(9+1).(11-1)^2=?
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H9
HT.Phong (9A5)
CTVHS
24 tháng 9 2023
a) \(2\dfrac{4}{9}+6\dfrac{7}{11}+7\dfrac{5}{9}+13\dfrac{4}{11}\)
\(=2+\dfrac{4}{9}+6+\dfrac{7}{11}+7+\dfrac{5}{9}+13+\dfrac{4}{11}\)
\(=\left(2+6+7+13\right)+\left(\dfrac{4}{9}+\dfrac{5}{9}\right)+\left(\dfrac{7}{11}+\dfrac{4}{11}\right)\)
\(=28+1+1\)
\(=30\)
b) \(\dfrac{3}{4}+\dfrac{1}{2}\times\dfrac{12}{5}-1\dfrac{1}{2}\)
\(=\dfrac{3}{4}+\dfrac{1}{2}-1-\dfrac{1}{2}\)
\(=\dfrac{3}{4}-1\)
\(=\dfrac{3}{4}-\dfrac{4}{4}\)
\(=-\dfrac{1}{4}\)
22 tháng 5 2022
Bài 1:
\(=\dfrac{-1}{2}+\dfrac{3}{5}-\dfrac{1}{9}+\dfrac{1}{131}+\dfrac{2}{7}+\dfrac{4}{35}-\dfrac{7}{18}\)
\(=\left(-\dfrac{1}{2}-\dfrac{1}{9}-\dfrac{7}{18}\right)+\left(\dfrac{3}{5}+\dfrac{2}{7}+\dfrac{4}{35}\right)+\dfrac{1}{131}\)
\(=\dfrac{-9-2-7}{18}+\dfrac{21+10+4}{35}+\dfrac{1}{131}\)
=1/131
Bài 2:
b: \(B=\dfrac{1}{99}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{98\cdot99}\right)\)
\(=\dfrac{1}{99}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{98}-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{99}-\dfrac{98}{99}=-\dfrac{97}{99}\)
16 tháng 7 2015
\(\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)....\left(\frac{1}{121}-1\right)=-\frac{3}{4}.-\frac{8}{9}....-\frac{120}{121}=\frac{-1.3}{2.2}\cdot\frac{-2.4}{3.3}....-\frac{10.12}{11.11}=\frac{-1}{2}.-\frac{12}{11}=\frac{6}{11}\)
16 tháng 7 2015
\(\frac{\frac{2}{3}+\frac{2}{5}+\frac{2}{7}+\frac{2}{9}}{\frac{11}{3}+\frac{11}{5}+\frac{11}{7}+\frac{11}{9}}=\frac{2\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}\right)}{11\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}\right)}=\frac{2}{11}\)
HT
16 tháng 7 2015
\(\frac{\frac{2}{3}+\frac{2}{5}+\frac{2}{7}+\frac{2}{9}}{\frac{11}{3}+\frac{11}{5}+\frac{11}{7}+\frac{11}{9}}\)
=\(\frac{2\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}\right)}{11\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}\right)}\)
=\(\frac{2}{11}\)
16 tháng 7 2015
\(\frac{\frac{2}{3}+\frac{2}{5}+\frac{2}{7}+\frac{2}{9}}{\frac{11}{3}+\frac{11}{5}+\frac{11}{7}+\frac{11}{9}}=\frac{2\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}\right)}{11\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}\right)}=\frac{2}{11}\)
(9+1)(11-1)^2=10.10^2=10.100=1000
\(\left(9+1\right).\left(11-1\right)^2\)
\(=10.10^2\)
\(=10.100\)
\(=1000\)