\(M=\frac{1}{3.4}+\frac{7}{3.4}+\frac{2}{3.5}+\frac{14}{5.9}-\frac{4}{9.13}=?\)
Tính bằng cách hợp lý
a) \(\frac{-1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-...-\frac{1}{99.100}\)
b)\(\frac{-4}{1.5}-\frac{4}{5.9}-\frac{4}{9.13}-...-\frac{4}{96.100}\)
1.Tính:
a,\(A=\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-......-\frac{1}{\left(n-1\right).n}\)\(n\in N\)
b,\(\frac{4}{1.5}-\frac{4}{5.9}-\frac{4}{9.13}-....-\frac{4}{\left(n-4\right).n}\)\(n\in N\)
c\(C=1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-.....-\frac{1}{2^{10}}\)
1 Tính :
a) \(A=\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-...-\frac{1}{\left(n-1\right).n}\)
\(=\frac{1}{1.2}-\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{n}\right)\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{n}\)
\(=\frac{1}{n}\)
b) \(B=\frac{4}{1.5}-\frac{4}{5.9}-\frac{4}{9.13}-...-\frac{4}{\left(n-4\right).n}\)
\(=\frac{4}{1.5}-\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{\left(n-4\right).n}\right)\)
\(=\frac{4}{5}-\left(\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{\left(n-4\right).n}\right)\)
\(=\frac{4}{5}-\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{n-4}-\frac{1}{n}\right)\)
\(=\frac{4}{5}-\left(\frac{1}{5}-\frac{1}{n}\right)\)
\(=\frac{4}{5}-\frac{1}{5}+\frac{1}{n}\)
\(=\frac{3}{5}+\frac{1}{n}\)
c) \(C=1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{10}}\)
\(=1-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
Đặt \(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
\(\Rightarrow C=1-B\left(1\right)\)
\(\Rightarrow2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
Lấy 2B trừ B ta có :
\(2B-B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
\(B=1-\frac{1}{2^{10}}\left(2\right)\)
Thay (2) vào (1) ta có :
\(C=1-\left(1-\frac{1}{10}\right)\)
\(=1-1+\frac{1}{10}\)
\(=\frac{1}{10}\)
Vậy \(C=\frac{1}{10}\)
bài 1 tính
A = \(\frac{2}{3.5}+\frac{2}{5.7}+..............+\frac{2}{37.39}\); B = \(\frac{4}{5.7}+\frac{4}{7.9}+..........+\frac{4}{59.61}\) ; C = \(\frac{4}{5.9}+\frac{4}{9.13}+.................+\frac{4}{41.45}\) Bài 2 chứng minh : \(\frac{m}{b.\left(b+m\right)}=\frac{1}{b}-\frac{1}{b+m}\)
Bài 1:
Chứng tỏ rằng phân số \(\frac{n+1}{2n+1}\)với n \(\varepsilon\)N và n \(\notin\)0
Bài 2:
Tìm n\(\in\)N để \(\frac{n+7}{n-2}\)\(\in\)Z
Bài 3:
a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}< 1\)
b) \(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+\frac{4}{17.21}< 1\)
c) \(\frac{4}{3.5}+\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{37.39}>\frac{7}{13}\)
Bài 4:
Tính:
A = \(\frac{\frac{2}{3}+\frac{2}{5}-\frac{2}{9}}{\frac{4}{3}+\frac{4}{5}-\frac{4}{9}}\)
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Tính A = \(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)-\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\right)+\left(-2-4-6-...-100\right)+\)\(\left(-1.2-2.3-3.4-...-99.100\right)\)
tính hợp lí
\(\frac{11}{125}-\frac{17}{18}-\frac{5}{7}+\frac{4}{9}+\frac{17}{14}\)
\(\frac{42}{46}+\frac{250}{286}+\frac{-2121}{2323}+\frac{-125125}{143143}\)
tính
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2003.2004}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{2003.2005}\)
a) B = \(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+.......+\frac{4}{99.101}\)
b) C = \(\frac{5}{1.5}+\frac{5}{5.9}+\frac{5}{9.13}+......+\frac{5}{101.105}\)
a, \(\frac{1}{2}.B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(\frac{1}{2}.B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
\(\frac{1}{2}.B=1-\frac{1}{101}=\frac{100}{101}\)
\(B=\frac{100}{101}.2=\frac{200}{101}\)
b, \(\frac{4}{5}.C=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{101.105}\)
