Tính
Q= 3/4.7 + 3/7.10 + . . . + 3/64.67
M= 22/1.3 + 22/3.5 + . . . + 22/101.103
Các bạn giúp mk nha
K
Khách
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TT
3
DP
22 tháng 7 2017
\(Q=\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{64.67}\)
\(Q=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{64}-\frac{1}{67}\)
\(Q=\frac{1}{4}-\frac{1}{67}=\frac{63}{268}\)
\(M=\frac{22}{1.3}+\frac{22}{3.5}+...+\frac{22}{101.103}\)
\(M=\frac{22}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{103}\right)\)
\(M=11\cdot\left(1-\frac{1}{103}\right)\)
\(M=11\cdot\frac{102}{103}=\frac{1122}{103}\)
MC
22 tháng 7 2017
\(Q=\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{64.67}\)
\(\Leftrightarrow Q=3\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{64}-\frac{1}{67}\right)\)
\(\Leftrightarrow Q=3\left(\frac{1}{4}-\frac{1}{67}\right)\)
\(\Leftrightarrow Q=3.\frac{63}{268}\)
\(\Leftrightarrow Q=\frac{189}{268}\)
Câu b) bạn làm tương tự nhé :)
27 tháng 4 2021
a)\(\dfrac{1}{2^2}<\dfrac{1}{1.2}\)
\(\dfrac{1}{3^3}<\dfrac{1}{2.3}\)
\(...\)
\(\dfrac{1}{8^2}<\dfrac{1}{7.8}\)
Vậy ta có biểu thức:
\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{8^2}<\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{7.8}\)
\(B= 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}\)
\(B<1-\dfrac{1}{8}=\dfrac{7}{8}<1\)
Vậy B < 1 (đpcm)
27 tháng 4 2021
Giải:
a) Ta có:
1/22=1/2.2 < 1/1.2
1/32=1/3.3 < 1/2.3
1/42=1/4.4 < 1/3.4
1/52=1/5.5 < 1/4.5
1/62=1/6.6 < 1/5.6
1/72=1/7.7 < 1/6.7
1/82=1/8.8 <1/7.8
⇒B<1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8
B<1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8
B<1/1-1/8
B<7/8
mà 7/8<1
⇒B<7/8<1
⇒B<1
b)S=3/1.4+3/4.7+3/7.10+...+3/40.43+3/43.46
S=1/1-1/4+1/4-1/7+1/7-1/10+...+1/40-1/43+1/43-1/46
S=1/1-1/46
S=45/46
Vì 45/46<1 nên S<1
Vậy S<1
Chúc bạn học tốt!
12 tháng 4 2023
b: 6B=2*4*6+4*6*6+6*8*6+...+46*48*6+48*50*6
=2*4*6-2*4*6+4*6*8-4*6*8+...-44*46*48+46*48*50-46*48*50+48*50*52
=48*50*52
=>B=20800
d: 9D=1*4*9+4*7*9+...+46*49*9
=1*4*2+1*4*7-1*4*7+1*7*10-1*7*10+...+46*49*52-46*49*43
=1*2*4+46*49*52
=117216
=>D=13024
a: 
TB
2
H
26 tháng 2 2023
\(\dfrac{2^2}{1\times3}\times\dfrac{3^2}{2.4}\times\dfrac{4^2}{3.5}\times\dfrac{5^2}{4.6}=\dfrac{2^2.3^2.4^2.5^2}{1.3.2.4.3.5.4.6}=\dfrac{2^2.3^2.4^2.5^2}{1.2.3.3.4.4.5.2.3}=\dfrac{2^2.3^2.4^2.5^2}{3^3.2^2.4^2.5.1}=\dfrac{5}{3.1}=\dfrac{5}{3}\)
26 tháng 2 2023
\(\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot\dfrac{5^2}{4.6}\\ =\dfrac{2^2\cdot3^2\cdot4^2\cdot5^2}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot4\cdot6}\\ =\dfrac{2^2\cdot3^2\cdot4^2\cdot5^2}{1\cdot2\cdot4^2\cdot4^2\cdot5\cdot6}\\ =\dfrac{2\cdot5}{6}=\dfrac{5}{3}\)
