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1)
A = \(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{132}\)
= \(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{11.12}\)
= \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{11}-\frac{1}{12}\)
= \(\frac{1}{5}-\frac{1}{12}\)
= \(\frac{7}{60}\)
B = \(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right).....\left(1+\frac{1}{99}\right)\)
= \(\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}\)
= \(\frac{3.4.5.....100}{2.3.4....99}\)
= \(\frac{100}{2}=50\)
C = \(\frac{1}{4^{2-1}}+\frac{1}{6^{2-1}}+\frac{1}{8^{2-1}}...+\frac{1}{30^{2-1}}\)
= \(\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...+\frac{1}{30}\)
= \(\frac{1}{2.2}+\frac{1}{2.3}+\frac{1}{2.4}+...+\frac{1}{2.15}\)
= \(\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{2}.\frac{1}{4}+...+\frac{1}{2}.\frac{1}{15}\)
= \(\frac{1}{2}.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{15}\right)\)
\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
\(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)
\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
\(A=\frac{1}{5}+\left(\frac{1}{6}-\frac{1}{6}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+\left(\frac{1}{8}-\frac{1}{8}\right)+\left(\frac{1}{9}-\frac{1}{9}\right)+\left(\frac{1}{10}-\frac{1}{10}\right)+\left(\frac{1}{11}-\frac{1}{11}\right)-\frac{1}{12}\)
\(A=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)
~ Hok tốt ~
Nên đợi ai đó giải hết 2 3 bài xong rồi mới đăng tiếp những bài còn lại, chứ dài vậy giải hơi nản =)))
Bài 1:
1, \(13\frac{2}{5}-\left(\frac{18}{32}-2\frac{6}{10}\right)\)
\(=\frac{67}{5}-\left(\frac{9}{16}-\frac{13}{5}\right)\)(Chuyển hỗn số thành p/số và rút gọn hai số trong ngoặc luôn)
\(=\frac{67}{5}-\left(\frac{-163}{80}\right)\)
\(=\frac{246}{16}\)
2, \(22.4\frac{5}{7}-\left(8.91+1,09\right)\)(Phần 2 viết vầy có đúng không vậy ? Nếu sai thì kêu chị sửa nhé)
\(=22.\frac{33}{7}-10\)
\(=\frac{726}{7}-10\)
\(=\frac{656}{7}\)
3, Chỗ ''3 phần 10 phần 2'' là sao :v ?
4, \(5\frac{2}{7}.\frac{8}{11}+5\frac{2}{7}.\frac{5}{11}-5\frac{2}{7}.\frac{2}{11}\)
\(=\frac{37}{7}.\frac{8}{11}+\frac{37}{7}.\frac{5}{11}-\frac{37}{7}.\frac{2}{11}\)(Chuyển hỗn số thành p/số)
\(=\frac{37}{7}.\left(\frac{8}{11}+\frac{5}{11}-\frac{2}{11}\right)\)(Dùng tính chất phân phối)
\(=\frac{37}{7}.\frac{11}{11}\)
\(=\frac{37}{7}.1=\frac{37}{7}\)
A<1-1/2+1/2-1/3+...+1/8-1/9=1-1/9=8/9
A>1/2-1/3+1/3-1/4+...+1/9-1/10=1/2-1/10=2/5
=>2/5<A<8/9
\(A=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{100^2}-1\right)=\frac{-3}{2^2}.\frac{-8}{3^2}...\frac{-9999}{100^2}\)
\(=-\frac{3.8...9999}{2^2.3^2...100^2}=-\frac{1.3.2.4...99.101}{2.2.3.3...100.100}=-\frac{\left(1.2....99\right).\left(3.4...101\right)}{\left(2.3...100\right).\left(2.3...100\right)}=-\frac{1.101}{100.2}=-\frac{101}{200}\)
\(< -\frac{100}{200}=\frac{1}{2}=B\)
=> A < B
A =\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{20^2}=\frac{1}{2^2}\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{20^2}\right)\)
\(< \frac{1}{2^2}\left(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{19.20}\right)=\frac{1}{4}\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\right)\)
\(=\frac{1}{4}\left(1+1-\frac{1}{20}\right)=\frac{1}{4}\left(2-\frac{1}{20}\right)=\frac{1}{2}-\frac{1}{80}< \frac{1}{2}\left(\text{đpcm}\right)\)
a, \(\left(\dfrac{1}{2}+1\right).\left(\dfrac{1}{3}+1\right).\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{999}+1\right)\)
\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{1000}{999}\)
\(=\dfrac{3.4.5...1000}{2.3.4...999}\)
\(=\dfrac{1000}{2}\)\(=500\)
b, \(\left(\dfrac{1}{2}-1\right).\left(\dfrac{1}{3}-1\right).\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{1000}-1\right)\)
\(=\dfrac{-1}{2}.\dfrac{-2}{3}.\dfrac{-3}{4}...\dfrac{-999}{1000}\)
\(=\dfrac{\left(-1\right).\left(-2\right).\left(-3\right)...\left(-999\right)}{2.3.4...1000}\)
\(=\dfrac{-1}{1000}\)