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a)(-2,5 x 0,38 x 0,4) - [0,125 x 3,15 x (-8)]
=[(-2,5 x 0,4) x 0,38] - [(-8 x 0,125) x 3,15]
=(-1 x 0,38) - (-1 x 3,15)
=(-0,38) - (-3,15)
=2,77
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) \(\frac{1}{5}.\frac{x}{3}=\frac{2}{3}:0,25\)
\(=>\frac{x}{15}=\frac{4}{3}\)
\(=>x=\frac{4.15}{3}=20\)
1)\(x+0,5+x+1,5+x+2,5=33\)
\(\Leftrightarrow3x=33-0,5-1,5-2,5=28,5\)
\(\Leftrightarrow x=9,5\)
2)\(\left(x+0,9\right)\left(1-0,4\right)=2412\)
\(\Leftrightarrow\left(x+0,9\right)\cdot0,6=2412\)
\(\Leftrightarrow x+0,9=4020\)
\(\Leftrightarrow x=1019,1\)
a/|x|-2,5=27,5
=>|x|=27,5+2,5=30
=>x=30 hoặc x=-30
b/\(\dfrac{3}{4}+\dfrac{2}{5}.x=\dfrac{29}{60}\)
=>\(\dfrac{2}{5}.x\)=\(\dfrac{29}{60}-\dfrac{3}{4}\)=\(\dfrac{-4}{15}\)
=>x=\(\dfrac{-4}{15}:\dfrac{2}{5}\)=\(\dfrac{-2}{3}\)
c/(x-1)\(^5\)=-32
=>x-1=-2 vì (-2)\(^5\)=-32
=>x=-2+1=-1
d/\(\dfrac{4}{5}.x+0,5=4.5\)
=>\(\dfrac{4}{5}.x+0,5=20\)
=>\(\dfrac{4}{5}.x=20-0,5=19,5\)
=>\(x=19,5:\dfrac{4}{5}\)=\(\dfrac{195}{8}\)
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: \(1,8:1,3=-2,7:5x\)
=>\(5x=-2,7:\dfrac{18}{13}=-2,7\cdot\dfrac{13}{18}=-1,95\)
=>\(x=-1,95:5=-0,39\)
\(0,4.0,5.\left(-2,5\right)+\left(1,2-0,45\right):\frac{3}{4}\)
\(=\frac{2}{5}.\frac{1}{2}.\left(\frac{-5}{2}\right)+\left(\frac{6}{5}-\frac{9}{20}\right):\frac{3}{4}\)
\(=\frac{1}{5}.\left(\frac{-5}{2}\right)+\frac{3}{4}:\frac{3}{4}\)
\(=\frac{-1}{2}+-=\frac{1}{2}\)
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: \(\left[\left(-20.83\right)\cdot0.2+\left(-9.17\right)\cdot0.2\right]:\left[2.47\cdot0.5-\left(-3.53\right)\cdot0.51\right]\)
\(=\dfrac{0.2\left(-20.83-9.17\right)}{3.0353}\)
\(=\dfrac{0.2\cdot\left(-30\right)}{3.0353}=\dfrac{-60000}{30353}\)
a) 10
b) 5
c) 1
d) -10
\(\left(-2,5\right).\left(-4\right)=10\)
\(\left(-2,5\right).0,5.\left(-2\right).2=\left[\left(-2,5\right).\left(-2\right).\left(0,5.2\right)\right]=5.1=5\)
\(\left(-0,5\right).0,5.\left(-2\right).2=\left[\left(-0,5\right).\left(-2\right).\left(0,5.2\right)\right]=1.1=1\)
\(25.\left(-5\right).\left(-0,4\right).\left(-0,2\right)=\left[25.\left(-0,4\right).\left(-5\right).\left(-0,2\right)\right]=-10.1=-10\)