\(\frac{4}{5}.C=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{101}-\frac{1}{105}\)
\(\frac{4}{5}.C=1-\frac{1}{105}=\frac{104}{105}\)
\(C=\frac{104}{105}.\frac{5}{4}=\frac{26}{21}\)
\(B=\frac{2}{2}\cdot\left(\frac{4}{1\cdot3}+\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+....+\frac{4}{99\cdot101}\right)\)
\(=\frac{4}{2}\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\right)\)
\(=2\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=2\cdot\left(1-\frac{1}{101}\right)\)
\(=2\cdot\frac{100}{101}\)
\(=1\frac{99}{101}\)
a) \(B=\frac{4}{1\cdot3}+\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+...+\frac{4}{99\cdot101}\)
\(\Rightarrow\frac{2}{4}B=\frac{2}{4}\left(\frac{4}{1\cdot3}+\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+...+\frac{4}{99\cdot101}\right)\)
\(\Leftrightarrow\frac{2}{4}B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+....+\frac{2}{99\cdot101}\)
\(\Leftrightarrow\frac{2}{4}B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{99}-\frac{1}{101}\)
\(\Leftrightarrow\frac{2}{4}B=1-\frac{1}{101}=\frac{100}{101}\)
\(\Leftrightarrow B=\frac{100}{101}:\frac{2}{4}=\frac{100\cdot4}{101\cdot2}=\frac{200}{101}\)
1
a, \(\frac{1}{3}\) +(\(\frac{1}{5}\) -\(\frac{1}{7}\))
b, \(\frac{3}{5}\)-(\(\frac{3}{4}\) - \(\frac{1}{2}\))
C=\(\frac{4}{7}\) - (\(\frac{2}{5}\)+ \(\frac{1}{3}\))
2
a, S =-\(\frac{1}{1.2}\) -\(\frac{1}{2.3}\)-\(\frac{1}{3.4}\)- ... - \(\frac{1}{\left(n_{ }-1\right).n}\)
b, S= \(\frac{-4}{1.5}\) - \(\frac{4}{5.9}\)-\(\frac{4}{9.13}\)-...-\(\frac{4}{\left(n-4\right).n}\)
1.
a.
\(\frac{1}{3}+\left(\frac{1}{5}-\frac{1}{7}\right)\)
\(=\frac{1}{3}+\frac{1}{5}-\frac{1}{7}\)
\(=\frac{35-21-15}{105}\)
\(=-\frac{1}{105}\)
b.
\(\frac{3}{5}-\left(\frac{3}{4}-\frac{1}{2}\right)\)
\(=\frac{3}{5}-\frac{3}{4}+\frac{1}{2}\)
\(=\frac{12-15+10}{20}\)
\(=\frac{7}{20}\)
c.
\(\frac{4}{7}-\left(\frac{2}{5}+\frac{1}{3}\right)\)
\(=\frac{4}{7}-\frac{2}{5}-\frac{1}{3}\)
\(=\frac{60-42-35}{105}\)
\(=-\frac{17}{105}\)
2.
a.
\(S=-\frac{1}{1\times2}-\frac{1}{2\times3}-\frac{1}{3\times4}-...-\frac{1}{\left(n-1\right)\times n}\)
\(S=-\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{\left(n-1\right)\times n}\right)\)
\(S=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\right)\)
\(S=-\left(1-\frac{1}{n}\right)\)
\(S=-1+\frac{1}{n}\)
b.
\(S=-\frac{4}{1\times5}-\frac{4}{5\times9}-\frac{4}{9\times13}-...-\frac{4}{\left(n-4\right)\times n}\)
\(S=-\left(\frac{4}{1\times5}+\frac{4}{5\times9}+\frac{4}{9\times13}+...+\frac{4}{\left(n-4\right)\times n}\right)\)
\(S=-\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{n-4}-\frac{1}{n}\right)\)
\(S=-\left(1-\frac{1}{n}\right)\)
\(S=-1+\frac{1}{n}\)
Chúc bạn học tốt
Tìm x biết:
a)\(\frac{x-1}{21}=\frac{3}{x+1}\)
b)\(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\)\(\frac{4}{41.45}=\frac{29}{45}\)
c)\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\)\(\frac{1}{\left(2x+1\right).\left(2x+3\right)}=\frac{15}{93}\)
\(a,\frac{x-1}{21}=\frac{3}{x+1}\)
\(\Leftrightarrow\left[x-1\right]\left[x+1\right]=63\)
\(\Leftrightarrow x^2-1=63\)
\(\Leftrightarrow x^2=64\)
\(\Leftrightarrow x^2=8^2\)
\(\Leftrightarrow x=\pm8\)
\(b,\frac{7}{x}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left[\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}\right]=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left[\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right]=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left[\frac{1}{5}-\frac{1}{45}\right]=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}=\frac{21}{45}\)
\(\Leftrightarrow\frac{7}{x}=\frac{7}{15}\)
\(\Leftrightarrow x=15\)
Vậy x = 15
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