4 2/5 . 0,5-1 3/7 . 14% +(-0,8)
giúp mình với mình cảm ơn
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a) \(\dfrac{5}{9}:\left(\dfrac{13}{7}+\dfrac{13}{9}\right)-\dfrac{5}{3}\)(chỗ này mk lười chép lại đề)
=\(\dfrac{5}{9}:\dfrac{208}{63}-\dfrac{5}{3}\)
=\(\dfrac{5}{9}.\dfrac{63}{208}-\dfrac{5}{3}\)
=\(\dfrac{5.63}{9.208}-\dfrac{5}{3}\)
=\(\dfrac{5.7}{1.208}-\dfrac{5}{3}\)
=\(\dfrac{36}{208}-\dfrac{5}{3}\)
=\(\dfrac{108}{624}-\dfrac{1040}{624}\)
=\(\dfrac{-932}{624}\)
=\(\dfrac{233}{156}\)
còn câu b mk chưa học nên mk chịu
Giải:
5/9:13/7+5/9:13/9 -1 2/3
=5/9.7/13+5/9.9/13-5/3
=5/9.(7/13+9/13)-5/3
=5/9.16/13-5/3
=80/117-5/3
=-115/117
4 2/5 : 0,5% -1 3/7 .14% +(-0,5)
=22/5:1/200-10/7.7/50 +(-1/2)
=880-1/5-1/2
=8793/10
Ta có: \(A=\dfrac{1-0.5\cdot\left(3.84-2.4\right):0.8}{\dfrac{4}{5}-\left(1\dfrac{1}{3}-2\dfrac{1}{6}\right)-1.5}\)
\(=\dfrac{1-0.5\cdot1.44:0.8}{\dfrac{4}{5}-\left(\dfrac{4}{3}-\dfrac{13}{6}\right)-\dfrac{3}{2}}\)
\(=\dfrac{1-0.9}{\dfrac{4}{5}+\dfrac{5}{6}-\dfrac{3}{2}}=\dfrac{0.1}{\dfrac{2}{15}}=\dfrac{3}{4}\)
Giải:
A=1-0,5.(3,84-2,4):0,8 / 4/5-(1 1/3 - 2 1/6)-1,5
A=1-0,5.1,44:0.8 / 4/5-(4/3-13/6)-3/2
A=1-0.9 / 4/5-(-5/6)-3/2
A=0.1 / 2/15
A= 1/10 : 2/15
A=3/4
Chúc bạn học tốt!
a, \(\dfrac{1}{2}\) - ( - \(\dfrac{1}{3}\) ) + \(\dfrac{1}{23}\) + \(\dfrac{1}{6}\)
= \(\dfrac{5}{6}\) + \(\dfrac{1}{23}\) + \(\dfrac{1}{6}\)
= 1 + \(\dfrac{1}{23}\)
= \(\dfrac{24}{23}\)
b, \(\dfrac{11}{24}\) - \(\dfrac{5}{41}\) + \(\dfrac{13}{24}\) + 0,5 - \(\dfrac{36}{41}\)
= (\(\dfrac{11}{24}\) + \(\dfrac{13}{24}\)) - ( \(\dfrac{5}{41}\) + \(\dfrac{36}{41}\)) + 0,5
= 1 - 1 + 0,5
= 0,5
c,\(-\dfrac{1}{12}-\left(\dfrac{1}{6}-\dfrac{1}{4}\right)\)
=\(-\dfrac{1}{12}-\left(-\dfrac{1}{12}\right)\)
=0
d, \(\dfrac{1}{6}-\left[\dfrac{1}{6}-\left(\dfrac{1}{4}+\dfrac{9}{12}\right)\right]\)
= \(\dfrac{1}{6}-\left[\dfrac{1}{6}-1\right]\)
= \(\dfrac{1}{6}-\left(-\dfrac{5}{6}\right)\)
= 1
`@` `\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)`
\(75\%-1\dfrac{1}{2}+0,5\div\dfrac{5}{12}\)
`=`\(-\dfrac{3}{4}+\dfrac{6}{5}=\dfrac{9}{20}\)
