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DN
1
12 tháng 6 2016
(-2/11+ 3/12 +-4/13 +5/14) : (-4/11 + 6/12+ -8/13 +10/14)0000000000000000000000000000000000000= 0
ML
3 tháng 8 2017
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(B=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{107.111}\)
\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{107}-\frac{1}{111}\)
\(B=\frac{1}{3}-\frac{1}{111}\)
\(B=\frac{12}{37}\)
\(C=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(C=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(C=7\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(C=7.\frac{3}{35}\)
\(C=\frac{3}{5}\)
TP
3 tháng 8 2017
Ta có:
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{1}-\frac{1}{100}=\frac{99}{100}\)
\(B=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{107.111}\)
\(B=4.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}\right)\)
\(B=4.\left(\frac{1}{3}-\frac{1}{111}\right)=4.\frac{12}{37}=\frac{48}{37}\)
\(C=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(C=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)
\(C=7.\left(\frac{1}{10}-\frac{1}{70}\right)=7.\frac{3}{35}=\frac{3}{5}\)
25 tháng 9 2020
a.
\(\left(1\frac{1}{4}+\frac{3}{5}\right):\left(-\frac{11}{12}\right)+\left(\frac{3}{8}-1\frac{2}{5}\right):\left(-\frac{11}{12}\right)\)
\(=\left(\frac{5}{4}+\frac{3}{5}+\frac{3}{8}-\frac{7}{5}\right):\left(-\frac{11}{12}\right)\)
\(=\left(\frac{13}{8}-\frac{4}{5}\right):\left(-\frac{11}{12}\right)\)
\(=\frac{33}{40}:\left(-\frac{11}{12}\right)\)
\(=\frac{33}{40}\cdot\left(-\frac{12}{11}\right)\)
\(=\frac{-9}{10}\)
b.
\(\left(\frac{3}{8}-1\frac{2}{5}\right):\left(-\frac{11}{15}\right)+\left(1\frac{1}{4}+\frac{3}{5}\right):\left(-\frac{11}{15}\right)\)
\(=\left(\frac{3}{8}-\frac{7}{5}+\frac{5}{4}+\frac{3}{5}\right):\left(-\frac{11}{15}\right)\)
\(=\left(\frac{13}{8}-\frac{4}{5}\right):\left(-\frac{11}{15}\right)\)
\(=\frac{33}{40}:\left(-\frac{11}{15}\right)\)
\(=\frac{33}{40}\cdot\left(-\frac{15}{11}\right)\)
\(=\frac{-9}{8}\)
SB
3
24 tháng 7 2023
\(-\dfrac{3}{4}.\dfrac{4}{11}+\left(-\dfrac{5}{4}\right).\dfrac{4}{11}=\dfrac{4}{11}\left(-\dfrac{3}{4}-\dfrac{4}{4}\right)=\dfrac{4}{11}.\left(-\dfrac{4}{7}\right)=-\dfrac{16}{77}\)
\(\dfrac{7}{12}-\left(-\dfrac{1}{5}\right)-\dfrac{5}{6}+\dfrac{2}{3}+\left(-\dfrac{1}{5}\right)\)
\(=\dfrac{7}{12}+\dfrac{1}{5}-\dfrac{5}{6}+\dfrac{2}{3}-\dfrac{1}{5}\)
\(=\dfrac{7}{12}-\dfrac{5}{6}+\dfrac{2}{3}=\dfrac{7}{12}-\dfrac{10}{12}+\dfrac{8}{12}=\dfrac{5}{12}\)
O
5 tháng 7 2023
\(\left(\dfrac{3}{7}\right)^2\cdot\left(-7\right)^4=\dfrac{9}{49}\cdot49^2=9\cdot49=441\)
\(\left(-11\right)^{12}\cdot\left(\dfrac{4}{11}\right)^4=11^{12}\cdot\dfrac{4^4}{11^4}=11^8\cdot4^4=54875873536\)
\(\left(-6\right)^8\cdot\left(\dfrac{5}{6}\right)^7=6^8\cdot\dfrac{5^7}{6^7}=6\cdot5^7=469750\)
H9
HT.Phong (9A5)
CTVHS
5 tháng 7 2023
4) \(\left(\dfrac{3}{7}\right)^2\cdot\left(-7\right)^4\)
\(=\left(\dfrac{3}{7}\right)^2\cdot\left[\left(-7\right)^2\right]^2\)
\(=\left(\dfrac{3}{7}\right)^2\cdot49^2\)
\(=\left(\dfrac{3}{7}\cdot49\right)^2\)
\(=\left(\dfrac{147}{7}\right)^2\)
\(=21^2\)
\(=441\)
5) \(\left(-11\right)^{12}\cdot\left(\dfrac{4}{11}\right)^6\)
\(=\left[\left(-11\right)^2\right]^6\cdot\left(\dfrac{4}{11}\right)^6\)
\(=121^6\cdot\left(\dfrac{4}{11}\right)^6\)
\(=\left(121\cdot\dfrac{4}{11}\right)^6\)
\(=44^6\)
6) \(6^8\cdot\left(\dfrac{5}{7}\right)^7\)
\(=6^8\cdot\dfrac{5^7}{6^7}\)
\(=\dfrac{6^8\cdot5^7}{6^7}\)
\(=6\cdot5^7\)
\(=469750\)
DB
0
H
2
\(\left(\dfrac{11}{4}\right)^{12}:\left(-\dfrac{11}{4}\right)^{11}\)
\(=\left(-\dfrac{11}{4}\right)^{12}:\left(-\dfrac{11}{4}\right)^{11}\)
\(=\left(-\dfrac{11}{4}\right)^{12-11}\)
\(=\left(-\dfrac{11}{4}\right)^1\)
\(=-\dfrac{11}{4}\)
(11/4)^12:(-11/4)^11