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\(B=\dfrac{1}{11}+\dfrac{1}{11^2}+\dfrac{1}{11^3}+...+\dfrac{1}{11^{99}}+\dfrac{1}{11^{100}}\\ 11B=1+\dfrac{1}{11}+\dfrac{1}{11^2}+...+\dfrac{1}{11^{98}}+\dfrac{1}{11^{99}}\\ 11B-B=1+\dfrac{1}{11}+\dfrac{1}{11^2}+...+\dfrac{1}{1^{99}0}-\dfrac{1}{11}-\dfrac{1}{11^2}-\dfrac{1}{11^3}-...-\dfrac{1}{11^{100}}\\ 10B=1-\dfrac{1}{11^{99}}\\ B=\dfrac{1-\dfrac{1}{11^{99}}}{10}\)
có : `1-1/(11^99)<1`
\(\Rightarrow\dfrac{1-\dfrac{1}{11^{99}}}{10}< \dfrac{1}{10}\)
hay `B<1/10`
\(\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)....\left(\frac{1}{121}-1\right)=-\frac{3}{4}.-\frac{8}{9}....-\frac{120}{121}=\frac{-1.3}{2.2}\cdot\frac{-2.4}{3.3}....-\frac{10.12}{11.11}=\frac{-1}{2}.-\frac{12}{11}=\frac{6}{11}\)
\(\frac{\frac{2}{3}+\frac{2}{5}+\frac{2}{7}+\frac{2}{9}}{\frac{11}{3}+\frac{11}{5}+\frac{11}{7}+\frac{11}{9}}=\frac{2\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}\right)}{11\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}\right)}=\frac{2}{11}\)
\(\frac{\frac{2}{3}+\frac{2}{5}+\frac{2}{7}+\frac{2}{9}}{\frac{11}{3}+\frac{11}{5}+\frac{11}{7}+\frac{11}{9}}\)
=\(\frac{2\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}\right)}{11\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}\right)}\)
=\(\frac{2}{11}\)
\(\frac{\frac{2}{3}+\frac{2}{5}+\frac{2}{7}+\frac{2}{9}}{\frac{11}{3}+\frac{11}{5}+\frac{11}{7}+\frac{11}{9}}=\frac{2\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}\right)}{11\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}\right)}=\frac{2}{11}\)
đáp án là 18 nhé
TL
= 18
Hok tốt