x}{7}=\dfrac{1}{2}-1\\ \dfrac{3x}{7}=\dfrac{-1}{2}\\ 3x=\dfrac{-1}{2}\cdot7\\ 3x=\dfrac{-7}{2}\\ x=\dfrac{-7}{2}:3\\ x=\dfrac{-7}{6}\)

c)

\(x+\dfrac{7}{12}=\dfrac{17}{18}-\dfrac{1}{9}\\ x+\dfrac{7}{12}=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{12}\\ x=\dfrac{1}{4}\)

d)

\(0,5x-\dfrac{2}{3}x=\dfrac{7}{12}\\ \dfrac{1}{2}x-\dfrac{2}{3}x=\dfrac{7}{12}\\ x\cdot\left(\dfrac{1}{2}-\dfrac{2}{3}\right)=\dfrac{7}{12}\\ \dfrac{-1}{6}x=\dfrac{7}{12}\\ x=\dfrac{7}{12}:\dfrac{-1}{6}\\ x=\dfrac{-7}{2}\)

e)

\(\dfrac{29}{30}-\left(\dfrac{13}{23}+x\right)=\dfrac{7}{46}\\ \dfrac{29}{30}-\dfrac{13}{23}-x=\dfrac{7}{46}\\ \dfrac{277}{690}-x=\dfrac{7}{46}\\ x=\dfrac{277}{690}-\dfrac{7}{46}\\ x=\dfrac{86}{345}\)

f)

\(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(2+\dfrac{1}{6}-\dfrac{1}{4}\right)=\dfrac{7}{46}\\ \left(x-\dfrac{1}{12}\right):\dfrac{23}{12}=\dfrac{7}{46}\\ x-\dfrac{1}{12}=\dfrac{7}{46}\cdot\dfrac{23}{12}\\ x-\dfrac{1}{12}=\dfrac{7}{24}\\ x=\dfrac{7}{24}+\dfrac{1}{12}\\ x=\dfrac{3}{8}\)

g)

\(\dfrac{13}{15}-\left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{13}{15}-\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{1}{6}\\ \dfrac{13}{21}+x=\dfrac{1}{6}:\dfrac{7}{12}\\ \dfrac{13}{21}+x=\dfrac{2}{7}\\ x=\dfrac{2}{7}-\dfrac{13}{21}\\ x=\dfrac{-1}{3}\)

h)

\(2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|-\dfrac{3}{2}=\dfrac{1}{4}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}+\dfrac{3}{2}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}:2\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{8}\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\end{matrix}\right.\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\ \dfrac{1}{2}x=\dfrac{7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{29}{24}\\ x=\dfrac{29}{24}:\dfrac{1}{2}\\ x=\dfrac{29}{12}\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\\ \dfrac{1}{2}x=\dfrac{-7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{-13}{24}\\ x=\dfrac{-13}{24}:\dfrac{1}{2}\\ x=\dfrac{-13}{12}\)

i)

\(3\cdot\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=0-\dfrac{1}{9}\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}:3\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{27}\\ \left(3x-\dfrac{1}{2}\right)^3=\left(\dfrac{-1}{3}\right)^3\\ \Leftrightarrow3x-\dfrac{1}{2}=\dfrac{-1}{3}\\ 3x=\dfrac{-1}{3}+\dfrac{1}{2}\\ 3x=\dfrac{1}{6}\\ x=\dfrac{1}{6}:3\\ x=\dfrac{1}{18}\)

27 tháng 6 2019

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

=> \(\frac{2}{3}x=\frac{4}{5}-\frac{1}{2}=\frac{3}{10}\)

=> \(x=\frac{3}{10}:\frac{2}{3}=\frac{9}{20}\)

Vậy \(x\in\left\{\frac{9}{20}\right\}\)

\(b,x+\frac{1}{4}=\frac{4}{3}\)

=> \(x=\frac{4}{3}-\frac{1}{4}=\frac{13}{12}\)

Vậy \(x\in\left\{\frac{13}{12}\right\}\)

\(c,\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)

=> \(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}=\frac{5}{14}\)

=> \(x=\frac{5}{14}:\frac{3}{5}=\frac{25}{42}\)

Vậy \(x\in\left\{\frac{25}{42}\right\}\)

\(d,\left|x+5\right|-6=9\)

=> \(\left|x+5\right|=9+6=15\)

=> \(\left[{}\begin{matrix}x+5=15\\x+5=-15\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=15-5=10\\x=-15-5=-20\end{matrix}\right.\)

Vậy \(x\in\left\{10;-20\right\}\)

\(e,\left|x-\frac{4}{5}\right|=\frac{3}{4}\)

