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Cao Minh Tâm
5 tháng 4 2018
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H9
HT.Phong (9A5)
CTVHS
17 tháng 1
a) \(\left(\dfrac{3}{29}-\dfrac{1}{5}\right)\cdot\dfrac{29}{3}\)
\(=\dfrac{3}{29}\cdot\dfrac{29}{3}-\dfrac{1}{5}\cdot\dfrac{29}{3}\)
\(=1-\dfrac{29}{15}\)
\(=\dfrac{15-29}{15}\)
\(=-\dfrac{14}{15}\)
b) \(\dfrac{3}{4}\cdot\dfrac{7}{9}+\dfrac{1}{4}\cdot\dfrac{7}{9}\)
\(=\dfrac{7}{9}\cdot\left(\dfrac{1}{4}+\dfrac{3}{4}\right)\)
\(=\dfrac{7}{9}\cdot1\)
\(=\dfrac{7}{9}\)
c) \(\dfrac{1}{7}\cdot\dfrac{5}{9}+\dfrac{5}{9}\cdot\dfrac{1}{7}+\dfrac{5}{9}\cdot\dfrac{3}{7}\)
\(=\dfrac{5}{9}\cdot\left(\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{3}{7}\right)\)
\(=\dfrac{5}{9}\cdot\dfrac{5}{7}\)
\(=\dfrac{25}{63}\)
d) \(4\cdot11\cdot\dfrac{3}{4}\cdot\dfrac{9}{121}\)
\(=\left(4\cdot\dfrac{3}{4}\right)\cdot\left(11\cdot\dfrac{9}{121}\right)\)
\(=3\cdot\dfrac{9}{11}\)
\(=\dfrac{27}{11}\)
T
12 tháng 8 2023
a) \(\dfrac{3}{7}-\dfrac{4}{11}-\dfrac{3}{7}\)
=\(\left(\dfrac{3}{7}-\dfrac{3}{7}\right)-\dfrac{4}{11}\)
\(=-\dfrac{4}{11}\)
b) \(-\dfrac{5}{11}+\dfrac{3}{8}-\dfrac{6}{11}\)
\(=\left(-\dfrac{5}{11}-\dfrac{6}{11}\right)+\dfrac{3}{8}\)
\(=-1+\dfrac{3}{8}=-\dfrac{8}{8}+\dfrac{3}{8}=-\dfrac{5}{8}\)
c) \(\dfrac{3}{4}-\dfrac{1}{3}-\dfrac{11}{4}\)
\(=\left(\dfrac{3}{4}-\dfrac{11}{4}\right)-\dfrac{1}{3}\)
\(=-2-\dfrac{1}{3}=-\dfrac{6}{3}-\dfrac{1}{3}=-\dfrac{7}{3}\)
13 tháng 1 2018
Mình làm mẫu 1 bài rùi bạn tự giải những bài còn lại nha
1, 7A = 7+7^2+7^3+....+7^2008
6A = 7A - A = (7+7^2+7^3+....+7^2008)-(1+7+7^2+....+7^2007) = 7^2008-1
=> A = (7^2008-1)/6
Tk mk nha
13 tháng 1 2018
\(A=1+7+7^2+7^3+...+7^{2007}\)
\(\Rightarrow7A=7+7^2+7^3+7^4+...+7^{2008}\)
\(\Rightarrow7A-A=\left(7+7^2+7^3+...+7^{2008}\right)-\left(1+7+7^2+...+7^{2007}\right)\)
\(\Rightarrow6A=7^{2008}-1\)
\(\Rightarrow A=\frac{7^{2008}-1}{6}\)
KV
25 tháng 9 2023
2) -3(4 - 7) + 5(-3 + 2)
= -3.4 + 3.7 - 5.3 + 5.2
= -12 + 21 -15 + 10
= 31 - 27
= 4
4) -5(2 - 7) + 4(2 - 5)
= -5.2 + 5.7 + 4.2 - 4.5
= -10 + 35 + 8 - 20
= 38 - 30
= 8
DD
Đoàn Đức Hà
Giáo viên
4 tháng 7 2021
\(A=1+7+7^2+7^3+...+7^{2007}\)
\(7A=7+7^2+7^3+7^4+...+7^{2008}\)
\(7A-A=\left(7+7^2+7^3+7^4+...+7^{2008}\right)-\left(1+7+7^2+7^3+...+7^{2007}\right)\)
\(6A=7^{2008}-1\)
\(A=\frac{7^{2008}-1}{6}\)
Tương tự, \(B=\frac{4^{101}-1}{3},C=\frac{3^{101}-1}{2}\).
