\(\frac{5}{11}-\left(\frac{3}{5}-\frac{6}{11}\right)\) ko quy đồng và nêu cách thực hiện phép tính
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a)
i.Ta có: BCNN(12, 30) = 60
60 : 12 = 5; 60 : 30 = 2. Do đó:
\(\frac{5}{{12}} = \frac{{5.5}}{{12.5}} = \frac{{25}}{{60}}\) và \(\frac{7}{{30}} = \frac{{7.2}}{{30.2}} = \frac{{14}}{{60}}.\)
ii.Ta có: BCNN(2, 5, 8) = 40
40 : 2 = 20; 40 : 5 = 8; 40 : 8 = 5. Do đó:
\(\frac{1}{2} = \frac{{1.20}}{{2.20}} = \frac{{20}}{{40}}\)
\(\frac{3}{5} = \frac{{3.8}}{{5.8}} = \frac{{24}}{{40}}\)
\(\frac{5}{8} = \frac{{5.5}}{{8.5}} = \frac{{25}}{{40}}\).
b)
i.Ta có: BCNN(6, 8) = 24
24 : 6 = 4; 24: 8 = 3. Do đó
\(\begin{array}{l}\frac{1}{6} + \frac{5}{8} = \frac{{1.4}}{{6.4}} + \frac{{5.3}}{{8.3}}\\ = \frac{4}{{24}} + \frac{{15}}{{24}} = \frac{{19}}{{24}}.\end{array}\)
ii. Ta có: BCNN(24, 30) = 120
120: 24 = 5; 120: 30 = 4. Do đó:
\(\begin{array}{l}\frac{{11}}{{24}} - \frac{7}{{30}} = \frac{{11.5}}{{24.5}} - \frac{{7.4}}{{30.4}}\\ = \frac{{55}}{{120}} - \frac{{28}}{{120}} = \frac{{27}}{{120}} = \frac{9}{{40}}\end{array}\)
a. \(\frac{2}{3}+\frac{1}{3}.\left(\frac{-4}{9}+\frac{5}{6}\right):\frac{7}{12}\)
\(=1.\frac{7}{12}:\frac{7}{12}\)
\(=1\)
b.
\(\frac{5}{9}.\frac{8}{11}+\frac{5}{9}.\frac{9}{11}-\frac{5}{9}.\frac{6}{11}\)
\(=\frac{5}{9}.\left(\frac{8}{11}+\frac{9}{11}-\frac{6}{11}\right)\)
\(=\frac{5}{9}.1\)
\(=\frac{5}{9}\)
Tk mk nha!
b) \(=\frac{5}{9}.\left(\frac{8}{11}+\frac{9}{11}-\frac{6}{11}\right)\)
\(=\frac{5}{9}.1\)
\(=\frac{5}{9}\)
\(\frac{\left(-6\right).5.\left(-15\right).\left(11\right)}{11.7.34.\left(-30\right)}\)
\(\frac{4950}{-78540}\)
\(\frac{-495}{7854}\)
m) (\(\frac{-5}{12}\)+\(\frac{6}{11}\))+(\(\frac{7}{17}\)+\(\frac{5}{11}\)+\(\frac{5}{12}\))
= \(\frac{-5}{12}\)+\(\frac{6}{11}\)+\(\frac{7}{17}\)+\(\frac{5}{11}\)+\(\frac{5}{12}\)
= (\(\frac{-5}{12}\)+\(\frac{5}{12}\))+(\(\frac{6}{11}\)+\(\frac{5}{11}\))+\(\frac{7}{17}\)
= 0+1+\(\frac{7}{17}\)
= \(\frac{24}{17}\)
n) (\(\frac{9}{16}\)+\(\frac{8}{-27}\))+(1+\(\frac{7}{16}\)+\(\frac{-19}{27}\))
= \(\frac{9}{16}\)+\(\frac{8}{-27}\)+1+\(\frac{7}{16}\)+\(\frac{-19}{27}\)
= (\(\frac{9}{16}\)+\(\frac{7}{16}\))+(\(\frac{8}{-27}\)+\(\frac{-19}{27}\))+1
= 1+(-1)+1
= 0+1
= 1
o) (6-2\(\frac{4}{5}\)).3\(\frac{1}{8}\)-1\(\frac{3}{5}\):\(\frac{1}{4}\)
= (6-\(\frac{14}{5}\)).\(\frac{25}{8}\)-\(\frac{8}{5}\):\(\frac{1}{4}\)
= \(\frac{16}{5}\).\(\frac{25}{8}\)-\(\frac{8}{5}\):\(\frac{1}{4}\)
= 10-\(\frac{8}{5}\):\(\frac{1}{4}\)
= 10-\(\frac{32}{5}\)
= \(\frac{18}{5}\)
CHÚC BẠN HỌC TỐT
Số cây cam là:
120 : ( 2 + 3 ) x 2 = 48 (cây)
Số cây xoài là:
( 1 + 5 ) = 20 ( cây )
Số cây chanh là:
120 - ( 48 + 20 ) = 52 ( cây )
Đáp số : cam : 48 cây
xoài : 20 cây
chanh : 52 cây.
