Bài 1: Thực hiện phép tính
S= \(\frac{3}{\left(1\cdot2\right)^2}+\frac{5}{\left(2\cdot3\right)^2}+...+\frac{61}{\left(30\cdot31\right)^2}\)
Thực hiện phép tính:
S = \(\frac{3}{\left(1\cdot2\right)^2}\)+\(\frac{5}{\left(2\cdot3\right)^2}\)+.......+\(\frac{61}{\left(30\cdot31\right)^2}\)
\(S=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+...+\frac{61}{\left(30.31\right)^2}\)
\(S=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+...+\frac{61}{30^2.31^2}\)
\(S=\frac{3}{1.4}+\frac{5}{4.9}+...+\frac{61}{900.961}\)
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{1}{900}-\frac{1}{961}\)
\(S=1-\frac{1}{961}\)
\(S=\frac{960}{961}\)
thực hiện phép tính
A=\(\frac{5\cdot\left(2^2\cdot3^2\right)^9\cdot\left(2^2\right)^6-2\cdot\left(2^2\cdot3\right)^{14}\cdot3^4}{5\cdot2^{28}\cdot3^{18}-7\cdot2^{29}\cdot3^{18}}\)
A =\(\frac{5\cdot\left(2^2\cdot3^2\right)^9\cdot\left(2^2\right)^6-2\cdot\left(2^2\cdot3\right)^{14}}{5\cdot2^{28}\cdot3^{18}-7\cdot2^{29}\cdot3^{18}}\cdot3^4\)
Hãy thực hiện phép tính
Tính tổng :
a) \(A=\frac{5}{2\cdot1}+\frac{4}{1\cdot11}+\frac{3}{11\cdot14}+\frac{1}{14\cdot15}+\frac{13}{15\cdot28}\)
b) \(B=\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)
c) \(C=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
d) \(D=\frac{1}{1\cdot2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4\cdot5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
e) \(E=\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{37\cdot38\cdot39}\right)\cdot1482\cdot185\cdot8\)
Tính tổng :
a) \(A=\frac{5}{2\cdot1}+\frac{4}{1\cdot11}+\frac{3}{11\cdot14}+\frac{1}{14\cdot15}+\frac{13}{15\cdot28}\)
b) \(B=\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)
c) \(C=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
d) \(D=\frac{1}{1\cdot2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4\cdot5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
e) \(E=\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{37\cdot38\cdot39}\right)\cdot1482\cdot185\cdot8\)
\(A=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{A}{7}=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{A}{7}=\frac{7-2}{2.7}+\frac{11-7}{7.11}+\frac{14-11}{11.4}+\frac{15-14}{14.15}+\frac{28-15}{15.28}\)
\(\frac{A}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}=\frac{1}{2}-\frac{1}{28}=\frac{13}{28}\)
\(A=7.\frac{13}{28}\)
\(A=\frac{13}{4}\)
Bài 1:
a) \(\frac{1}{1}\cdot2+\frac{1}{2}\cdot3+\frac{1}{3}\cdot4+...+\frac{1}{n}\cdot\left(n+1\right)\)
b) \(\frac{1}{1}\cdot2\cdot3+\frac{1}{2}\cdot3\cdot4+\frac{1}{3}\cdot4\cdot5+...+\frac{1}{a}\cdot\left(a+1\right)\cdot\left(a+2\right)\)
Thực hiện phép tính:
\(\left(1-\frac{2}{2\cdot3}\right)\left(1-\frac{2}{3\cdot4}\right)\left(1-\frac{2}{4\cdot5}\right)...\left(1-\frac{2}{99\cdot100}\right)\)
Bài 3: Thực hiện phép tính a) \(\left(\frac{1}{25}-0,6\right)^{^2}:\frac{49}{125}+\left[\left(3\frac{1}{4}-6\frac{5}{9}\right)\cdot2\frac{2}{17}\right]\)
Bài 3 :
a) \(\left(\frac{1}{25}-0,6\right)^2:\frac{49}{125}+\left[\left(3\frac{1}{4}-6\frac{5}{9}\right)\cdot2\frac{2}{17}\right]\)
\(=\left(\frac{1}{25}-\frac{3}{5}\right)^2\cdot\frac{125}{49}+\left[\left(\frac{13}{4}-\frac{59}{9}\right)\cdot\frac{36}{17}\right]\)
\(=\left(-\frac{14}{25}\right)^2\cdot\frac{125}{49}+\left[\left(-\frac{119}{36}\right)\cdot\frac{36}{17}\right]\)
\(=-\frac{196}{625}\cdot\frac{125}{49}+\left(-7\right)=-\frac{4}{5}+\left(-7\right)=-\frac{39}{5}\)
Trả lời :
\(\left(\frac{1}{25}-0,6\right)^2\div\frac{49}{125}+\left[\left(3\frac{1}{4}-6\frac{5}{9}\right)\times2\frac{2}{17}\right]\)
\(=\left(\frac{1}{25}-\frac{3}{5}\right)^2\div\frac{49}{125}+\left[\frac{-119}{36}\times\frac{36}{17}\right]\)
\(=\left(\frac{-14}{25}\right)^2\div\frac{49}{125}-7\)
\(=\frac{4}{5}-7\)
\(=\frac{-31}{5}\)
\(\left(\frac{1}{25}-0,6\right)^2:\frac{49}{125}+\left[\left(3\frac{1}{4}-6\frac{5}{9}\right).2\frac{2}{17}\right]\)
\(=\left(-\frac{14}{25}\right)^2.\frac{125}{49}+\left[\left(\frac{13}{4}-\frac{59}{9}\right).\frac{36}{17}\right]\)
\(=\frac{196}{625}.\frac{125}{49}+\left(-\frac{119}{36}.\frac{36}{17}\right)\)
\(=\frac{4}{5}+\left(-7\right)\)
\(=-\frac{31}{5}\)
Tính tổng sau
\(S=\frac{3}{\left(1\cdot2\right)^2}+\frac{5}{\left(2\cdot3\right)^2}+...+\frac{4017}{\left(2008\cdot2009\right)^2}\)
Với \(n\ge1\)thì \(\frac{2n+1}{n^2\left(n+1\right)^2}=\frac{n^2+2n+1-n^2}{n^2\left(n+1\right)^2}=\frac{\left(n+1\right)^2-n^2}{n^2\left(n+1\right)^2}=\frac{\left(n+1\right)^2}{n^2\left(n+1\right)^2}-\frac{n^2}{n^2\left(n+1\right)^2}\)
Do đó \(S=\frac{3}{\left(1\cdot2\right)^2}+\frac{5}{\left(2\cdot3\right)^2}+...+\frac{4017}{\left(2008\cdot2009\right)^2}=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{1}{2008^2}-\frac{1}{2009^2}\)
\(=1-\frac{1}{2009^2}\)
sao bạn hôm đăng bài lớp 8 hôm thì đăng bài lớp 6 vậy