=> \(\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{4}\\x-\frac{4}{5}=-\frac{3}{4}\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{3}{4}+\frac{4}{5}=\frac{31}{20}\\x=-\frac{3}{4}+\frac{4}{5}=\frac{1}{20}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{31}{20};\frac{1}{20}\right\}\)

\(f,\frac{1}{2}-\left|x\right|=\frac{1}{3}\)

=> \(\left|x\right|=\frac{1}{2}-\frac{1}{3}\)

=> \(\left|x\right|=\frac{1}{6}\)

=> \(\left[{}\begin{matrix}x=\frac{1}{6}\\x=-\frac{1}{6}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{1}{6};-\frac{1}{6}\right\}\)

\(g,x^2=16\)

=> \(\left|x\right|=\sqrt{16}=4\)

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

vậy \(x\in\left\{4;-4\right\}\)

\(h,\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)

=> \(x-\frac{1}{2}=\sqrt[3]{\frac{1}{27}}=\frac{1}{3}\)

=> \(x=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\)

Vậy \(x\in\left\{\frac{5}{6}\right\}\)

\(i,3^3.x=3^6\)

\(x=3^6:3^3=3^3=27\)

Vậy \(x\in\left\{27\right\}\)

\(J,\frac{1,35}{0,2}=\frac{1,25}{x}\)

=> \(x=\frac{1,25.0,2}{1,35}=\frac{5}{27}\)

Vậy \(x\in\left\{\frac{5}{27}\right\}\)

\(k,1\frac{2}{3}:x=6:0,3\)

=> \(\frac{5}{3}:x=20\)

=> \(x=\frac{5}{3}:20=\frac{1}{12}\)

Vậy \(x\in\left\{\frac{1}{12}\right\}\)

29 tháng 3 2017

Tìm x:

a, x+30%=-1,3

<=> x+0,3=-1,3

<=> x=-1,3-0,3

<=> x=-1,6

Vậy x=-1,6

b, 0,5x-\(\dfrac{2}{3}x\) =\(\dfrac{7}{12}\)

<=>\(\dfrac{-1}{6}x\)=\(\dfrac{7}{12}\)

<=> x=\(\dfrac{-7}{2}\)

Vậy x=\(\dfrac{-7}{2}\)

29 tháng 6 2017

\(\left|x-3,5\right|=7,5\)

\(\Leftrightarrow x-3,5=7,5\Rightarrow x=11\)

\(\Leftrightarrow x-3,5=-7,5\Rightarrow x=-4\)

\(\left|x+\dfrac{4}{5}\right|-\dfrac{1}{2}=0\)

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

\(\Leftrightarrow x+\dfrac{4}{5}=\dfrac{1}{2}\Rightarrow x=\dfrac{-3}{10}\)

\(\Leftrightarrow x+\dfrac{4}{5}=-\dfrac{1}{2}\Rightarrow x=\dfrac{-13}{10}\)

\(3,6-\left|x-0,4\right|=0\)

\(\Leftrightarrow\left|x-0,4\right|=3,6\)

\(\Leftrightarrow x-0,4=3,6\Rightarrow x=4\)

\(\Leftrightarrow x-0,4=-3,6\Rightarrow x=-3,2\)

7 tháng 8 2017

Bài 1 :

a) \(x+2\dfrac{3}{4}=5\dfrac{2}{3}\)

\(x+\dfrac{11}{4}=\dfrac{17}{3}\)

\(x=\dfrac{17}{3}-\dfrac{11}{4}\)

\(x=\dfrac{35}{12}\)

Vậy .........................

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

\(x.\dfrac{7}{2}=\dfrac{19}{4}\)

\(x=\dfrac{19}{4}:\dfrac{7}{2}\)

\(x=\dfrac{19}{14}\)

Vậy .................

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

\(x:\dfrac{7}{2}=\dfrac{19}{4}\)

\(x=\dfrac{19}{4}.\dfrac{7}{2}\)

\(x=\dfrac{133}{8}\)

Vậy ...................

e) \(x-\dfrac{3}{4}=6.\dfrac{3}{8}\)

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

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

\(x=3\)

Vậy .............

f) \(\dfrac{7}{8}:x=3-\dfrac{1}{2}\)

\(\dfrac{7}{8}:x=\dfrac{5}{2}\)

\(x=\dfrac{7}{8}:\dfrac{5}{2}\)

\(x=\dfrac{7}{20}\)

Vậy ................

