Viết các biểu thức sau dưới dạng \(a^n\)(a thuộc Q, n thuộc N)
a) \(9\times3^3\times\frac{1}{81}\times3^2\)
b) \(4\times2^5\div\left(2^3\times\frac{1}{16}\right)\)
c) \(3^2\times2^5\times\left(\frac{2}{3}\right)^2\)
d) \(\left(\frac{1}{3}\right)^2\times\frac{1}{9}\times9^2\)
Chứng minh \(1\times2\times3\times...\times2012\times\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}\right)\)là một số tự nhiên chia hết cho 2013.
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Tính C=\(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+....+\frac{1}{n\times\left(n+1\right)\times\left(n+2\right)}\)
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Tính giá trị biểu thức: A=\(\frac{\left(1+17\right)\times\left(1+\frac{17}{2}\right)\times\left(1+\frac{17}{3}\right)....\left(1+\frac{17}{19}\right)}{\left(1+19\right)\times\left(1+\frac{19}{2}\right)\times\left(1+\frac{19}{3}\right)....\left(1+\frac{19}{17}\right)}\)
Thực hiện phép tính:
a)\(1\times2\times3\times......\times9-1\times2\times3\times......\times8-1\times2\times3\times.....\times8\times8\)
b)\(B=\frac{\left(3\times4\times2^{16}\right)^2}{11\times2^{13}\times4^{11}-16^9}\)
c)\(C=70\times\left(\frac{131313}{565656}+\frac{131313}{727272}+\frac{131313}{909090}\right)\)
d)\(B=\frac{1}{4\times9}+\frac{1}{9\times14}+\frac{1}{14\times19}+...+\frac{1}{64\times69}\)
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a)1.2.3.4...9-1.2.3.4...8-1.2.3.4...8.8
=1.2.3.4...8(9-1-8)
=1.2.3.4...8.0
=0
b)(3.4.216)2/11.123.411-169=(3.22.216)2/11.213.222-236=32.24.232/11.235-236=32.226/235.(11-2)
=32.236/235.9=32.236/235.32=2
c)70.(131313/565656+131313/727272+131313/909090
=70.(13/56+13/72+13/90)
=70.39/70=39
d)1/4.9+1/9.14+1/14.19+...+1/64.69
=4/4.9.4+4/9.4.14+4/14.19.4+...+4/64.69.4.
=1/4.(4/4.9+4/9.14+4/14.19+...+4/64.69)
=1/4.(1/4-1/9+1/9-1/14+1/14-1/19+...+1/64-1/69)
=1/4.(1/4-1/69)
=1/4.65/276=65/1104
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d)\(\frac{1}{3}\times\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{64}-\frac{1}{69}\right)\)
\(\frac{1}{3}\times\left(\frac{1}{4}-\frac{1}{69}\right)\)
\(\frac{1}{3}\times\frac{65}{276}\)
Chứng minh rằng số tự nhiên \(A=1\times2\times3\times...\times69\times70\times\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{70}\right)⋮71\)
Ta có:
\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{70}=\left[1+\frac{1}{70}\right]+\left[\frac{1}{2}+\frac{1}{69}\right]+\left[\frac{1}{3}+\frac{1}{68}\right]+...+\left[\frac{1}{35}+\frac{1}{36}\right]\)
\(=\frac{71}{1.70}+\frac{71}{2.69}+\frac{71}{3.68}+...+\frac{71}{35.36}\)
\(=71\left[\frac{1}{1.70}+\frac{1}{2.69}+\frac{1}{3.68}+...+\frac{1}{35.36}\right]⋮71\)
=> \(A=1\times2\times3\times4\times...\times70\times\left[1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{70}\right]⋮71\)=> ĐPCM
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Xét \(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{70}=\left(1+\frac{1}{70}\right)+\left(\frac{1}{2}+\frac{1}{69}\right)+...+\left(\frac{1}{35}+\frac{1}{36}\right)\)
\(=\frac{71}{1.70}+\frac{71}{2.69}+...+\frac{71}{35.36}=71\left(\frac{1}{1.70}+\frac{1}{2.69}+...+\frac{1}{35.36}\right)\)
=>\(A=1.2.3.4...71.\left(\frac{1}{1.70}+\frac{1}{2.69}+...+\frac{1}{35.36}\right)⋮71\)
Vậy A chia hết cho 71
\(A=\left[1-\left(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+......+\frac{1}{98\times99\times100}\right)\right]\times\frac{14851}{19800}\)
Tính giá trị biểu thức sau một cách hợp lí:
\(A=\left(1+\frac{1}{1+2}\right)\times\left(1+\frac{1}{1+2+3}\right)\times...\times\left(1+\frac{1}{1+2+...+997}\right)\)
Tính giá trị của biểu thức
A =\(\left(\frac{1}{2}+1\right)\times\left(\frac{1}{3}+1\right)\times\left(\frac{1}{4}+1\right)\times....\times\left(\frac{1}{99}+1\right)\)
Chứng tỏ
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{97.99}>32\%\)
A =(1/2 +1)×(1/3 +1)×(1/4 +1)×....×(1/99 +1)
=3/2x4/3x...............x100/99
=2-1/99
=197/99
A= \(\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot.....\cdot\frac{100}{99}\)
A=\(\frac{\left(3\cdot4\cdot5\cdot....\cdot99\right)\cdot100}{2\cdot\left(3\cdot4\cdot5\cdot...\cdot99\right)}\)
A=\(\frac{100}{2}=50\)
\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\)
\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
=> \(\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)>\(\frac{32}{100}\)=32%
Câu đầu tiên:
\(A=\left(\frac{1}{2}+1\right)\cdot\left(\frac{1}{3}+1\right)\cdot...\cdot\left(\frac{1}{99}+1\right)\)
\(A=\frac{3}{2}\cdot\frac{4}{3}\cdot...\cdot\frac{100}{99}=\frac{3\cdot4\cdot5\cdot...\cdot99\cdot100}{3\cdot4\cdot5\cdot...\cdot99\cdot2}=\frac{100}{2}=50\)
Câu thứ 2:
\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}>\frac{32}{100}\)