DD
Đoàn Đức Hà
Giáo viên
4 tháng 7 2021
\(D=7+7^3+7^5+7^7+...+7^{99}\)
\(7^2.D=7^3+7^5+7^7+7^9+...+7^{101}\)
\(\left(7^2-1\right)D=\left(7^3+7^5+7^7+7^9+...+7^{101}\right)-\left(7+7^3+7^5+7^7+...+7^{99}\right)\)
\(48D=7^{101}-7\)
\(D=\frac{7^{101}-7}{48}\)
Tương tự, \(E=\frac{2^{9011}-2}{3}\)
18 tháng 4 2023
\(1,-\dfrac{4}{7}+\dfrac{2}{3}\times\dfrac{-9}{14}\)
\(=\dfrac{-4}{7}+\dfrac{-18}{42}\)
\(=\dfrac{-4\times6}{7\times6}+\dfrac{-18}{42}\)
\(=\dfrac{-20}{42}+\dfrac{-18}{42}\)
\(=-\dfrac{38}{42}\)
\(=-\dfrac{19}{21}\)
\(2,\dfrac{17}{13}-\left(\dfrac{4}{13}-11\right)\)
\(=\dfrac{17}{13}-\dfrac{4}{13}+11\)
\(=\dfrac{13}{13}+11\)
\(=1+11\)
\(=12\)
\(3,8\dfrac{2}{7}-\left(3\dfrac{4}{9}+4\dfrac{2}{7}\right)\)
\(=\dfrac{58}{7}-\left(\dfrac{31}{9}+\dfrac{30}{7}\right)\)
\(=\dfrac{58}{7}-\dfrac{31}{9}-\dfrac{30}{7}\)
\(=\dfrac{58}{7}-\dfrac{30}{7}-\dfrac{31}{9}\)
\(=\dfrac{28}{7}-\dfrac{31}{9}\)
\(=\dfrac{28\times9}{7\times9}-\dfrac{31\times7}{9\times7}\)
\(=\dfrac{252}{63}-\dfrac{217}{63}\)
\(=\dfrac{35}{63}\)
\(=\dfrac{5}{9}\)
\(5,\left(\dfrac{2}{3}-1\dfrac{1}{2}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\left(\dfrac{2}{3}-\dfrac{3}{2}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\left(\dfrac{2\times2}{3\times2}-\dfrac{3\times3}{2\times3}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\left(\dfrac{4}{6}-\dfrac{9}{6}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\dfrac{-5}{6}:\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\dfrac{-5}{6}\times\dfrac{3}{4}+\dfrac{1}{2}\)
\(=\dfrac{-15}{24}+\dfrac{1}{2}\)
\(=\dfrac{-15}{24}+\dfrac{1\times12}{2\times12}\)
\(=\dfrac{-15}{24}+\dfrac{12}{24}\)
\(=\dfrac{-3}{24}\)
\(=-\dfrac{1}{8}\)
\(6,\dfrac{-5}{13}+\dfrac{2}{5}+\dfrac{-8}{13}+\dfrac{3}{5}-\dfrac{3}{7}\)
\(=\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)+\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{3}{7}\)
\(=\dfrac{-13}{13}+\dfrac{5}{5}-\dfrac{3}{7}\)
\(=-1+1-\dfrac{3}{7}\)
\(=-\dfrac{3}{7}\)
\(7,\dfrac{6}{5}\times\dfrac{3}{7}+\dfrac{6}{5}:\dfrac{7}{10}+\dfrac{6}{5}\)
\(=\dfrac{6}{5}\times\dfrac{3}{7}+\dfrac{6}{5}\times\dfrac{10}{7}+\dfrac{6}{5}\)
\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+1\right)\)
\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+\dfrac{1\times7}{1\times7}\right)\)
\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+\dfrac{7}{7}\right)\)
\(=\dfrac{6}{5}\times\dfrac{20}{7}\)
\(=\dfrac{120}{35}\)
\(=\dfrac{24}{7}\)
24 tháng 6 2018
a) \(\frac{7}{2}-\frac{14}{3}+\left(\frac{3}{4}-\frac{7}{3}\right)-\left(\frac{5}{6}-\frac{7}{4}\right)+\frac{11}{2}-3\)
\(=\frac{7}{2}-\frac{14}{3}+\frac{3}{4}-\frac{7}{3}-\frac{5}{6}+\frac{7}{4}+\frac{11}{2}-3\)
\(=\left(\frac{7}{2}+\frac{11}{2}\right)-\left(\frac{14}{3}+\frac{7}{3}\right)-3+\left(\frac{3}{4}+\frac{7}{4}\right)-\frac{5}{6}\)
\(=9-7-3+(\frac{5}{2}-\frac{5}{6})\)
\(=-1+\frac{5}{3}\)
\(=\frac{2}{3}\)
b) \(\frac{7}{3}-\frac{7}{5}+\frac{11}{10}-\left(\frac{2}{5}-\frac{5}{6}\right)+\frac{4}{15}-\frac{4}{3}\)
\(=\left(\frac{7}{3}-\frac{4}{3}\right)-\left(\frac{7}{5}+\frac{2}{5}\right)+(\frac{11}{10}+\frac{5}{6}+\frac{4}{15})\)
\(=1-\frac{9}{5}+\frac{11}{5}\)
\(=1-\left(\frac{9}{5}-\frac{11}{5}\right)\)
\(=1-\left(\frac{-2}{5}\right)\)
\(=1\frac{2}{5}\)
24 tháng 6 2018
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mik ko biết
mong bn thông cảm
nha ................
