Tính:1/1*2*3+1/2*3*4+1/3*4*5+...+1/98*99*100
Những câu hỏi liên quan
tính nhanh (2/3+3/4+5/6+...+99/100).(1/2+2/3+3/4+...+98/99)-(1/2+1/3+...+99/100).(2/3+2/4+...+98/99)
Bài 4: Tính tổng 1) 1 + (-2) + 3 + (-4) + . . . + 19 + (-20) 2) 1 – 2 + 3 – 4 + . . . + 99 – 100 3) 2 – 4 + 6 – 8 + . . . + 48 – 50 4) – 1 + 3 – 5 + 7 - . . . . + 97 – 99 5) 1 + 2 – 3 – 4 + ... + 97 + 98 – 99 - 100
1. 1 + ( -2) +3 +(-4) + .........+ 19 + (-20)
= -1 + ( -1) +....+(-1)
= -1. 10
= -10
2. 1 – 2 + 3 – 4 + . . . + 99 – 100
= ( -1) + (-1) +....+(-1)
= -1. 50
= -50
3. 2 – 4 + 6 – 8 + . . . + 48 – 50
= (-2) + (-2) +....+ (-2)
= -2. 12 + 26
= -24 + 26
= 2
4. – 1 + 3 – 5 + 7 - . . . . + 97 – 99
= 2 + 2 +......+2
= 2.25
= 50
5. 1 + 2 – 3 – 4 + ... + 97 + 98 – 99 - 100
= (1+2-3-4) +......+ ( 97+98-99 -100)
= -4 . (-4).....(-4)
= -4. 25
= -100
Tính A:B
a)A=98+1/2+1/3+1/4+...+1/99
B=2/3+4/3+5/4+...+100/99
b)A=2018+1/2+1/3+1/4+...+1/2019
B=3/2+4/3+3/4+...+2020/2019
c)A=99/1+98/2+97/3+...+2/98+1/99
B=1/2+1/3+1/4+...+1/100
Giải đầy đủ
a) \(A=98+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\)(có 98 phân số nên ta cộng 1 vào mỗi phân số)
\(A=\left(\frac{1}{2}+1\right)+\left(\frac{1}{3}+1\right)+...+\left(\frac{1}{99}+1\right)\)
\(A=\frac{3}{2}+\frac{4}{3}+...+\frac{100}{99}\)
Và \(B=\frac{3}{2}+\frac{4}{3}+...+\frac{100}{99}\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{3}{2}+\frac{4}{3}+...+\frac{100}{99}}{\frac{3}{2}+\frac{4}{3}+...+\frac{100}{99}}=1\)
b) \(A=2018+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\)(có 2018 phân số nên ta cộng 1 vào mỗi phân số)
\(A=\left(\frac{1}{2}+1\right)+\left(\frac{1}{3}+1\right)+...+\left(\frac{1}{2019}+1\right)\)
\(A=\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}\)
Và \(B=\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}}{\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}}=1\)
c) \(A=\frac{99}{1}+\frac{98}{2}+...+\frac{1}{99}\)
\(A=99+\frac{98}{2}+...+\frac{1}{99}\)(có 98 phân số nên ta cộng 1 vào từng phân số)
\(A=\left(\frac{98}{2}+1\right)+\left(\frac{97}{3}+1\right)+...+\left(\frac{1}{99}+1\right)+1\)
\(A=\frac{100}{2}+\frac{100}{3}+...+\frac{100}{99}+1\)
\(A=100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)\)
Và \(B=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\)
\(\Rightarrow\frac{A}{B}=\frac{100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}}=100\)
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a)\(B=\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+...+\frac{100}{99}\)
\(B=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{4}\right)+...+\left(1+\frac{1}{99}\right)\)
\(\Rightarrow B=\left(1+1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}\right)\)
\(\Rightarrow B=98+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}\)
