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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
=>\(x=\dfrac{7}{4}\)
(0.25-0.3x)x1/3-1/4=-5/16
(0.25-0.3x)x1/3=-1/16 (làm tắt)
0.25-0.3x=-1/16:1/3
0.25-0.3x=-3/16
0.3x=0.4375 =>x=35/24
`(x-5/12)^2 : 1 1/3= 0,25`
`=> (x-5/12)^2 : 4/3 = 0,25`
`=> (x-5/12)^2 =0,25 . 4/3`
`=> (x-5/12)^2 =1/3`
`=> (x-5/12)^2 =` \(\pm\left(\dfrac{1}{\sqrt{3}}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{5}{12}=\dfrac{1}{\sqrt{3}}\\x-\dfrac{5}{12}=-\dfrac{1}{\sqrt{3}}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{\sqrt{3}}+\dfrac{5}{12}\\x=-\dfrac{1}{\sqrt{3}}+\dfrac{5}{12}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5+4\sqrt{3}}{12}\\x=\dfrac{5-4\sqrt{3}}{12}\end{matrix}\right.\)
`@ Yamada-Akiro`
Câu 1: 2 bài đầu hình như đề sai.
\(-\dfrac{5}{6}.\dfrac{4}{17}+-\dfrac{7}{12}.\dfrac{4}{7}\)
\(=\dfrac{4}{7}.\left(-\dfrac{5}{6}+-\dfrac{7}{12}\right)\)
\(=\dfrac{4}{7}.\dfrac{-17}{12}\)
\(=-\dfrac{1}{3}\)
Câu 2:
\(\dfrac{4}{5}.x=125\%-0,25\)
\(\dfrac{4}{5}.x=\dfrac{5}{4}-\dfrac{1}{4}\)
\(\dfrac{4}{5}.x=1\)
\(\Rightarrow\) \(x=1:\dfrac{4}{5}\)
\(\Rightarrow\) \(x=\dfrac{5}{4}\)
Vậy \(x=\dfrac{5}{4}\)
\(\Rightarrow\)\(\left[3x+\dfrac{3}{2}+3-5x\right].\dfrac{1}{2}=-x:\dfrac{31}{5}\)
\(\Rightarrow\)\(\left[-2x+\dfrac{9}{2}\right].\dfrac{1}{2}=-x.\dfrac{5}{31}\)
\(\Rightarrow\)\(\dfrac{-2x}{2}+\dfrac{9}{4}=-x.\dfrac{5}{31}\)
\(\Rightarrow\)\(-x+\dfrac{9}{4}=-x.\dfrac{5}{31}\)
\(\Rightarrow\dfrac{9}{4}=\dfrac{-x5}{31}+x\)
\(\Rightarrow\dfrac{9}{4}=\dfrac{-x5+31x}{31}\)
\(\Rightarrow\dfrac{9}{4}=\dfrac{x\left(-5+31\right)}{31}\)
\(\Rightarrow\dfrac{9}{4}=\dfrac{x.26}{31}\)
\(\Rightarrow\dfrac{9}{4}.31=x.26\)\(\Rightarrow\dfrac{279}{4}=x.26\)
\(\Rightarrow x=\dfrac{279}{4}:26=\dfrac{279}{4}.\dfrac{1}{26}\)
\(\Rightarrow x=\dfrac{279}{104}\)
Vậy x=\(\dfrac{279}{104}\)
\(\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}\) \(\frac{3}{5}.\frac{2}{8}+\frac{-6}{16}.\frac{2}{5}+\frac{-6}{15}:\left(-16\right)\)
\(=\frac{-5}{7}\left(\frac{2}{11}+\frac{9}{11}\right)\) \(=\frac{3}{20}+\frac{-3}{20}+\frac{1}{40}\)
\(=\frac{-5}{7}.1=\frac{-5}{7}\) \(=0+\frac{1}{40}=\frac{1}{40}\)
\(x-\frac{2}{5}=0,24\) \(\left(\frac{7}{3}x-0,6\right):3\frac{2}{5}=1\)
\(\Rightarrow x=0,24+\frac{2}{5}=\frac{16}{25}\) \(\Rightarrow\left(\frac{7}{3}x-0,6\right):\frac{17}{5}=1\)
vậy x = 16/25 \(\Rightarrow\frac{7}{3}x-0,6=\frac{17}{5}\)
\(\Rightarrow\frac{7}{3}x=\frac{17}{5}+0,6=4\)
\(\Rightarrow x=4:\frac{7}{3}=\frac{12}{7}\)
vậy x = 12/7
[ 2x - 0,6 ] + 0,25 = 5/6
[ 2x - 0,6 ] = 5/6 - 0,25
2x - 0,6 = 7/12
2x = 7/12 + 0,6
2x = 71/60
x = 71/60 : 2
x = 71/120
\(\left(2x-0,6\right)+0,25=\frac{5}{6}\)
\(\left(2x-0,6\right)=\frac{5}{6}-0,25\)
\(2x-0,6=\frac{7}{12}\)
\(2x=\frac{7}{12}+0,6\)
\(2x=\frac{71}{60}\)
\(x=\frac{71}{60}:2\)
\(x=\frac{71}{120}\)