1/5÷8/10=?
Giải giúp mk và kb vs mk nha!🤗
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3/10 + 4/12 + 2/10 + 2/12
= ( 3/10 + 2/10 ) + ( 4/12 + 2/12)
= 5/10 + 6/12
= 1/2 + 1/2
= 2/2 = 1
10/15 -7/5 * 10/15 - 5/5
= 10/15 * ( 7/5 -5/5)
= 2/3 * 2/5
= 4/15
Cbht!!!
Xét: \(\frac{\left(17^{2017}+16^{2017}\right)^{2018}}{17^{2017.2018}}=\left(\frac{17^{2017}+16^{2017}}{17^{2017}}\right)^{2018}=\left(1+\left(\frac{16}{17}\right)^{2017}\right)^{2018}\)
\(\frac{\left(17^{2018}+16^{2018}\right)^{2017}}{17^{2017.2018}}=\left(\frac{17^{2018}+16^{2018}}{17^{2018}}\right)^{2017}=\left(1+\left(\frac{16}{17}\right)^{2018}\right)^{2017}\)
Ta có: \(0< \frac{16}{17}< 1\)
=> \(\left(\frac{16}{17}\right)^{2017}>\left(\frac{16}{17}\right)^{2018}\)
=> \(1+\left(\frac{16}{17}\right)^{2017}>1+\left(\frac{16}{17}\right)^{2018}>1\)
=> \(\left(1+\left(\frac{16}{17}\right)^{2017}\right)^{2018}>\left(1+\left(\frac{16}{17}\right)^{2018}\right)^{2017}\)
=> \(\left(17^{2017}+16^{2017}\right)^{2018}>\left(17^{2018}+16^{2018}\right)^{2017}\)
\(-\dfrac{2}{3}=\dfrac{x}{-6}\Rightarrow x=\left(-\dfrac{2}{3}\right)\left(-6\right)=4\)
\(-\dfrac{2}{3}=\dfrac{10}{-y}\Rightarrow y=\left(-10\right):\left(-\dfrac{2}{3}\right)=15\)
\(-\dfrac{2}{3}=\dfrac{z}{9}\Rightarrow z=\left(-\dfrac{2}{3}\right).9=-6\)
\(\dfrac{-2}{3}=\dfrac{x}{-6}=\dfrac{10}{-y}=\dfrac{z}{9}\)
\(x=\left(-6.-2\right):3=4;y=\left(-6.10\right):-4=15;z=\left(10.9\right):-15=-6\)
\(\frac{1}{5}\div\frac{8}{10}\)
\(=\frac{1}{5}\times\frac{10}{8}\)
\(=\frac{10}{40}\)
\(=\frac{1}{4}\)
\(\frac{1}{5}:\frac{8}{10}=\frac{1}{5}.\frac{10}{8}=\frac{1}{5}.\frac{5}{4}=\frac{1.1}{1.4}=\frac{1}{4}\)