BN
19 tháng 3 2017
\(A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2015.2017}\)
\(A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\)
\(A=1-\dfrac{1}{2017}=\dfrac{2016}{2017}\)
\(B=\dfrac{3}{1.4}+\dfrac{3}{5.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}\)
\(B=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\)
\(B=1-\dfrac{1}{103}=\dfrac{102}{103}\)
\(C=\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{62.65}\)
\(3C=3\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{62.65}\right)\)
\(3C=\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{62.65}\)
\(3C=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{62}-\dfrac{1}{65}\)
\(3C=\dfrac{1}{2}-\dfrac{1}{65}\)
\(3C=\dfrac{63}{130}\)
\(C=\dfrac{63}{130}:3=\dfrac{21}{130}\)
31 tháng 1
\(A=1\cdot2+2\cdot3+...+151\cdot152\)
\(=1\left(1+1\right)+2\left(1+2\right)+...+151\left(1+151\right)\)
\(=\left(1+2+3+...+151\right)+\left(1^2+2^2+...+151^2\right)\)
\(=\dfrac{151\left(151+1\right)}{2}+\dfrac{151\left(151+1\right)\left(2\cdot151+1\right)}{6}\)
\(=151\cdot76+\dfrac{151\cdot152\cdot303}{6}\)
\(=151\cdot76+151\cdot7676=1170552\)
\(C=2\cdot4+4\cdot6+...+2024\cdot2026\)
\(=2\cdot2\left(1\cdot2+2\cdot3+...+1012\cdot1013\right)\)
\(=4\left[1\left(1+1\right)+2\left(1+2\right)+...+1012\left(1+1012\right)\right]\)
\(=4\left[\left(1+2+...+1012\right)+\left(1^2+2^2+...+1012^2\right)\right]\)
\(=4\left[1012\cdot\dfrac{1013}{2}+\dfrac{1012\left(1012+1\right)\left(2\cdot1012+1\right)}{6}\right]\)
\(=4\left[506\cdot1013+345990150\right]\)
\(=1386010912\)
\(M=1^2+2^2+...+2024^2\)
\(=\dfrac{2024\left(2024+1\right)\cdot\left(2\cdot2024+1\right)}{6}\)
\(=2024\cdot2025\cdot\dfrac{4049}{6}\)
=2765871900
\(N=1^3+2^3+...+100^3\)
\(=\left(1+2+3+...+100\right)^2\)
\(=\left[\dfrac{100\left(100+1\right)}{2}\right]^2\)
\(=\left[50\cdot101\right]^2=5050^2\)
\(Q=1^3+2^3+...+2024^3\)
\(=\left(1+2+3+...+2024\right)^2\)
\(=\left[\dfrac{2024\left(2024+1\right)}{2}\right]^2\)
\(=\left[1012\left(2024+1\right)\right]^2\)
\(=2049300^2\)
13 tháng 2 2015
a) 1/1 - 1/3 +1/3 - 1/5 +........+1/49 - 1/51
=1/1 - 1/51 (các số liền kề nhau cộng lại bằng 0)
=50/51
còn câu b bạn tự giải
nhớ thank mik nha!!!!!
14 tháng 2 2015
b,khoảng cách của nó là 3 mà tử của nó bằng 3 chứng tỏ nó là dạng đủ
1/1-1/4+1/4-1/7+...+1/97-1/100
1-1/100=99/100
Q=3/4.7+3/7.10+...+3/64+67
Q=1/4-1/7+1/7-1/10+.....+1/64-1/67
Q=1/4-1/67
Q=63/268
M=22/1.3+22/3.5+..+22/101.103
M=2.(2/1.3+2/3.5+...+2/101+103)
M=2.(1-1/3+1/3-1/5+1/5-......+1/101-1/103)
M=2.(1-1/103)
M=2.102/103
M=204/103
K MÌNH NHA