`\text{Tính}`
`a , \frac{3}{1.6} + \frac{3}{5.11} + ... + \frac{3}{26.31}`
`b , 2 + 4 + 6 + ... + 100`
`c , 7 + 7^2 + 7^3 + ... + 7^100`
1.So sánh A và B:
\(A=\frac{10^{11}-1}{10^{12}-1}\)
\(B=\frac{10^{10}+1}{10^{11}+1}\)
2.tính tổng
A=1+2+22+...+2100
B=3-32+33-3+...+399-3100
C1+52+54+56+...+5200
D=7-74+77+...+7301
E=\(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\)
F=\(0-\frac{4}{\text{5}}+\frac{4}{\text{5}^2}-\frac{4}{\text{5}^3}+...+\frac{4}{\text{5}^{200}}\)
Tính các tổng sau:
a) \(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}.\)
b) \(-\frac{4}{5}+\frac{4}{5^2}-\frac{4}{5^3}+...+\frac{4}{5^{200}}.\)
c)\(\frac{-1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{100}}-\frac{1}{3^{101}}\)
Đăt A = \(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+......+\frac{1}{7^{100}}\)
\(\Rightarrow7A=1+\frac{1}{7}+\frac{1}{7^2}+.....+\frac{1}{7^{100}}\)
\(\Rightarrow7A-A=1-\frac{1}{7^{100}}\)
\(\Rightarrow6A=1-\frac{1}{7^{100}}\)
\(\Rightarrow A=\frac{1-\frac{1}{7^{100}}}{6}\)
Tính
A = \(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\)
B = \(\frac{4}{5}+\frac{4}{5^2}-\frac{4}{5^3}+...+\frac{4}{5^{200}}\)
A=\(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\)
\(\Rightarrow7A=(1+\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{99}})-\left(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+....+\frac{1}{7^{100}}\right)\)
\(\Rightarrow6A=\left(1-\frac{1}{7^{99}}\right)\)
\(\Rightarrow A=\left(1-\frac{1}{7^{99}}\right):6\)
Câu b tương tự nha
a) \(A=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...........+\frac{1}{7^{100}}\)
\(\Rightarrow7A=1+\frac{1}{7}+\frac{1}{7^2}+.........+\frac{1}{7^{99}}\)
\(\Rightarrow7A-A=6A=1-\frac{1}{7^{100}}\)
\(\Rightarrow A=\frac{1-\frac{1}{7^{100}}}{6}\)
\(A=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\)
\(\frac{A}{7}=\frac{1}{7^2}+\frac{1}{7^3}+\frac{1}{7^4}+...+\frac{1}{7^{101}}\)
\(A-\frac{A}{7}=\left(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\right)-\left(\frac{1}{7^2}+\frac{1}{7^3}+\frac{1}{7^4}+...+\frac{1}{7^{101}}\right)\)
\(\frac{6}{7}A=\frac{1}{7}-\frac{1}{7^{101}}\)
\(A=\left(\frac{1}{7}-\frac{1}{7^{101}}\right).\frac{7}{6}\)
\(A=\frac{1}{6}-\frac{1}{6.7^{100}}\)
\(B=\frac{4}{5}+\frac{4}{5^2}-\frac{4}{5^3}+...+\frac{4}{5^{200}}\)
\(=4.\left(\frac{1}{5}+\frac{1}{5^2}-\frac{1}{5^3}+...+\frac{1}{5^{200}}\right)\)
Gọi \(C=\frac{1}{5}+\frac{1}{5^2}-\frac{1}{5^3}+...+\frac{1}{5^{200}}\)
\(\frac{C}{5}=\frac{1}{5^2}+\frac{1}{5^3}-\frac{1}{5^4}+...+\frac{1}{5^{201}}\)
\(C-\frac{5}{C}=\left(\frac{1}{5}+\frac{1}{5^2}-\frac{1}{5^3}+...+\frac{1}{5^{200}}\right)-\left(\frac{1}{5^2}+\frac{1}{5^3}-\frac{1}{5^4}+...+\frac{1}{5^{201}}\right)\)
\(\frac{4}{5}C=\frac{1}{5}-\frac{1}{5^{201}}\)
\(C=\left(\frac{1}{5}-\frac{1}{5^{201}}\right).\frac{5}{4}\)
\(=\frac{1}{4}-\frac{1}{4.5^{200}}\)
