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\(\frac{8}{15}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{4242}+\frac{3333}{5656}\right)\)
= \(\frac{8}{15}.\left(\frac{11}{4}+\frac{33}{20}+\frac{11}{14}+\frac{33}{56}\right)\)
= \(\frac{8}{15}.\frac{231}{40}\)
= \(\frac{77}{25}\)
Ta có : A = \(\dfrac{7}{4}.\left(\dfrac{333}{1212}+\dfrac{3333}{2020}+\dfrac{3333}{3030}+\dfrac{3333}{4242}\right)\)
\(\Rightarrow A=\dfrac{7}{4}.\left(\dfrac{11}{4}+\dfrac{33}{20}+\dfrac{11}{10}+\dfrac{11}{14}\right)\)
\(\Rightarrow A=\dfrac{7}{4}.\dfrac{44}{7}\)
\(\Rightarrow A=11\)
Vậy A = 11
A=7/4(33/12+33/20+33/30+33/42)
=7/4.33.(1/12+1/20+1/30+1/42)
=7/4.33.4/21
=11
1/ \(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{8^2}\)
\(B< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{7.8}\)
\(B< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}\)
\(B< \dfrac{1}{1}-\dfrac{1}{8}< 1\)
\(B< 1\)
2/ \(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{20}\right)\)
\(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{19}{20}\)
\(B=\dfrac{1\times2\times3\times...\times19}{2\times3\times4\times...\times20}\)
\(B=\dfrac{1}{20}\)
3/ \(A=\dfrac{7}{4}\cdot\left(\dfrac{3333}{1212}+\dfrac{3333}{2020}+\dfrac{3333}{3030}+\dfrac{3333}{4242}\right)\)
\(A=\dfrac{7}{4}\cdot\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)
\(A=\dfrac{7}{4}\cdot\left(\dfrac{33}{3.4}+\dfrac{33}{4.5}+\dfrac{33}{5.6}+\dfrac{33}{6.7}\right)\)
\(A=\dfrac{7}{4}.33.\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)
\(A=\dfrac{231}{4}.\left(\dfrac{1}{3}-\dfrac{1}{7}\right)\)
\(A=\dfrac{231}{4}\cdot\dfrac{4}{21}\)
\(A=11\)
4/ A phải là \(\dfrac{2011+2012}{2012+2013}\)
Ta có : \(B=\dfrac{2011}{2012}+\dfrac{2012}{2013}>\dfrac{2011}{2013}+\dfrac{2012}{2013}=\dfrac{2011+2012}{2013}>\dfrac{2011+2012}{2012+2013}=A\)
\(\Rightarrow B>A\)
B=(1-\(\frac{1}{2}\))x(1-\(\frac{1}{3}\))x(1-\(\frac{1}{4}\))x...x(1-\(\frac{1}{20}\))
B=\(\frac{1}{2}\)X\(\frac{2}{3}\)X\(\frac{3}{4}\)X...X\(\frac{19}{20}\)
B=\(\frac{1.2.3.4.4.5.7.8.9.10.11.12.13.14.15.16.17.18.19}{2.3.4.5.6.7.8.9.10.11.12.13.14.15.16.17.18.19.20}\)
B=20
Vậy B=20
Không biết kết quả đúng ko nhưng cách làm thì đúng.
B= \(\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{3}\right)x\left(1-\frac{1}{4}\right)x...x\left(1-\frac{1}{20}\right)\)
=\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{19}{20}\)(Rút gọn cả tử xuống mẫu )
= \(\frac{1.2.3...19}{2.3.4...20}\)
=\(\frac{1}{20}\)
Vậy B= \(\frac{1}{20}\)
=> A=\(\frac{7}{4}\) . ( \(\frac{33}{12}\) + \(\frac{33}{20}\) + \(\frac{33}{30}\) + \(\frac{33}{42}\) ) => A= \(\frac{7}{4}\).33. ( \(\frac{1}{12}\) + \(\frac{1}{20}\) + \(\frac{1}{30}\) + \(\frac{1}{42}\) )
=> A=\(\frac{7}{4}\).33. ( \(\frac{1}{3.4}\) + \(\frac{1}{4.5}\) + \(\frac{1}{5.6}\) + \(\frac{1}{6.7}\) ) = \(\frac{7}{4}\).33.(\(\frac{1}{3}\) - \(\frac{1}{4}\) + \(\frac{1}{4}\) - \(\frac{1}{5}\) + \(\frac{1}{5}\) - \(\frac{1}{6}\) + \(\frac{1}{6}\) - \(\frac{1}{7}\) )
= \(\frac{7}{4}\) .33.(\(\frac{1}{3}\) - \(\frac{1}{7}\)) = \(\frac{7}{4}\) .33. \(\frac{4}{21}\) = 11. Vậy A=11
Bài 1:
\(A=\frac{3333}{101}\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)=\frac{3333}{101}\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(A=\frac{3333}{101}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(A=\frac{3333}{101}\left(\frac{1}{3}-\frac{1}{7}\right)=\frac{3333}{101}.\frac{4}{21}=\frac{1111.4}{101.7}=\frac{4444}{707}\)
Bài 2
\(A=\frac{2^{10}+1}{2^{10}-1}=\frac{2^{10}-1+2}{2^{10}-1}=1+\frac{2}{2^{10}-1}\)
\(B=\frac{2^{10}-1}{2^{10}-3}=\frac{2^{10}-3+4}{2^{10}-3}=1+\frac{4}{2^{10}-3}\)
Ta thấy \(2^{10}-1>2^{10}-3\Rightarrow\frac{2}{2^{10}-1}< \frac{2}{2^{10}-3}< \frac{4}{2^{10}-3}\)
Từ đó \(\Rightarrow1+\frac{2}{2^{10}-1}< 1+\frac{4}{2^{10}-3}\Rightarrow A< B\)
Bài 3\(P=\frac{\left(\frac{2}{3}-\frac{1}{4}\right)+\frac{5}{11}}{\frac{5}{12}+\left(1-\frac{7}{11}\right)}=\frac{\frac{5}{12}+\frac{5}{11}}{\frac{5}{12}+\frac{4}{11}}=\frac{\frac{55+60}{11.12}}{\frac{55+48}{12.11}}=\frac{115}{103}\)
\(\frac{7}{4}.\left(\frac{101.33}{101.12}+\frac{101.33}{101.20}+\frac{101.33}{101.30}+\frac{101.33}{101.42}\right)\)
\(=\frac{7.33}{4}\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\\ =\frac{7.33}{4}\left(\frac{35+21+14+1}{420}\right)\)
\(=\frac{7.3.11}{4}.\frac{71}{420}=\frac{7.3.11.71}{4.4.5.3.7}=\frac{781}{100}\)
mk lm chak vớ vẩn rồi
bao quynh Cao bạn ơi hình như bn làm sai đề ạ 7/4 mà sao lại 4/7 ạ
\(=\dfrac{8}{15}\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{42}+\dfrac{33}{56}\right)\)
\(=\dfrac{8}{15}\cdot33\cdot\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\right)\)
\(=8\cdot\dfrac{11}{5}\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{8}\right)\)
\(=\dfrac{88}{5}\cdot\dfrac{7}{40}=\dfrac{77}{25}\)