A=1.2+2.3+3.4+...+100.101.
các bạn giúp mình nhé :=)
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A = 1.2 + 2.3 + 3.4 + ... + 100.101
3A = 1.2.(3-0) + 2.3.(4-1) + 3.4.(5-2) + ... + 100.101.(102-99)
3A = 1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 100.101.102 - 99.100.101
3A = (1.2.3 + 2.3.4 + 3.4.5 + ... + 100.101.102) - (0.1.2 + 1.2.3 + 2.3.4 + ... + 99.100.101)
3A = 100.101.102 - 0.1.2
3A = 100.101.102
A = 100.101.34
A = 343400
3A=1.2.3+2.3.3+3.4.3+...+19.20.3
3A=1.2.3+2.3.(4-1)+3.4.(5-2)+...+19.20.(21-18)
3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+19.20.21-18.19.20
3A=19.20.21
=> \(A=\frac{19.20.21}{3}=2660\)
mk dùng cách của lớp 8 nha bạn ;
ta có công thức xích ma như sau x(x+1)
nhập vào xích ma ta có kết quả 2660
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{2019.2020}\)
\(=9\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\right)\)
\(=9\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\right)\)
\(=9\left(1-\frac{1}{2020}\right)\)
\(=9.\frac{2019}{2020}\)
\(=\frac{18171}{2020}\)
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{2019.2020}\)
\(A=9.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\right)\)
\(A=9\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\right)\)
\(A=9\left(1-\frac{1}{2020}\right)=\frac{9.2019}{2020}=\frac{18171}{2020}\)
...
1.Tính
A= (1-1/22).(1-1/32)...(1-1/1002)
B= -1/1.2-1/2.3-1/3.4-...-1/100.101
C= 1.2+2.3+3.4+...+100.101
Lời giải :
Đặt S=1.2+2.3+3.4+4.5+…+99.100+100.101
3S=1.2.3+2.3.3+3.4.3+4.5.3+…+99.100.3+100.101.3
=1.2(3−0)+2.3(4−1)+3.4(5−2)+4.5(6−3)+…+99.100(101−98)+100.101(102−99)
=0.1.2-1.2.3+1.2.3-2.3.4+...+99.100.101-100.101.102
=100.101.102
S=100.101.34=343400
1.Tính
a) Ta có:
A=(1-1/22).(1-1/32)...(1-1/1002)
=>A=3/22.8/32.....9999/1002
=>A=(1.3/2.2).(2.4/3.3).....(99.101/100.100)
=>A=(1.2.3.....99/2.3.4.....100).(3.4.5.....101/2.3.4.....100)
=>A=1/100.101/2
=>A=101/200
b) Ta có:
B=-1/1.2-1/2.3-1/3.4-...-1/100.101
=>B=-(1/1.2+1/2.3+1/3.4+...+1/100.101)
=>B=-(1-1/2+1/2-1/3+1/3-1/4+...+1/100-1/101)
=>B=-(1-1/101)
=>B=-100/101
c) Ta có:
C=1.2+2.3+3.4+...+100.101
=>3C=1.2.3+2.3.3+3.4.3+...+100.101.3
=>3C=1.2.3+2.3.(4-1)+3.4.(5-2)+...+100.101.(102-99)
=>3C=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-3.4.5+...+100.101.102
=>3C=100.101.102
=>3C=1030200
=>C=343400
Chúc bạn hok tốt nhé >:)!!!!!
A = 1.2 + 2.3 + 3.4 + ...... + 100.101
3A = 1.2.3 + 2.3.3 + 3.4.3 + ...... + 100.101.3
3A = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ..... + 100.101.(102 - 99)
3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ...... + 100.101.102 - 99.100.101
3A = 100.101.102
A = 100.101.34
A = 343400
A = 1.2 + 2.3 + 3.4 + ... + 99.100 + 100.101
⇒ 3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 99.100.3 + 100.101.3
= 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 99.100.(101 - 98) + 100.101.(102 - 99)
= 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + 3.4.5 + ... - 98.99.100 + 99.100.101 - 99.100.101 + 100.101.102
= 100.101.102
= 1030200
⇒ A = 1030200 : 3
= 343400
Ta có: \(A=1.2+2.3+3.4+...+100.101\)
\(\Rightarrow3A=1.2.3+2.3.3+3.4.3+...+100.101.3\)
\(\Rightarrow3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+100.101\left(102-99\right)\)
\(\Rightarrow3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+100.101.102-99.100.101\)
\(\Rightarrow3A=\left(1.2.3-1.2.3\right)+...+\left(99.100.101-99.100.101\right)+100.101.102\)
\(\Rightarrow3A=\) \(100.101.102\)
\(\Rightarrow A=\dfrac{100.101.102}{3}=343400\)
Vậy \(A=343400.\)
A = 1.2+2.3+3.4+.....+100.101
3A = 1.2.3+2.3.4+3.4.3+...........+99.100.3
3A= 1.2.(3-0)+ 2.3.(4-1)+ 3.4.(5-2)....... . 99.100.(101-98)
3A=(1.2.3+2.3.4+3.4.5+......+99.100.101)-(0.1.2 + 1.2.3 + 2.3.4 +........+98.99.100)
3A = 99.100.101 - 0.1.2
3A = 999900 - 0
3A = 999900
=> A = 999900 : 3
=> A = 333300
A = 1.2 + 2.3 + 3.4 + 4.5 + ... + 99.100 + 100.101
3.A = 1.2.3 + 2.3.3 +3.4.3 + ... + 100.101.3
3A= 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 100.101.(102 - 99)
3A = 1.2.3 + 2.3.4 - 1.2.3 + 2.3.4 -3.4.5 + ... +99.100.101 -100.101.102
3A = 99.100.101
A = 99.100.101 : 3
A = 33.100.101
Vậy A = 33. 100 .101 (Tự tính)
\(A = 1 . 2 + 2 . 3 + 3 . 4 + ... + 100 . 101\)
\(3A=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+...+100\cdot101\cdot\left(102-99\right)\)
\(3A=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+...+100\cdot101\cdot102-99\cdot100\cdot101\)
\(3A=100\cdot101\cdot102\)
\(3A=1030200\)
\(A=1030200\text{ : }3\)
\(A=343400\)
\(A = 1 . 2 + 2 . 3 + 3 . 4 + ... + 100 . 101\)
\(3A=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+...+100\cdot101\cdot\left(102-99\right)\)
\(3A=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+...+100\cdot101\cdot102-99\cdot100\cdot101\)
\(3A=100\cdot101\cdot102\)
\(3A=1030200\)
\(A=1030200\text{ : }3\)
\(A=343400\)