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

\(x+\dfrac{1}{6}=\dfrac{3}{4}\)

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

\(x=\dfrac{7}{12}\)

Vậy .................

h) \(x+17,67=100-63,2\)

\(x+17,67=36,8\)

\(x=36,8-17,67\)

\(x=19,13\)

Vậy ................

i) \(x:0,01=10\)

\(x=10.0,01\)

\(x=0,1\)

Vậy ...............

k) \(8,01-x=1,99\)

\(x=8,01-1,99\)

\(x=6,02\)

Vậy ............

l) \(x.0,5=2,2\)

\(x=2,2:0,5\)

\(x=4,4\)

Vậy ............

m) \(x:7,5=3,7+4,1\)

\(x:7,5=7,8\)

\(x=7,8.7,5\)

\(x=58,5\)

Vậy ............

7 tháng 8 2017

bạn ơi làm giúp mình bài toán đố

16 tháng 3 2017

Bài 4: 26 \(\dfrac{1}{4}\)= 26, 25 . Quãng đương là 26,25 . 2,4 = 63 km.

Thời gian đi từ B đến A là : 63 : 30 = 2,1 h

11 tháng 4 2017

bài 1:

a) \(4\dfrac{1}{2}x:\dfrac{5}{12}=0,5\) ; b)\(1,5+1\dfrac{1}{4}x=\dfrac{2}{3}\)

\(\dfrac{9}{2}x:\dfrac{5}{12}=\dfrac{1}{2}\) \(\dfrac{3}{2}+\dfrac{5}{4}x=\dfrac{2}{3}\)

\(\dfrac{9}{2}x\) \(=\dfrac{1}{2}.\dfrac{5}{12}\) \(\dfrac{5}{4}x=\dfrac{2}{3}-\dfrac{3}{2}\)

\(\dfrac{9}{2}x\) \(=\dfrac{5}{24}\) \(\dfrac{5}{4}x=\dfrac{-5}{6}\)

\(x\) \(=\dfrac{5}{24}:\dfrac{9}{2}\) \(x=\dfrac{-5}{6}:\dfrac{5}{4}\)

\(x\) \(=\dfrac{5}{108}\) \(x=\dfrac{-2}{3}\)

c) Cho mình hỏi x ở đâu vậy ???

d)\(\left(x-5\right):\dfrac{1}{3}=\dfrac{2}{5}\) e)\(\left(4,5-2x\right):\dfrac{3}{4}=1\dfrac{1}{3}\)

\(\left(x-5\right)\) \(=\dfrac{2}{5}.\dfrac{1}{3}\) \(\left(\dfrac{9}{2}-2x\right):\dfrac{3}{4}=\dfrac{4}{3}\)

\(x-5\) \(=\dfrac{2}{15}\) \(\dfrac{9}{2}-2x\) =\(\dfrac{4}{3}.\dfrac{3}{4}\)

\(x\) \(=\dfrac{2}{15}+5\) \(\dfrac{9}{2}-2x=1\)

\(x\) \(=\dfrac{77}{15}\) \(2x=\dfrac{9}{2}-1\)

f) \(\left(2,7x-1\dfrac{1}{2}x\right):\dfrac{2}{7}=\dfrac{-21}{7}\) \(2x=\dfrac{7}{2}\)

\(\left(\dfrac{27}{10}x-\dfrac{3}{2}x\right):\dfrac{2}{7}=-3\) \(x=\dfrac{7}{2}:2\)

\(\left[x\left(\dfrac{27}{10}-\dfrac{3}{2}\right)\right]=-3.\dfrac{2}{7}\) \(x=\dfrac{7}{4}\)

\(x.\dfrac{6}{5}=\dfrac{-6}{7}\)

\(x=\dfrac{-6}{7}:\dfrac{6}{5}\)

\(x=\dfrac{-5}{7}\)

bài 2:

Theo bài ra ta có :\(\dfrac{a}{27}=\dfrac{-5}{9}=\dfrac{-45}{b}\)

\(\Rightarrow9a=27.\left(-5\right)\Rightarrow a=\dfrac{27.\left(-5\right)}{9}=-15\)

\(\Rightarrow\left(-5\right)b=\left(-45\right).9\Rightarrow b=\dfrac{\left(-45\right).9}{-5}=81\)

Vậy \(a=-15;b=81\)

15 tháng 4 2018

c)

\(\left|\dfrac{2}{3}x-\dfrac{1}{2}\right|-1=\dfrac{5}{6}\)

\(\Leftrightarrow\left|\dfrac{4x-3}{6}\right|=1+\dfrac{5}{6}=\dfrac{11}{6}\)

\(\Leftrightarrow\left|4x-3\right|=11\Leftrightarrow\left[{}\begin{matrix}4x-3=11=>x=\dfrac{11+3}{4}=\dfrac{7}{2}\\4x-3=-11=>x=\dfrac{-8}{2}=-4\end{matrix}\right.\)