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18 tháng 5 2018
1) P = 2/3.5 + 2/5.7 + 2/7.9 + 2/9.11 + 2/11.13 + 2/13.15
P= (1/3-1/5) + (1/5-1/7) + (1/7-1/9) + (1/9-1/11) + (1/11-1/13) + (1/13-1/15)
P=1/3-1/15= 4/15
2) a/ 0,2:1+3/5+80%
= 2/10:8/5+8/10
= 2/10.5/8+8/10
= 1/8 + 4/5 = 5/40 + 32/40 = 37/40
b/ 0,5:5/4-2+1/5
= 5/10:5/4-11/5
= 5/10.4/5-11/5
=2/5-11/5 = -9/5
26 tháng 2 2022
a: \(=\dfrac{14-2+9}{32}\cdot\dfrac{4}{5}=\dfrac{21}{5}\cdot\dfrac{1}{8}=\dfrac{21}{40}\)
b: \(=10+\dfrac{2}{9}+2+\dfrac{3}{5}+6+\dfrac{2}{9}=18+\dfrac{47}{45}=\dfrac{857}{45}\)
c: \(=\dfrac{3}{10}-\dfrac{12}{5}+\dfrac{1}{10}=\dfrac{4}{10}-\dfrac{12}{5}=\dfrac{2}{5}-\dfrac{12}{5}=-2\)
d: \(=\dfrac{-25}{30}\left(\dfrac{37}{44}+\dfrac{13}{44}-\dfrac{6}{44}\right)=\dfrac{-25}{30}\cdot1=-\dfrac{5}{6}\)
LD
2
7 tháng 4 2017
\(N=-2\frac{2}{3}-0,2+1,75-\frac{5}{6}+0,7\)
\(N=\frac{-8}{3}-\frac{1}{5}+\frac{7}{4}-\frac{5}{6}+\frac{7}{10}\)
\(N=\frac{-160}{60}-\frac{12}{60}+\frac{105}{60}-\frac{50}{60}+\frac{42}{60}\)
\(N=\frac{\left[\left(-160\right)-12+105-50=42\right]}{60}\)
\(N=\frac{-75}{60}=\frac{-5}{4}\)
\(\Rightarrow N=\frac{-5}{4}\)
7 tháng 4 2017
= -8/3 - 1/5 + 7/4 - 5/6 + 7/10 =
= -43/15 + 11/12 + 7/10 =
= -39/20 + 7/10 =
= -5/4
tk cho mình nhé bạn
TQ
6
NN
27 tháng 7 2016
a)\(0,2:1\frac{3}{5}+80\%=\frac{1}{8}+80\%=\frac{3205}{32}\)
b)\(0,5:\frac{5}{4}-2\frac{1}{5}=\frac{2}{5}-\frac{11}{5}=-\frac{9}{5}\)
27 tháng 7 2016
\(0,2:1\frac{3}{5}+80\%=\left(\frac{2}{10}+\frac{8}{5}\right)+80\%=\left(\frac{20}{100}+\frac{160}{100}\right)+80\%=\frac{180}{100}+80\%=180\%+80\%=260\%\)
30 tháng 4 2019
b)=(2/3 +2/7 - 2/28)/(-3/3 -3/7 + 3/28)
=[2(1/3+1/7-1/28)]/[(-3)(1/3+1/7-1/28)]
=2/-3
=-2/3
26 tháng 6 2023
`@` `\text {Ans}`
`\downarrow`
`a.`
\(0,3-\dfrac{4}{9}\div\dfrac{4}{3}\cdot\dfrac{6}{5}+1\)
`=`\(0,3-\dfrac{1}{3}\cdot\dfrac{6}{5}+1\)
`=`\(0,3-0,4+1\)
`= -0,1 + 1`
`= 0,9`
`b.`
\(1+2\div\left(\dfrac{2}{3}-\dfrac{1}{6}\right)\cdot\left(-2,25\right)\)
`=`\(1+2\div\dfrac{1}{2}\cdot\left(-2,25\right)\)
`=`\(1+4\cdot\left(-2,25\right)\)
`= 1+ (-9) = -8`
`c.`
\(\left[\left(\dfrac{1}{4}-0,5\right)\cdot2+\dfrac{8}{3}\right]\div2\)
`=`\(\left(-\dfrac{1}{4}\cdot2+\dfrac{8}{3}\right)\div2\)
`=`\(\left(-\dfrac{1}{2}+\dfrac{8}{3}\right)\div2\)
`=`\(\dfrac{13}{6}\div2\)
`=`\(\dfrac{13}{12}\)
`d.`
\(\left[\left(\dfrac{3}{8}-\dfrac{5}{12}\right)\cdot6+\dfrac{1}{3}\right]\cdot4\)
`=`\(\left(-\dfrac{1}{24}\cdot6+\dfrac{1}{3}\right)\cdot4\)
`=`\(\left(-\dfrac{1}{4}+\dfrac{1}{3}\right)\cdot4\)
`=`\(\dfrac{1}{12}\cdot4=\dfrac{1}{3}\)
`e.`
\(\left(\dfrac{4}{5}-1\right)\div\dfrac{3}{5}-\dfrac{2}{3}\cdot0,5\)
`=`\(-\dfrac{1}{5}\div\dfrac{3}{5}-\dfrac{1}{3}\)
`=`\(-\dfrac{1}{3}-\dfrac{1}{3}=-\dfrac{2}{3}\)
`f.`
\(0,8\div\left\{0,2-7\left[\dfrac{1}{6}+\left(\dfrac{5}{21}-\dfrac{5}{14}\right)\right]\right\}\)
`=`\(0,8\div\left[0,2-7\left(\dfrac{1}{6}-\dfrac{5}{42}\right)\right]\)
`=`\(0,8\div\left(0,2-7\cdot\dfrac{1}{21}\right)\)
`=`\(0,8\div\left(0,2-\dfrac{1}{3}\right)\)
`= 0,8 \div (-2/15)`
`=-6`
`@` `yHGiangg.`
a) \(\frac{-2}{5}:\left(-0,2\right)=\frac{-2}{5}:\frac{-2}{10}=\frac{-2}{5}\cdot\left(-5\right)=\frac{\left(-2\right)\cdot\left(-5\right)}{5}=2\)
b) \(0,2:\frac{-2}{5}=\frac{2}{10}:\frac{-2}{5}=\frac{1}{5}\cdot\frac{-5}{2}=\frac{1\cdot\left(-5\right)}{5\cdot2}=\frac{-1}{2}\)
#Tokitou-Muichirou
a) \(-\frac{2}{5}:\left(-0,2\right)=-\frac{2}{5}\cdot\left(-5\right)=2\)
b) \(0,2:\left(-\frac{2}{5}\right)=\frac{2}{10}\cdot\left(-\frac{5}{2}\right)=-\frac{1}{2}\)