ai trên 10 điểm thì mình nha
\(=\frac{-1}{4}+\frac{7}{33}-\frac{5}{3}-\left(-\frac{5}{4}\right)-\frac{6}{11}+\frac{48}{49}\)\(\frac{48}{49}\)
\(=\frac{-1}{4}+\frac{7}{33}-\frac{5}{3}+\frac{5}{4}-\frac{18}{33}+\frac{48}{49}\)
\(=\left(\frac{-1}{4}+\frac{5}{4}\right)+\left(\frac{7}{33}-\frac{18}{33}\right)-\frac{5}{3}\)\(+\frac{48}{48}\)
\(=1+\frac{-1}{3}-\frac{5}{3}+\frac{48}{49}\)
\(=1+\left(-2\right)+\frac{48}{49}\)
\(=-1+\frac{48}{49}\)
\(=\frac{-49}{49}+\frac{48}{49}\)
\(=\frac{-1}{49}\)
\(\frac{5}{9}:\left(\frac{1}{11}-\frac{5}{22}\right)+\frac{5}{9}:\left(\frac{1}{15}-\frac{2}{3}\right)\)
\(=\frac{5}{9}:\left(\frac{-3}{22}\right)+\frac{5}{9}:\left(\frac{-3}{5}\right)\)
\(=\frac{5}{9}.\left(-\frac{22}{3}\right)+\frac{5}{9}.\left(-\frac{5}{3}\right)\)
\(=\frac{5}{9}.\left[-\frac{22}{3}+\left(-\frac{5}{3}\right)\right]\)
\(=\frac{5}{9}.\left(-9\right)\)
\(=-5\)
\(\frac{5}{9}:\left(\frac{1}{11}-\frac{5}{22}\right)+\frac{5}{9}:\left(\frac{1}{15}-\frac{2}{3}\right)\)
\(=\frac{5}{9}:\left[\left(\frac{1}{11}-\frac{5}{22}\right)+\left(\frac{1}{15}-\frac{2}{3}\right)\right]\)
\(=\frac{5}{9}:\left[-\frac{3}{22}-\frac{3}{5}\right]\)
\(=\frac{5}{9}:\frac{-81}{110}=\frac{5}{9}.\frac{-110}{81}\)
\(=-\frac{550}{729}\)
a,\(\frac{-2}{5}+\frac{7}{21}=\frac{-2}{5}+\frac{1}{3}=\frac{-6}{15}+\frac{5}{15}=\frac{-1}{15}\)
b,\(\left(\frac{1}{3}\right)^5.3^5-2020^0=\left(\frac{1}{3}.3\right)^5-1=1^5-1=1-1=0\)
c,\(\left(-\frac{1}{4}\right).6\frac{2}{11}+3\frac{9}{11}.\left(-\frac{1}{4}\right)\)
\(=\left(-\frac{1}{4}\right).\left(6\frac{2}{11}+3\frac{9}{11}\right)=\left(-\frac{1}{4}\right).\left[\left(6+3\right)+\left(\frac{2}{11}+\frac{9}{11}\right)\right]\)
\(=\left(-\frac{1}{4}\right).\left[9+1\right]=\frac{-1}{4}.10=\frac{\left(-1\right).10}{4}=\frac{\left(-1\right).5}{2}=\frac{-5}{2}\)