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11) \(\dfrac{5}{7}\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{4}{7}\right)+\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{4}{7}\right):\dfrac{7}{5}\)
= \(\dfrac{5}{7}\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{4}{7}\right)+\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{4}{7}\right)\cdot\dfrac{5}{7}\)
= \(\dfrac{5}{7}\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{4}{7}+\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{4}{7}\right)\)
= \(\dfrac{5}{7}\cdot0\)
=0
12) \(\dfrac{43}{5}\left(\dfrac{17}{3}-\dfrac{16}{9}+2\right)-\dfrac{43}{5}\left(\dfrac{17}{3}-\dfrac{16}{9}\right)\)
= \(\dfrac{43}{5}\left(\dfrac{17}{3}-\dfrac{16}{9}+2-\dfrac{17}{3}+\dfrac{16}{9}\right)\)
= \(\dfrac{43}{5}\cdot2=\dfrac{43}{10}\)
11, 5/7( 1/2-1/3+1/4)+ (1/3-1/2-1/4):7/5
= 5/7.(1/2 - 1/3 + 1/4 )+( 1/3 - 1/2 - 1/4). 5/7
= 5/7.(1/2 - 2/3 + 1/4 + 1/3 - 1/2 - 1/4)
= 5/7 . -1/3
= -5/21
12, 43/5.(17/3 - 16/9 + 2)- 43/5. (17/3 - 16/9)
= 43/5.( 17/3 - 16/9 + 2 - 17/3 + 16/9)
= 43/5 . 2
= 86/5
a) `1/3 - 1/4 : 2/5 = 1/3 - 5/8 = -7/24`
b) `6/7-(5/6+1/3)-(2/3+1/7) = 6/7-5/6-1/3-2/3-1/7`
`=(6/7-1/7)-(1/3+2/3)-5/6`
`=5/7-1-5/6`
`=-47/42`
c) `-5/9 . 2/5 + 4 5/9 + 5/9 . (-3/5)`
`= -5/9 . 2/5 + 4 + 5/9 + (-5/9) . 3/5`
`=-5/9 . (2/5 + 3/5-1) + 4`
`=-5/9 . 0 +4`
`=4`
d) 3 1/2 - (5 4/7 - 1 1/2) : 0,75`
`=7/2 - (39/7 - 3/2) : 3/4`
`= 7/2 - 57/14 : 3/4`
`=7/2 - 38/7`
`=-27/14`
`A=(8 2/7-4 2/7)-3 4/9`
`=8+2/7-4-2/7-3-4/9`
`=4-3-4/9`
`=1-4/9=5/9`
`B=(10 2/9-6 2/9)+2 3/5`
`=10+2/9-6-2/9+2+3/5`
`=4+2+3/5`
`=6+3/5=33/5`
Bài 2:
`a)5 1/2*3 1/4`
`=11/2*13/4`
`=143/8`
`b)6 1/3:4 2/9`
`=19/3:38/9`
`=19/3*9/38=3/2`
`c)4 3/7*2`
`=31/7*2`
`=62/7`
Bài 1:
\(A=\left(8\dfrac{2}{7}-4\dfrac{2}{7}\right)-3\dfrac{4}{9}\)
\(A=\left(\dfrac{58}{7}-\dfrac{30}{7}\right)-\dfrac{31}{9}\)
\(A=4-\dfrac{31}{9}\)
\(A=\dfrac{5}{9}\)
\(B=\left(10\dfrac{2}{9}-6\dfrac{2}{9}\right)+2\dfrac{3}{5}\)
\(B=\left(\dfrac{92}{9}-\dfrac{56}{9}\right)+\dfrac{13}{5}\)
\(B=4+\dfrac{13}{5}\)
\(B=\dfrac{33}{5}\)
\(1)A=\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{9}-\frac{2}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{9}-\frac{4}{11}}\)
\(=\frac{2\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}\)
\(=\frac{2}{4}=\frac{1}{2}\)
\(2)B=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}\)
\(=\frac{1.1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}.\frac{4.4}{4.5}\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}\)
\(=\frac{1.2.3.4}{2.3.4.5}=\frac{1}{5}\)
\(3)C=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}\)
\(=\frac{2.2.3.3.4.4.5.5}{1.3.2.4.3.5.4.6}\)
\(=\frac{2.5}{1.6}=\frac{2.5}{1.3.2}=\frac{5}{3}\)
\(4)D=\left(\frac{150}{1111}+\frac{5}{75}-\frac{14}{77}\right)\left(\frac{1}{5}-\frac{1}{6}-\frac{1}{30}\right)\)
\(=\left(\frac{150}{1111}+\frac{5}{75}-\frac{14}{77}\right)\left(\frac{6}{30}-\frac{5}{30}-\frac{1}{30}\right)\)
\(=\left(\frac{150}{1111}+\frac{5}{75}-\frac{14}{77}\right).0=0\)
