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MH
2
AH
Akai Haruma
Giáo viên
10 tháng 9 2023
Lời giải:
a.
$\frac{7}{4}+\frac{5}{-6}+\frac{21}{8}=\frac{42}{24}+\frac{-20}{24}+\frac{63}{24}=\frac{85}{24}$
b.
$\frac{4}{9}+\frac{-7}{12}+\frac{8}{15}$
$=\frac{80}{180}+\frac{-105}{180}+\frac{96}{180}=\frac{71}{180}$
c.
$=\frac{-1}{3}+\frac{5}{6}+\frac{19}{12}+2$
$=\frac{-4}{12}+\frac{10}{12}+\frac{19}{12}+2=\frac{25}{12}+2=\frac{49}{12}$
24 tháng 4 2019
a,\(\frac{7}{10}\cdot\frac{4}{9}+\frac{3}{10}\cdot\frac{4}{9}-1\frac{7}{9}\)
\(=\frac{14}{45}+\frac{2}{15}-\frac{16}{9}\)
\(=\frac{14}{45}+\frac{6}{45}-\frac{80}{45}\)
\(=\frac{-60}{45}=\frac{-4}{3}\)
b,\(\frac{-5}{6}+\frac{4}{9}\cdot\left(\frac{5}{4}-\frac{2}{3}\right)\cdot\left(-3\right)^2+\frac{5}{9}\cdot30\%\)
\(=\frac{-5}{6}+\frac{4}{9}\cdot\left(\frac{7}{12}\right)\cdot9+\frac{5}{9}\cdot\frac{3}{10}\)
\(=\frac{-5}{6}+\frac{7}{3}+\frac{1}{6}\)
\(=\frac{-5}{6}+\frac{14}{6}+\frac{1}{6}\)
=\(=\frac{10}{6}=\frac{5}{3}\)
8 tháng 4 2023
1)\(\dfrac{2}{9}+\dfrac{-3}{4}+\dfrac{5}{30}\)
\(=\dfrac{2.20}{9.20}+\dfrac{-3.45}{4.45}+\dfrac{5.6}{30.6}\)
\(=\dfrac{40}{180}+\dfrac{-135}{180}+\dfrac{30}{180}\)
\(=\dfrac{40+\left(-135\right)+30}{180}\)
\(=\dfrac{-65}{180}\)
\(=\dfrac{-13}{36}\)
2)\(\dfrac{-7}{12}-\dfrac{11}{18}\)
\(=\dfrac{-7.3}{12.3}-\dfrac{11.2}{18.2}\)
\(=\dfrac{-21}{36}-\dfrac{22}{36}\)
\(=\dfrac{-21-22}{36}\)
\(=\dfrac{-43}{36}\)
3)\(\dfrac{7}{8}-\dfrac{-5}{16}\)
\(=\dfrac{7.2}{8.2}-\dfrac{-5}{16}\)
\(=\dfrac{14}{16}-\dfrac{-5}{16}\)
\(=\dfrac{14-\left(-5\right)}{16}\)
\(=\dfrac{19}{16}\)
4)\(\dfrac{3}{8}-\dfrac{-9}{10}-\dfrac{5}{16}\)
\(=\dfrac{3.10}{8.10}-\dfrac{-9.8}{10.8}-\dfrac{5.5}{16.5}\)
\(=\dfrac{30}{80}-\dfrac{-72}{80}-\dfrac{25}{80}\)
\(=\dfrac{30-\left(-72\right)-25}{80}\)
\(=\dfrac{77}{80}\)
23 tháng 7 2023
`@` `\text {Ans}`
`\downarrow`
`a)`
\(\left(\dfrac{7}{8}-\dfrac{3}{4}\right)\cdot1\dfrac{1}{3}-\dfrac{2}{3}\cdot0,5\)
`=`\(\dfrac{1}{8}\cdot\dfrac{4}{3}-\dfrac{1}{3}\)
`=`\(\dfrac{1}{6}-\dfrac{1}{3}=-\dfrac{1}{6}\)
`b)`
\(\left(2+\dfrac{5}{6}\right)\div1\dfrac{1}{5}+\left(-\dfrac{7}{12}\right)\)
`=`\(\dfrac{17}{6}\div1\dfrac{1}{5}-\dfrac{7}{12}\)
`=`\(\dfrac{85}{36}-\dfrac{7}{12}=\dfrac{16}{9}\)
`c)`
\(75\%-1\dfrac{1}{2}+0,5\div\dfrac{5}{12}\)
`=`\(-\dfrac{3}{4}+\dfrac{6}{5}=\dfrac{9}{20}\)
VN
Võ Ngọc Phương
VIP
23 tháng 7 2023
a) \(\left(\dfrac{7}{8}-\dfrac{3}{4}\right).1\dfrac{1}{3}-\dfrac{2}{3}.0,5\)
\(=\left(\dfrac{7}{8}-\dfrac{6}{8}\right).\dfrac{4}{3}-\dfrac{2}{3}.\dfrac{1}{2}\)
\(=\dfrac{1}{8}.\dfrac{4}{3}-\dfrac{2}{3}.\dfrac{1}{2}\)
\(=\dfrac{1}{6}-\dfrac{1}{3}\)
\(=\dfrac{-1}{6}\)
b) \(\left(2+\dfrac{5}{6}\right):1\dfrac{1}{5}+\dfrac{-7}{12}\)
\(=\left(\dfrac{12}{6}+\dfrac{5}{6}\right):\dfrac{6}{5}+\dfrac{-7}{12}\)
\(=\dfrac{17}{6}.\dfrac{5}{6}+\dfrac{-7}{12}\)
\(=\dfrac{85}{36}+\dfrac{-7}{12}\)
\(=\dfrac{16}{9}\)
c) \(75\%-1\dfrac{1}{2}+0,5:\dfrac{5}{12}\)
\(=\dfrac{3}{4}-\dfrac{3}{2}+\dfrac{1}{2}.\dfrac{12}{5}\)
\(=\dfrac{3}{4}-\dfrac{6}{4}+\dfrac{6}{5}\)
\(=\dfrac{-3}{4}+\dfrac{6}{5}\)
\(=\dfrac{9}{20}\)
25 tháng 4 2019
\(\frac{4}{5}\)+ \(\frac{1}{5}\): \(\frac{-2}{15}\)= \(\frac{4}{5}\)+ \(\frac{1}{5}\)* \(\frac{-15}{2}\)=\(\frac{4}{5}\) + \(\frac{-3}{2}\)= \(\frac{-7}{10}\)
25 tháng 4 2019
b = \(\frac{-7}{12}\). (\(\frac{3}{11}\)+ \(\frac{-14}{11}\)) + \(\frac{5}{6}\)= \(\frac{-7}{12}\). (-1) + \(\frac{5}{6}\)= \(\frac{7}{12}\)+ \(\frac{10}{12}\)= \(\frac{17}{12}\)
giải.
A=(-1/5) - (-3/12) - 3/12
A=(-1/5) - 0
A= -1/5.
Hok tốt~~~!