2 Chứng minh:
a) \(\dfrac{1}{n}.\dfrac{1}{n+4}=\dfrac{1}{4}.(\dfrac{1}{n}-\dfrac{1}{n+4})\) b)Tính A=\(\dfrac{4}{3}.\dfrac{4}{7}+\dfrac{4}{7}.\dfrac{4}{11}+...+\dfrac{4}{95}.\dfrac{4}{99}\)
Rút gọn bt: A = \(\dfrac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)
B = \(\left(\dfrac{n-1}{1}+\dfrac{n-2}{2}+\dfrac{n-3}{3}+..+\dfrac{2}{n-2}+\dfrac{1}{n-1}\right):\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{n}\right)\)
A = \(\dfrac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)
Xét: n4 + 4 = (n2+2)2 - 4n2 = (n2-2n+2)(n2+2n+2) = [(n-1)2+1][(x+1)2+1] nên: A = \(\dfrac{\left(0^2+1\right)\left(2^2+1\right)}{\left(2^2+1\right)\left(4^2+1\right)}.\dfrac{\left(4^2+1\right)\left(6^2+1\right)}{\left(6^2+1\right)\left(8^2+1\right)}.....\dfrac{\left(20^2+1\right)\left(22^2+1\right)}{\left(22^2+1\right)\left(24^2+1\right)}=\dfrac{1}{24^2+1}=\dfrac{1}{577}\)
B = \(\left(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{2}{n-2}+\dfrac{1}{n-1}\right):\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{n}\right)\)
Đặt C = \(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{n-\left(n-2\right)}{n-2}+\dfrac{n-\left(n-1\right)}{n-1}\)
= \(\dfrac{n}{1}+\dfrac{n}{2}+...+\dfrac{n}{n-2}+\dfrac{n}{n-1}-1-1-...-1\)
= \(n+\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}-\left(n-1\right)\)
= \(\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}+\dfrac{n}{n}\)
= \(n\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{n}\right)\)
Vậy ...
3. Tính :
a/ \(\dfrac{-1}{2}\) + \(\dfrac{5}{6}\) + \(\dfrac{1}{3}\) b/ \(\dfrac{-3}{8}\) + \(\dfrac{7}{4}\) - \(\dfrac{1}{12}\) c/ \(\dfrac{3}{5}\) : (\(\dfrac{1}{4}\) . \(\dfrac{7}{5}\)) d/ \(\dfrac{10}{11}\) + \(\dfrac{4}{11}\) : 4 - \(\dfrac{1}{8}\)
3.a)\(\dfrac{-1}{2}+\dfrac{5}{6}+\dfrac{1}{3}=\dfrac{-3}{6}+\dfrac{5}{6}+\dfrac{2}{6}=\dfrac{-3+5+2}{6}=\dfrac{4}{6}=\dfrac{2}{3}\)
b)\(\dfrac{-3}{8}+\dfrac{7}{4}-\dfrac{1}{12}=\dfrac{-9}{24}+\dfrac{42}{24}-\dfrac{2}{24}=\dfrac{-9+42-2}{24}=\dfrac{31}{24}\)
c)\(\dfrac{3}{5}:\left(\dfrac{1}{4}.\dfrac{7}{5}\right)=\dfrac{3}{5}:\dfrac{7}{20}=\dfrac{3}{5}.\dfrac{20}{7}=\dfrac{12}{7}\)
d)\(\dfrac{10}{11}+\dfrac{4}{11}:4-\dfrac{1}{8}=\dfrac{10}{11}+\dfrac{4}{11}.\dfrac{1}{4}-\dfrac{1}{8}=\dfrac{10}{11}+\dfrac{1}{11}-\dfrac{1}{8}=1-\dfrac{1}{8}=\dfrac{8}{8}-\dfrac{1}{8}=\dfrac{7}{8}\)
Cho \(A=1+\dfrac{1}{4}+\dfrac{1}{9}+...+\dfrac{1}{n^2}\) với n là số tự nhiên. Chứng minh rằng \(A< \dfrac{7}{4}\).
