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b: \(x+\dfrac{5}{6}=\dfrac{3}{8}\)
=>\(x=\dfrac{3}{8}-\dfrac{5}{6}\)
=>\(x=\dfrac{9}{24}-\dfrac{20}{24}=-\dfrac{11}{24}\)
c: \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{5}{6}\)
=>\(\dfrac{2}{3}x=\dfrac{5}{6}+\dfrac{1}{2}=\dfrac{8}{6}=\dfrac{4}{3}\)
=>2x=4
=>x=4/2=2
d: \(\dfrac{x}{7}=\dfrac{6}{-21}\)
=>\(\dfrac{x}{7}=\dfrac{-2}{7}\)
=>x=-2
e: \(\left(\dfrac{7}{3}x-0,6\right):\dfrac{3^2}{5}=1\)
=>\(\dfrac{7}{3}x-0,6=\dfrac{3^2}{5}=1,8\)
=>\(\dfrac{7}{3}x=2,4\)
=>\(x=2,4:\dfrac{7}{3}=2.4\cdot\dfrac{3}{7}=\dfrac{7.2}{7}=\dfrac{36}{35}\)
f: \(\dfrac{x}{45}=\dfrac{5}{6}+\dfrac{-29}{30}\)
=>\(\dfrac{x}{45}=\dfrac{25}{30}-\dfrac{29}{30}=-\dfrac{4}{30}=-\dfrac{2}{15}\)
=>\(x=-\dfrac{2}{15}\cdot45=-6\)
g: \(\left(4,5-2x\right)\cdot\left(-\dfrac{1^4}{7}\right)=\dfrac{11}{14}\)
=>\(4,5-2x=\dfrac{11}{14}:\dfrac{-1}{7}=\dfrac{-11}{2}\)
=>\(2x=4,5+\dfrac{11}{2}=\dfrac{20}{2}=10\)
=>x=10/2=5
h: \(-\dfrac{2}{7}+\dfrac{4}{7}x=\dfrac{5}{7}\)
=>\(\dfrac{4}{7}x=\dfrac{5}{7}+\dfrac{2}{7}=\dfrac{7}{7}\)
=>4x=7
=>\(x=\dfrac{7}{4}\)
a, \(x\) \(\times\) \(\dfrac{1}{2}\) - \(\dfrac{3}{4}\) = \(\dfrac{5}{6}\)
\(x\) \(\times\) \(\dfrac{1}{2}\) = \(\dfrac{5}{6}\) + \(\dfrac{3}{4}\)
\(x\) \(\times\) \(\dfrac{1}{2}\) = \(\dfrac{19}{12}\)
\(x\) = \(\dfrac{19}{12}\) : \(\dfrac{1}{2}\)
\(x\) = \(\dfrac{19}{6}\)
b, \(x\) : \(\dfrac{1}{2}\) - \(\dfrac{3}{4}\) = \(\dfrac{5}{6}\)
\(x\): \(\dfrac{1}{2}\) = \(\dfrac{5}{6}\) + \(\dfrac{3}{4}\)
\(x\) : \(\dfrac{1}{2}\) = \(\dfrac{19}{12}\)
\(x\) = \(\dfrac{19}{12}\) \(\times\) \(\dfrac{1}{2}\)
\(x\) = \(\dfrac{19}{24}\)
c, \(x\) \(\times\) \(\dfrac{3}{4}\) + \(x\) \(\times\) \(\dfrac{1}{4}\) = \(\dfrac{7}{8}\)
\(x\) \(\times\) ( \(\dfrac{3}{4}\) + \(\dfrac{1}{4}\)) = \(\dfrac{7}{8}\)
\(x\) \(\times\) 1 = \(\dfrac{7}{8}\)
\(x\) = \(\dfrac{7}{8}\)
d, \(x\times\) \(\dfrac{3}{4}\) - \(x\) \(\times\) \(\dfrac{1}{4}\) = \(\dfrac{7}{8}\)
\(x\) \(\times\) ( \(\dfrac{3}{4}\) - \(\dfrac{1}{4}\)) = \(\dfrac{7}{8}\)
\(x\) \(\times\) \(\dfrac{1}{2}\) = \(\dfrac{7}{8}\)
\(x\) = \(\dfrac{7}{8}\) : \(\dfrac{1}{2}\)
\(x\) = \(\dfrac{7}{4}\)
a) Ta có: \(x+\dfrac{1}{3}=\dfrac{2}{6}\)
\(\Leftrightarrow x+\dfrac{1}{3}=\dfrac{1}{3}\)
hay x=0
Vậy: x=0
b) Ta có: \(x-\dfrac{1}{4}=\dfrac{1}{-2}\)
\(\Leftrightarrow x-\dfrac{1}{4}=\dfrac{-1}{2}\)
\(\Leftrightarrow x=\dfrac{-1}{2}+\dfrac{1}{4}=\dfrac{-2}{4}+\dfrac{1}{4}=\dfrac{-1}{4}\)
Vậy: \(x=-\dfrac{1}{4}\)
c) Ta có: \(\dfrac{-1}{6}=\dfrac{3}{2}x\)
\(\Leftrightarrow x=\dfrac{-1}{6}:\dfrac{3}{2}=\dfrac{-1}{6}\cdot\dfrac{2}{3}\)
hay \(x=\dfrac{-1}{9}\)
Vậy: \(x=\dfrac{-1}{9}\)
A= 1/2 x 3 + 1/3 x 4 + 1/4 x 5 + 1/5 x 6 + 1/6 x 7 + ... + 1/19 x 20
--> A= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 +...+ 1/19 - 1/20
= 1/2 - 1/20 = 10/20 - 1/20 = 9/20
a: =>4x-4-3x+6=-5
=>x+2=-5
=>x=-7
b: =>\(-4x-4+8x-24=24\)
=>4x-28=24
=>4x=52
=>x=13
c: \(\Leftrightarrow12x-48+6x-12-16x-48=7\cdot4=28\)
=>2x-108=28
=>2x=136
=>x=68
d: =>2|x-6|=2
=>|x-6|=1
=>x=7 hoặc x=5
e: =>|x+2|=20-6x+6x-24=-4(loại)
Bài 1:
\(A=\frac{8}{7}+\frac{4}{11}(\frac{-6}{7}-\frac{5}{11})=\frac{8}{7}+\frac{-404}{847}=\frac{564}{847}\)
\(B=\frac{1}{5}.10-\frac{1}{3}.\frac{-21}{20}-\frac{1}{8}=2+\frac{7}{20}-\frac{1}{8}=\frac{89}{40}\)
Bài 2:
a.
$\frac{3}{4}+\frac{1}{4}:x=-3$
$\frac{1}{4}:x =-3-\frac{3}{4}=\frac{-15}{4}$
$x=\frac{1}{4}: \frac{-15}{4}=\frac{-1}{15}$
b.
$(x-\frac{1}{3})^2=1-\frac{5}{9}=\frac{4}{9}=(\frac{2}{3})^2=(\frac{-2}{3})^2$
$\Rightarrow x-\frac{1}{3}=\frac{2}{3}$ hoặc $x-\frac{1}{3}=\frac{-2}{3}$
$\Rightarrow x=1$ hoặc $x=\frac{-1}{3}$
\(x-\dfrac{4}{6}=\dfrac{1}{2}\)
\(x=\dfrac{1}{2}+\dfrac{4}{6}\)
\(x=\dfrac{7}{6}\)
x-4/6=1/2
x=1/2+ 4/6
x=7/6
Vậy x= 7/6