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Tính:
(-2)2.3 -(110+8):(-3)2
=4.3-(1+8):9
=12-9:9
=12-1
=11
A>1
B<1
bvaif này dễ lần sau sẽ có bài khó hơn là nó ko CMR đc a lớn hơn hay bé hơn 1
\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\)
Ta có:
\(\frac{1}{2^2}< \frac{1}{1\times2}\)
\(\frac{1}{3^2}< \frac{1}{2\times3}\)
\(\frac{1}{4^2}< \frac{1}{3\times4}\)
\(...\)
\(\frac{1}{10^2}< \frac{1}{9\times10}\)
\(\rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}< \frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{9\times10}\)
\(\Rightarrow S< \frac{9}{10}\)mà \(S>0\Rightarrow\left[S\right]=0\)
1.S1=1 - 2 + 3 - 4 + ... + 1997 - 1998 + 1999
= (1 - 2) + ...+(1997 - 1998) + 1999
= -1 + -1 + ...+-1 + 1999
SH:1998 : 2
= 999 . -1
= -999
TDS:-999 + 1999
= 1000
b.S2=1 - 4 + 7 - 10 + ...- 2998+3001
= (1 - 4) + (7 - 10) + ...+ (2995 - 2998) + 3001
= -3 + -3 + ...+-3 + 3001
= (2998 - 1) : 3 + 1
= 1000 . -3
= -3000 + 3001
= 1
câu b mình làm lộn :
S2=1000 : 2
= 500 . -3
=-1500 + 3001
= 1501
KẾT QUẢ RA 1501 NHA
\(\frac{1}{3}+....+\frac{2}{x.\left(x+1\right)}=\frac{1999}{2001}\)
=>\(\frac{1}{2}.\left(\frac{1}{3}+...+\frac{2}{x.\left(x+1\right)}\right)=\frac{1999}{2001}.\frac{1}{2}\)
\(\Rightarrow\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x.\left(x+1\right)}=\frac{1999}{4002}\)
\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}=\frac{1999}{4002}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{4002}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2001}\)
=> x=2000
Tìm stn biết: 1/3 + 1/6 + 1/10 + ...+2/x(x+1)=1999/2001
Bài giải: Gọi x là số tự nhiên cần tìm
Cho S= 1/3 + 1/6 +1/10 +...+ 1/x(x+1)
\(\Rightarrow\)S= 2/6 + 2/12+ 2/20 +...+ 2/2[x(x+1)]
\(\Rightarrow\)1/2S= 1/2.3 + 1/3.4 + 1/ 4.5 +...+1/2[x(x+1)]
\(\Rightarrow\)1/2S=1/2-1/3+1/3-1/4+...+1/(x-1) .(x+1)
\(\Leftrightarrow\)1/2S=1/2-1/x+1
Vì S = 1999 / 2001\(\Rightarrow\)1/2S=1/2-1 . (x+1)=1999/2001-1998-2001=1/2001
\(\Rightarrow\)1/x+1=1/2001
\(\Leftrightarrow\)x+1=2001
x =2001-1 =2000
Vậy số tự nhiên đó là: 2000
\(1^3=0+1\)
\(2^3=2+1.2.3\)
.......................................
\(10^3=10+9.10.11\)
\(=1+2+1.2.3+3+2.3.4+....+10+9.10.11\)
\(=\left(1+2+....+10\right)+\left(1.2.3+....+9.10.11\right)\)
Đặt (1 + 2 + ... + 10) là A ; (1.2.3 + 2 .3.4 + .... + 9.10.11) là B . Ta có :
\(A=\left(1+2+...+10\right)\)
\(A=\frac{\left(10+1\right).10}{2}\)
\(A=55\)
\(B=1.2.3+2.3.4+....+9.10.11\)
\(4B=1.2.3.4+2.3.4.4+...+9.10.11.4\)
\(4B=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)+....+9.10.11.\left(12-8\right)\)
\(4B=1.2.3.4+2.3.4.5+....+9.10.11.12\)
\(4B=9.10.11.12=11880\)
\(\Rightarrow B=\frac{11880}{4}=2970\)
\(\Rightarrow1^3+2^3+....+10^3=A+B=55+2970=3025\)
Đặt \(S_1\) với \(S_2\) cho dễ ha ~.~
\(S_1=1+2+2^2+...+2^{10}\)
\(2S_1=2+2^2+2^3+...+2^{11}\)
\(2S_1-S_1=\left(2+2^2+2^3+...+2^{11}\right)-\left(1+2+2^2+...+2^{10}\right)\)
\(S_1=2^{11}-1\)
\(S_2=1+10+10^2+...+10^{10}\)
\(10S_2=10+10^2+10^3+...+10^{11}\)
\(10S_2-S_2=\left(10+10^2+10^3+...+10^{11}\right)-\left(1+10+10^2+...+10^{10}\right)\)
\(9S_2=10^{11}-1\)
\(S_2=\frac{10^{11}-1}{9}\)
Chúc bạn học tốt ~
\(S=1+2+2^2+....+2^{10}\)
\(\Rightarrow2S=2+2^2+2^3+...+2^{11}\)
\(\Rightarrow2S-S=\left(2+2^2+2^3+...+2^{11}\right)-\left(1+2+2^2+...+2^{10}\right)\)
\(\Rightarrow S=2^{11}-1\)
\(b,S=1+10+10^2+....+10^{10}\)
\(\Rightarrow10S=10+10^2+10^3+...+10^{11}\)
\(\Rightarrow10S-S=\left(10+10^2+10^3+...+10^{11}\right)-\left(1+10+10^2+...+10^{10}\right)\)
\(\Rightarrow9S=10^{11}-1\Rightarrow S=\frac{10^{11}-1}{9}\)