Thực hiện phép tính: a) (6\(\dfrac{4}{9}\) + 3 \(\dfrac{7}{11}\)) - 4\(\dfrac{4}{9}\) b) \(\dfrac{1}{7}\). \(\dfrac{1}{5}\) + \(\dfrac{1}{7}\) . \(\dfrac{2}{5}\) + \(\dfrac{1}{7}\) .\(\dfrac{4}{5}\)
c) 25% - 1\(\dfrac{1}{2}\) . (-2019)o + 0,5 . \(\dfrac{12}{5}\)
1)Cho B=\(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{19}\).Chứng minh B>1
2)Tính nhanh:M=\(\dfrac{\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{3}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{11}}\)
2,
\(M=\dfrac{\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{3}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{11}}\) =\(\dfrac{3\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\)
\(=\dfrac{3}{4}\)
tìm nghiệm của phân thức viết dưới dạng phân số
a.\(\dfrac{4}{\left(2+\dfrac{2}{1+\dfrac{4}{5}}\right)x-\left(1-\dfrac{4}{2+\dfrac{1}{1+\dfrac{7}{8}}}\right)}+\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{4}}}\)
= \(4+\dfrac{2}{1+\dfrac{8}{9}}\)
b.
\(\dfrac{1}{2+\dfrac{3}{4+\dfrac{5}{6+\dfrac{7}{8}}}}=\dfrac{1}{3+\dfrac{2}{5+\dfrac{3}{7+\dfrac{4}{9}}}}+x.\left(4+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{2}}}\right)\)
(giải bằng máy tính casio )
Tính tổng đại số
\(A=\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{3}{4}-\dfrac{1}{5}-\dfrac{2}{5}-\dfrac{3}{5}-\dfrac{4}{5}+...+\dfrac{1}{10}+\dfrac{2}{10}+...+\dfrac{9}{10}\)
\(B=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{3}{4}+...+\dfrac{1}{n}+\dfrac{2}{n}+...+\dfrac{n-1}{n}\)\(\left(n\in Z,n\ge2\right)\)
a, Tính: M = \(1+\dfrac{1}{5}+\dfrac{3}{35}+...+\dfrac{3}{9603}+\dfrac{3}{9999}\)
b, Chứng tỏ: S = \(\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+...+\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{4}\left(n\in N,n\ge2\right)\)
a: \(M=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}+\dfrac{2}{99\cdot101}\right)\)
\(=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{101}\right)\)
\(=\dfrac{6}{5}+\dfrac{3}{10}-\dfrac{3}{202}=\dfrac{150}{101}\)
b:
a) \(\dfrac{ 3}{ 4 }\)+\(\dfrac{ -7 }{ 5 }\)+\(\dfrac{ 1 }{ 4 }\)+\(\dfrac{ -3 }{ 5 }\) b) \(\dfrac{ 4 }{ 9 }.\dfrac{7 }{ 11 }-\dfrac{4 }{ 11 }.\dfrac{ 2 }{9 } + \dfrac{ 6 }{ 11 }.\dfrac{ 4 }{ 9 }\)
a) 3/4 + (-7/5) + 1/4 + (-3/5)
= (3/4 + 1/4) + (-7/5 - 3/5)
= 1 - 2
= -1
b) 4/9 . 7/11 - 4/11 . 2/9 + 6/11 . 4/9
= 4/9 . (7/11 - 2/11 + 6/11)
= 4/9 . 1
= 4/9
a)
\(\dfrac{3}{4}+\dfrac{-7}{5}+\dfrac{1}{4}+\dfrac{-3}{5}\)
\(=\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+\left(\dfrac{-7}{5}-\dfrac{3}{5}\right)\)
\(=1-2\)
\(=-1\)
b)
\(\dfrac{4}{9}.\dfrac{7}{11}-\dfrac{4}{11}.\dfrac{2}{9}+\dfrac{6}{11}.\dfrac{4}{9}\)
\(=\dfrac{28}{99}-\dfrac{8}{99}+\dfrac{24}{99}\)
\(=\dfrac{28-8+24}{99}\)
\(=\dfrac{44}{99}\)
\(=\dfrac{4}{9}\)