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a,=16.(64+17)+81.84
=16.81+81.84
=81(16+84)
=81.100=8100
b,=32.19+32
=32.(19+1)
=32.20=640
c,=12.19+12.37+44.12
=12.(19+37+44)
=12.100=1200
d, Khoảng cách là:
5-1=4;9-5=4
Số số hạng là:
(81-1):4+1=21(số)
Tổng dãy số là:
(81+1).21:2=861
a
\(\text{=16.(64+17)+81.84}\)
\(\text{=16.81+81.84}\)
\(\text{ =81.(16+84)}\)
\(\text{=81.100=8100}\)
b
\(\text{=32.19+32}\)
\(\text{ =32.(19+1)}\)
\(\text{ =32.20=640}\)
c
\(\text{=12.19+12.37+44.12}\)
\(\text{ =12.(19+37+44)}\)
\(\text{ =12.100}\)
\(=1200\)
d
Có tất cả số hạng là
\(\text{( 81 - 1 ) : 4 + 1 = 21 (số )}\)
Tổng là
\(\text{( 81 + 1 ) x 21 : 2 = 861}\)
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\(a;\left(9^3\right)^3.\left(27^4\right)^5:81^5.3\)
\(=9^9.3.\left(\frac{27^4}{81}\right)^5=3^{18}.\left(\frac{3^{12}}{3^4}\right)^5=3^{18}.\left(3^8\right)^5=3^{18}.3^{40}=3^{58}\)
\(b;2^{11^2}:16^5.\left(4^3.8^2\right)^5=2^{22}:2^{20}.\left(2^6.2^6\right)^5\)
\(=2^2.\left(2^{12}\right)^5=2^2.2^{60}=2^{62}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{81}+\frac{1}{100}\)
\(=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}+\frac{1}{10^2}\)
\(\Rightarrow A>\frac{1}{2^2}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}+\frac{1}{10.11}\)
\(\Rightarrow A>\frac{1}{4}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(\Rightarrow A>\frac{1}{4}+\frac{1}{3}-\frac{1}{11}\)
\(\Rightarrow A>\frac{65}{132}\left(đpcm\right)\)
Chúc bạn học tốt !!!!
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{81}+\frac{1}{100}\)
\(\Leftrightarrow A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}+\frac{1}{10^2}\)
\(\Leftrightarrow A>\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}\)
\(\Leftrightarrow A>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(\Leftrightarrow A>\frac{1}{2}-\frac{1}{11}\)
\(\Leftrightarrow A>\frac{9}{22}\)
Ta lại có:
\(\frac{9}{22}=\frac{9.11}{22\cdot11}=\frac{99}{132}\)
Ta thấy: 99>65
\(\Rightarrow\frac{99}{132}>\frac{65}{132}\)
\(\Rightarrow A>\frac{65}{132}\)
Vậy \(A>\frac{65}{132}\left(đpcm\right)\)
\(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{81}+\frac{1}{100}\)
\(A=\frac{1}{4}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}+\frac{1}{10^2}\)
\(A>\frac{1}{4}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}+\frac{1}{10.11}\)
\(A>\frac{1}{4}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)
\(A>\frac{1}{4}+\frac{1}{3}-\frac{1}{11}\)
\(A>\frac{33}{132}+\frac{44}{132}-\frac{12}{132}\)
\(A>\frac{65}{132}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=1/2*2+1/3*3+1/4*4+...+1/10*10.
A>1/1*2+1/2*3+1/3*4+...+1/9*10.
A>1-1/2+1/2-1/3+...+1/9-1/10.
A>1-1/10.
A>9/10.
=>A>1/2.
Mà 1/2=66/132>65/132.
=>A>65/132.
Vậy A>65/132.
A=1/2^2+1/3^2+1/4^2+......+1/9^2+1/10^2
=1/4+1/3×3+1/4×4+.....+1/9×9+1/10×10
=>A>1/4+(1/3×4+1/4×5+...+1/9×10+1/10×11)
=>A>1/4+(1/3-1/11)
=>A>1/4+8/33
=>A>65/132( đpcm)
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1 =1x1
4=2x2
9=3x3
.........
81 = 9x9
Dãy số viết đủ là: 1,4,9,16,25,36,49,64,81 =(1 + 9) +(4+16) +(36+64) +(49+81) +25 =285
TA có dãy số 1+4+9+16+25+36+49+64+81
=(81+9)+(64+16)+(49+1)+(4+25+36)
= 90+80+50+65
=285