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) \(75\%-1\dfrac{1}{2}+0,5:\dfrac{5}{12}\)
\(=\dfrac{3}{4}-\dfrac{3}{2}+\dfrac{1}{2}.\dfrac{12}{5}\)
\(=\dfrac{3}{4}-\dfrac{6}{4}+\dfrac{6}{5}\)
\(=\dfrac{-3}{4}+\dfrac{6}{5}\)
\(=\dfrac{9}{20}\)
\(\left(\dfrac{-5}{7}+\dfrac{14}{11}\right)+\left(\dfrac{-11}{14}+\dfrac{14}{-11}\right)\\ =\dfrac{-10}{14}+\dfrac{14}{11}-\dfrac{11}{14}+\dfrac{-14}{11}\\ =\left(\dfrac{-10}{14}-\dfrac{11}{14}\right)+\left(\dfrac{14}{11}+\dfrac{-14}{11}\right)\\ =\dfrac{-3}{2}+0\\ =\dfrac{-3}{2}\)
a) \(\dfrac{-15}{4}:\dfrac{21}{-10}=\dfrac{-15}{4}.\dfrac{-10}{21}=\dfrac{25}{14}\)
b) \(\dfrac{-7}{14}:\left(-0,14\right)=\dfrac{-7}{14}.\dfrac{-50}{7}=\dfrac{25}{7}\)
c) \(\left(\dfrac{-11}{15}\right):1\dfrac{1}{10}=\dfrac{-11}{15}.\dfrac{10}{11}=\dfrac{-2}{3}\)
d) \(2\dfrac{1}{7}:1\dfrac{1}{14}=\dfrac{15}{7}.\dfrac{14}{15}=2\)
\(a.-\dfrac{15}{4}:\left(\dfrac{21}{-10}\right)\)
\(=-\dfrac{15}{4}\cdot\left(-\dfrac{10}{21}\right)\)
\(=\dfrac{25}{14}\)
\(b.-\dfrac{7}{14}:\left(-0,14\right)\)
\(=-\dfrac{1}{2}:\left(-\dfrac{7}{50}\right)\)
\(=\dfrac{25}{7}\)
\(c.\left(-\dfrac{11}{15}\right):\left(1\dfrac{1}{10}\right)\)
\(=\left(-\dfrac{11}{15}\right):\dfrac{11}{10}\)
\(=-\dfrac{2}{3}\)
\(d.\left(2\dfrac{1}{7}\right):\left(1\dfrac{1}{14}\right)\)
\(=\dfrac{15}{7}:\dfrac{15}{14}\)
\(=2\)
\(\frac{1.3.5+2.6.10+4.12.20+7.21.35}{1.5.7+2.10.14+4.20.28+7.35.49}\)
\(=\frac{1.3.5\left(1+2+4+7\right)}{1.5.7\left(1+2+7+7\right)}=\frac{1.3.5}{1.5.7}=\frac{15}{35}=\frac{3}{7}\)
\(\frac{1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20+7\cdot21\cdot35}{1\cdot5\cdot7+2\cdot10\cdot14+4\cdot20\cdot28+7\cdot35\cdot49}\)
\(=\)\(\frac{1\cdot3\cdot5\cdot\left(1+2+4+7\right)}{1\cdot5\cdot7\cdot\left(1+2+7+7\right)}\)
\(=\frac{1\cdot3\cdot5}{1\cdot5\cdot7}\)\(=\frac{15}{35}=\frac{3}{7}\)
bn vội quá viết nhầm lun kìa
hj hj chúc bn làm bài tốt nha
b) \(\left(x-1\right)^3=\dfrac{1}{8}\)
\(\left(x-1\right)^3=\left(\dfrac{1}{2}\right)^3\)
\(x-1=\dfrac{1}{2}\)
\(x=\dfrac{1}{2}+1\)
\(x=\dfrac{3}{2}\)
Không đăng lại ạ!