Tính hợp lý (1-\(\frac{1}{2}\))(1-\(\frac{1}{3}\))(1-\(\frac{1}{4}\))...(1-\(\frac{1}{100}\))
Thực hiện phép tính bằng cách hợp lý nhất
\(\left(\frac{1}{125}-\frac{1}{2^3}\right).\left(\frac{1}{125}-\frac{1}{3^3}\right)...\left(\frac{1}{125}-\frac{1}{100^3}\right)\)
trong tích trên có 1 thừa số như thế này:
\(\left(\frac{1}{125}-\frac{1}{5^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{125}\right)\)
=0
=> tích trên bằng 0
BÀI 1:TÍNH:
\(B=1+\frac{3}{2^3}+\frac{4}{2^4}+\frac{5}{2^5}+....+\frac{100}{2^{100}}\)
BÀI 2: CHỨNG MINH RẰNG:
\(B=1-\frac{1}{2^2}-\frac{1}{3^2}-.....-\frac{1}{2004^2}>\frac{1}{2004}\)
BÀI 3:THỰC HIỆN PHÉP TÍNH BẰNG CÁCH HỢP LÝ:
\(B=\frac{1}{3}+\frac{1}{6}.\left(1+2\right)+\frac{1}{9}.\left(1+2+3\right)+.....+\frac{1}{6045}.\left(1+2+3+....+2015\right)\)
Tính bằng cách hợp lý nhất:
\(C=\frac{\frac{3}{101}+0,75-\frac{3}{100}}{\frac{4}{101}+1-\frac{1}{25}}\)
Tính hợp lý các tổng và tích sau:
1) \(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
2) \(\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)
3) \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{300}}\)
\(\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+....+\frac{10}{1400}\)
\(=\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+.....+\frac{5}{25.28}\)
\(=\frac{5}{3}.\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+....+\frac{3}{25.28}\right)\)
\(=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.....+\frac{1}{25}-\frac{1}{28}\right)\)
\(=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{28}\right)\)
\(=\frac{5}{3}.\frac{3}{14}=\frac{5}{14}\)
a)Đặt A=Tổng trên, ta có:
\(2A=2\left(1+\frac{1}{2}+...+\frac{1}{2^{100}}\right)\)
\(2A=2+1+...+\frac{1}{2^{99}}\)
\(2A-A=\left(2+1+...+\frac{1}{2^{99}}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^{100}}\right)\)
\(A=2-\frac{1}{2^{100}}\)
b)có đứa làm rồi
c)Đặt C=Tổng trên
\(3C=3\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{300}}\right)\)
\(3C=1+\frac{1}{3}+...+\frac{1}{3^{299}}\)
\(3C-C=\left(1+\frac{1}{3}+...+\frac{1}{3^{299}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{300}}\right)\)
\(2C=1-\frac{1}{3^{300}}\)
\(C=\frac{1-\frac{1}{3^{300}}}{2}\)
tính hợp lý:
a) \(\left(\frac{9}{10}-\frac{15}{16}\right)\left(\frac{5}{12}-\frac{11}{15}-\frac{7}{20}\right)\)
b) \(\left(\frac{1}{2}-1\right)\left(\frac{1}{3-1}\right)\left(\frac{1}{4}-1\right).......\left(\frac{1}{100}-1\right)\)
b)\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{100}-1\right)=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}....\frac{-99}{100}=\frac{-1.\left(-2\right).\left(-3\right)...\left(-99\right)}{2.3.4...100}=-\frac{1}{100}\)
Tính hợp lý
a, \(\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3}-1\right).\left(\frac{1}{4}-1\right)............\left(\frac{1}{100^2}-1\right)\)
b, \(1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{5}}}}}\)
làm được câu nào thì giúp mình nhé , nói cả cách làm nữa nhé ^_^
Bài 1 : Thực hiện phép tính hợp lý (nếu có thể)
\(b.\frac{1}{125}-\frac{17}{18}-\frac{5}{7}+\frac{4}{9}+\frac{17}{14}\)
\(c.1-\frac{1}{2}+2-\frac{2}{3}+3-\frac{3}{4}+4-\frac{1}{4}-3-\frac{1}{3}-2-\frac{1}{2}-1\)
Tính hợp lý \(\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{11}}{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}+\frac{\frac{1}{4}-\frac{1}{5}+\frac{1}{7}}{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}}\)
\(\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{11}}{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}+\frac{\frac{1}{4}-\frac{1}{5}+\frac{1}{7}}{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}}\)
\(=\frac{2\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}{3\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}+\frac{1\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}\right)}{3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}\right)}\)
\(=\frac{2}{3}+\frac{1}{3}\)
\(=1\)
Tính hợp lý
\(\frac{\frac{5}{3}+\frac{5}{8}-\frac{5}{7}}{\frac{-4}{3}-\frac{-4}{8}+\frac{4}{7}}:\frac{\frac{2}{3}-\frac{1}{6}+\frac{6}{7}}{\frac{-1}{3}+\frac{1}{6}-\frac{1}{7}}\)
\(=\frac{5\left(\frac{1}{3}+\frac{1}{8}-\frac{1}{7}\right)}{-4\left(\frac{1}{3}+\frac{1}{8}-\frac{1}{7}\right)}:\frac{2\left(\frac{1}{3}-\frac{1}{12}+\frac{3}{7}\right)}{ }\)
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