2,5:(4x)=0,5:0,2
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a) \(2,5:4x=0,5:0,2\)
\(2,5:4x=\frac{5}{2}\)
\(4x=2,5:\frac{5}{2}\)
\(4x=1\)
\(x=\frac{1}{4}\)
Vậy \(x=\frac{1}{4}\)
b) \(\frac{1}{5}.x:3=\frac{2}{3}:0,25\)
\(\frac{1}{5}.x:3=\frac{8}{3}\)
\(\frac{1}{5}.x=\frac{8}{3}.3\)
\(\frac{1}{5}.x=8\)
\(x=8:\frac{1}{5}\)
\(x=40\)
Vậy \(x=40\)
a) \(\frac{2,5}{4x}=\frac{0,5}{0,2}\)
\(=>4x=\frac{0,2.2,5}{0,5}=1\)
\(=>x=\frac{1}{4}\)
b) \(\frac{1}{5}.\frac{x}{3}=\frac{2}{3}:0,25\)
\(=>\frac{x}{15}=\frac{4}{3}\)
\(=>x=\frac{4.15}{3}=20\)
a: \(2,5:4x=0,5:0,2\)
=>\(2,5:4x=0,5\cdot5=2,5\)
=>4x=1
=>\(x=\dfrac{1}{4}\)
b: \(3,8:2x=\dfrac{1}{4}:2\dfrac{2}{3}\)
=>\(3,8:2x=\dfrac{1}{4}:\dfrac{8}{3}=\dfrac{1}{4}\cdot\dfrac{3}{8}=\dfrac{3}{32}\)
=>\(2x=3,8:\dfrac{3}{32}=\dfrac{19}{5}\cdot\dfrac{32}{3}=\dfrac{608}{15}\)
=>\(x=\dfrac{608}{15}:2=\dfrac{304}{15}\)
c: \(5,25:7x=3,6:2,4\)
=>\(5,25:7x=1,5\)
=>\(7x=5,25:1,5=3,5\)
=>\(x=\dfrac{3.5}{7}=0,5\)
d: \(1,8:1,3=-2,7:5x\)
=>\(5x=-2,7:\dfrac{18}{13}=-2,7\cdot\dfrac{13}{18}=-1,95\)
=>\(x=-1,95:5=-0,39\)
a) \(2,5:0,4x=0,5:0,2\)
\(\Rightarrow\frac{5}{2}:4x=\frac{1}{2}:\frac{1}{5}=\frac{5}{2}\)
\(\Rightarrow4x=\frac{5}{2}:\frac{5}{2}=1\)
\(\Rightarrow x=\frac{1}{4}\)
b) \(\frac{1}{5}x:3=\frac{2}{3}:0,25\)
\(\Rightarrow\frac{1}{5}x:3=\frac{8}{3}\)
\(\Rightarrow\frac{1}{5}x=\frac{8}{3}.3=8\Rightarrow x=40\)
a)2,5:4x=0,5:0,2
2,5:4x=2.5
4x=2,5:2,5
4x=1
x=1:4
x=0,25
a) Ta có: \(\left(-2.5\cdot0.38\cdot0.4\right)-\left[0.125\cdot3.15\cdot\left(-8\right)\right]\)
\(=\left(-1\cdot0.38\right)-\left[-1\cdot3.15\right]\)
\(=-0.38+3.15\)
\(=\dfrac{277}{100}\)
b) Ta có: \(\left[\left(-20.83\right)\cdot0.2+\left(-9.17\right)\cdot0.2\right]:\left[2.47\cdot0.5-\left(-3.53\right)\cdot0.51\right]\)
\(=\dfrac{0.2\left(-20.83-9.17\right)}{3.0353}\)
\(=\dfrac{0.2\cdot\left(-30\right)}{3.0353}=\dfrac{-60000}{30353}\)
2,5:(4.x)=0,5:0,2
2,5:(4.x)=2,5
(4.x)=2,5:2,5=1
x =1:4=0,25
Hok tốt!