Thay vào B ta có
\(B=4.\left(\frac{1}{4}-\frac{1}{4.5^{200}}\right)\)
=\(=1-\frac{1}{5^{200}}\)
a ) 4\(\frac{5}{9}\)\(\div\)\((-\frac{5}{7})\)\(+\frac{4}{9}\div\frac{\left(-5\right)}{7}\)
b ) \((-\frac{3}{5}+\frac{4}{9})\div\frac{7}{11}+(\frac{-2}{5}+\frac{5}{9})\div\frac{7}{11}\)
c ) \(\frac{6}{7}\div(\frac{3}{26}-\frac{3}{13})+\frac{6}{7}\div(\frac{1}{10}-\frac{8}{5})\)
d ) \(A=(2+4+6+........+100)\orbr{\begin{cases}3\\5\end{cases}}\div0,7+3(\frac{-2}{7})]\div(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+......+\frac{1}{100})\)
Giải cụ thể hộ mk vs
\(a,4\frac{5}{9}:\frac{\left(-5\right)}{7}+\frac{4}{9}:\frac{-5}{7}\)
\(=\frac{41}{9}.\frac{-7}{5}+\frac{4}{9}.\frac{-7}{5}\)
\(=\frac{-7}{5}.\left(\frac{41}{9}+\frac{4}{9}\right)\)
\(=-\frac{7}{9}.5\)
\(=-7\)
a)Bn Kaito Kid làm rùi!
B)Không viết lại đề
\(=\frac{11}{7}\cdot\left(-\frac{3}{5}+\frac{4}{9}-\frac{2}{5}+\frac{5}{9}\right)=\frac{11}{7}\cdot0=0\)
c)Không viết lại đề
\(A=\left(2+4+...+100\right)\left(\frac{3}{5}\cdot\frac{10}{7}-\frac{6}{7}\right):\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(2+4+6+...+100\right)\cdot0\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)=0\)
\(=\frac{7}{6}\cdot\left(\frac{3}{26}-\frac{3}{13}+\frac{1}{10}-\frac{8}{5}\right)=\frac{7}{6}\left(\frac{-3}{26}+\frac{-17}{10}\right)=\frac{7}{6}\cdot\frac{236}{130}=\frac{413}{195}\)
D)
Dòng trên chữ D) là phần c)
Dưới phần c) là phần d) nha!(Lỗi đánh máy)
Cho M =\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\) .Hãy chứng minh M<\(\frac{3}{16}\)
Câu 2 Chứng minh rằng :
\(\frac{1}{7^2}-\frac{1}{7^4}+...+\frac{1}{7^{98}}-\frac{1}{7^{100}}< \frac{1}{50}\)
Tham khảo nha bạn :
Câu hỏi của Trần Minh Hưng - Toán lớp | Học trực tuyến
Tính các tổng sau
\(A=7-7^4+7^7-...+7^{301}\)
\(B=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\)
\(C=\frac{4}{5}+\frac{4}{5^2}-\frac{4}{5^3}+...+\frac{4}{5^{200}}\)
AI TÍNH ĐƯỢC MÌNH CHO 10 TICK NHA
thực hiên các phép tính tính :
a) \(\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}}{\left(\frac{1}{14}+\frac{1}{7}-\frac{-3}{35}\right).\frac{-4}{3}}\)
b) \(\frac{\left(1+2+3+...+100\right).\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(6,3.12-21.3,6\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
a, Tính : \(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{10}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
b, Tính : \(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
c, Tính : \(\frac{\left(1+2+3+...+99+100\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}\)
1 Tính
A = 1+52+54+....................+5200 , B = 3-32+33-34+.....................+399-3100
C = \(\frac{1}{7}+\frac{1}{7^2}+.................+\frac{1}{7^{100}}\)
D = \(-\frac{4}{5}+\frac{4}{5^2}-\frac{4}{5^3}+..................+\frac{4}{5^{200}}\)
E = 63+64+..............................+6150