[ (0.125)^5 ×(2.4)^5] :[(-0.3)^5 ×( 0.01)^3]
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\(\frac{\left(-0,125\right)^5\times\left(2,4\right)^5}{\left(-0,3\right)^5\times\left(0,01\right)^3}=\frac{\left(-0,12\times2,4\right)^5}{\left(-0,3\right)^5\times0,000001}=\frac{\left(-0,3\right)^5}{\left(-0,3\right)^5\times0,000001}=\frac{1}{0,000001}\)
0.3 + 0.3 : 0.25 + 0.3 : 0.5 + 0.3 : 0.125=700.9875
k nha!
mình nhanh nhất!
0.3 + 0.3 : 0.25 + 0.3 : 0.5 + 0.3 : 0.125
= 0.3 : 1 + 0.3 : 0.25 + 0.3 : 0.5 + 0.3 : 0.125
= 0.3 : ( 1 + 0.25 + 0.5 +0.125 )
= 0.3 : 1.875
= 0.16
a: \(\Leftrightarrow2x=\dfrac{19}{5}:\dfrac{3}{32}=\dfrac{608}{15}\)
hay x=304/15
b: \(\Leftrightarrow0.25x=20\)
hay x=80
c: \(\Leftrightarrow x=\dfrac{1}{100}:\dfrac{5}{2}=\dfrac{2}{500}=\dfrac{1}{250}\)
d: \(\Leftrightarrow0.1x=\dfrac{2}{3}:\dfrac{5}{3}=\dfrac{2}{5}\)
hay \(x=\dfrac{2}{5}:\dfrac{1}{10}=\dfrac{20}{5}=4\)
Bài 1 :
Áp dụng t.c tỉ lệ thức là làm dc thôi bn
Bài 2 :
Đặt :
\(\dfrac{x}{4}=\dfrac{y}{7}=k\)\(\Leftrightarrow\left\{{}\begin{matrix}x=4k\\y=7k\end{matrix}\right.\)\(\left(1\right)\)
Thay \(xy=112\) vào \(\left(1\right)\) ta có :
\(4k.7k=112\)
\(\Leftrightarrow28k^2=112\)
\(\Leftrightarrow k^2=4\)
\(\Leftrightarrow\left[{}\begin{matrix}k^2=2^2\\k^2=\left(-2\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\)
+) \(k=2\)\(\Leftrightarrow\left\{{}\begin{matrix}x=4.2=8\\y=7.2=14\end{matrix}\right.\)
+) \(k=-2\)\(\Leftrightarrow\left\{{}\begin{matrix}x=4.\left(-2\right)=-8\\y=7.\left(-2\right)=-14\end{matrix}\right.\)
Vậy ..
\(0,25x:3=\frac{5}{6}:0,125\)
\(0,25x:3=\frac{20}{3}\)
\(0,25x=\frac{20}{3}\cdot3\)
\(0,25x=20\)
\(x=20:0,25\)
\(x=80\)
\(\frac{x-3}{x+5}=\frac{5}{7}\)
\(\left(x-3\right)\cdot7=5\cdot\left(x+5\right)\)
\(7x-21=5x+25\)
\(7x-5x=25+21\)
\(2x=46\)
\(x=46:2\)
\(x=23\)
\(\frac{0,125-\frac{1}{5}+\frac{1}{7}}{0,375-\frac{3}{5}+\frac{3}{7}}+\frac{\frac{1}{2}+\frac{1}{3}-\frac{1}{5}}{\frac{3}{4}+\frac{1}{2}-\frac{3}{10}}\)
\(=\frac{0,125-\frac{1}{5}+\frac{1}{7}}{3\left(0,125-\frac{1}{5}+\frac{1}{7}\right)}+\frac{\frac{1}{2}+\frac{1}{3}-\frac{1}{5}}{\frac{3}{2}\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{5}\right)}\)
\(=\frac{1}{3}+\frac{1}{\frac{3}{2}}\)
\(=\frac{1}{3}+\frac{2}{3}\)
\(=1\)