\(5)M=8\frac{2}{7}-\left(3\frac{4}{9}+3\frac{9}{7}\right)\) \(N=\left(10\frac{2}{9}+2\frac{3}{5}\right)-6\frac{2}{9}\)
\(=\frac{58}{7}-\left(\frac{31}{9}+\frac{30}{7}\right)\) \(=\left(\frac{92}{9}+\frac{13}{5}\right)-\frac{56}{9}\)
\(=\frac{58}{7}-\left(\frac{217}{63}+\frac{270}{63}\right)\) \(=\left(\frac{460}{45}+\frac{117}{45}\right)-\frac{280}{45}\)
\(=\frac{58}{7}-\frac{487}{63}\) \(=\frac{577}{45}-\frac{280}{45}\)
\(=\frac{522}{63}-\frac{487}{63}=\frac{5}{9}\) \(=\frac{33}{5}\)
\(P=M-N\)
\(\Rightarrow P=\frac{5}{9}-\frac{33}{5}\)
\(\Rightarrow P=\frac{25}{45}-\frac{297}{45}\)
\(\Rightarrow P=\frac{-272}{45}\)
Vậy P = \(\frac{-272}{45}\)
\(6)E=10101\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{3.7.11.13.37}\right)\)
\(=\frac{5}{11}+\frac{5}{22}-\left(10101.\frac{4}{111111}\right)\)
\(=\frac{10}{22}+\frac{5}{22}-\frac{4}{11}\)
\(=\frac{15}{22}-\frac{8}{22}=\frac{7}{22}\)
\(7)F=\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}+\frac{2}{7}-\frac{2}{13}}.\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{256}+\frac{3}{64}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)
\(=\frac{1\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{13}\right)}{2\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{13}\right)}.\frac{3\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{256}+\frac{1}{64}\right)}{1\left(1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}\right)}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{3\left(\frac{16}{64}-\frac{4}{64}+\frac{1}{64}-\frac{1}{256}\right)}{1\left(\frac{64}{64}-\frac{16}{64}+\frac{4}{64}-\frac{1}{64}\right)}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{3\left(\frac{13}{64}-\frac{1}{256}\right)}{1.\frac{51}{64}}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{3\left(\frac{52}{256}-\frac{1}{256}\right)}{\frac{51}{64}}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{3\left(\frac{51}{256}\right)}{\frac{51}{64}}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{\frac{153}{256}}{\frac{51}{64}}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{153}{256}:\frac{51}{64}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{3}{4}+\frac{5}{8}\)
\(=\frac{3}{8}+\frac{5}{8}=1\)
Xin lỗi tớ đã làm hết buổi tối mà chỉ có 7 bài mong bạn thông cảm cho mình nhé !
1. \(A=\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}=\dfrac{2}{4}=\dfrac{1}{2}\)
2. \(B=\dfrac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\dfrac{1}{5}\)
3.\(C=\dfrac{2^2.3^2.\text{4^2.5^2}.5^2}{1.2^2.3^2.4^2.5.6^2}=\dfrac{125}{36}\)
4.D=\(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right).\dfrac{4}{9}.\dfrac{1}{16}=\dfrac{19}{30}.\dfrac{1}{36}=\dfrac{19}{1080}\)
\(\dfrac{3}{10}+\dfrac{4}{7}.\dfrac{5}{4}-\dfrac{1}{70}=\dfrac{3}{10}+\dfrac{5}{7}-\dfrac{1}{70}=\dfrac{21}{70}+\dfrac{50}{70}-\dfrac{10}{70}=\dfrac{70}{70}=1\)
\(1+3+5+...+\left(2n-1\right)\)
\(=\frac{\left[\left(2n-1-1\right):2+1\right]\left(2n-1+1\right)}{2}\)
\(=\frac{\left[\left(2n-2\right):2+1\right]2n}{2}\)
\(=\frac{\left(n-1+1\right)2n}{2}\)
\(=\frac{n.2n}{2}\)
\(=\frac{2n^2}{2}\)
\(=n^2\)