\(\Rightarrow A:B=\frac{A}{B}=\frac{98+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}}{98+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}}=1.\)
Vậy \(A:B=1.\)
b)\(B=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{4}\right)+...+\left(1+\frac{1}{2019}\right)\)
\(\Rightarrow B=\left(1+1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right)\)
\(\Rightarrow B=2018+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\)
\(\Rightarrow A:B=\frac{A}{B}=\frac{2018+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}}{2018+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}}=1.\)
Vậy \(A:B=1.\)
c)\(A=\left(1+1+...+1\right)+\frac{98}{2}+\frac{97}{3}+...+\frac{2}{98}+\frac{1}{99}\)
\(A=\left(1+\frac{98}{2}\right)+\left(1+\frac{97}{3}\right)+...+\left(1+\frac{2}{98}\right)+\left(1+\frac{1}{99}\right)\)
\(A=\frac{100}{2}+\frac{100}{3}+...+\frac{100}{98}+\frac{100}{99}\)
\(A=100\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{98}+\frac{1}{99}\right)\)
\(\Rightarrow A:B=\frac{A}{B}=\frac{100\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{98}+\frac{1}{99}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{98}+\frac{1}{99}}=1.\)
Vậy \(A:B=1.\)
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S=1×2+2×3+3×4+4×5+...........+99×100
3S=1×2×3+2×3×(4-1)+3×4×(5-2)+4×5×(6-3)+............+99×100×(101-98)
3S=1×2×3+2×3×4-1×2×3+3×4×5-2×3×4+4×5×6-3×4×5+.............+99×100×101-98×99×100
3S=99×100×101
Tại sao 3S=99×100×101
Các bạn giải thích hộ mình với!
MÌNH CẢM ƠN MỌI NGƯỜI!
TÍNH TỔNG:
1. 1+(-2)+3+(-4)+...+19+(-20)
2. 1-2+3-4+...+99-100
3. 2-4+6-8+...+48-50
4. -1+3-5+7-...+97-99
5. 1+2-3-4+...+97+98-99-100
bài 1
A=1*2*3+2*3*4+3*4*5+...+99*100*101
B=1*3*5+3*5*7+...+95*97*99
C=2*4+4*6+..+98*100
D=1*2+3*4+5*6+...+99*100
E=1^2+2^2+3^2+...+100^2
G=1*3+2*4+3*5+4*6+...+99*101+100*102
H=1*2^2+2*3^2+3*4^2+...+99*100^2
I=1*2*3+3*4*5+5*6*7+7*8*9+...+98*99*100
K=1^2+3^2+5^2+...+99^2
A = 1*2*3 + 2*3*4 + 3*4*5 ... + 99*100*101
=> 4A = 1*2*3*4 + 2*3*4*4 + 3*4*5*4 + ... +99*100*101*4
=> 4A = 1*2*3*4 + 2*3*4*(5 - 1) + 3*4*5*( 6 - 2) + ... + 99*100*101*(102 - 98)
=> 4A = 1*2*3*4 + 2*3*4*5 - 1*2*3*4 + 3*4*5*6 - 2*3*4*5 + ... + 99*100*101*102 - 98*99*100*101
=> 4A = 99*100*101*102
=> 4A = 101989800
=> A = 25497450
1. Tính tổng:
1/ 1 + (-2) + 3 + (-4) +...+ 19 + (-20)
2/ 1 - 2 + 3 - 4 +..+ 99 - 100
3/ 2 - 4 + 6 - 8 +...+ 48 - 50
4/ -1 + 3 - 5 + 7 -...+ 97 - 99
5/ 1 + 2 - 3 - 4 +...+ 97 + 98 - 99 - 100
tính nhanh
e)A=1-2+3-4+...+99-100
g)B=1+3-5-7+9+11-...-397-399
h)C=1-2-3+4+5-6-7+..+97-98-99+100
i)D=2^100-2^99-2^98-..-2^2-2-1
BÀi 4: Tính tổng
1, 1+(-2)+3+(-4)+.....19+(-20)
2, 1-2+3-4+...+99-100
3, 2-4+6-8+...+48-50
4, -1+3-5+7-.....+97-99
5, 1+2-3-4+....+97+98-99-100
1) 1+(-2)+3+(-4)+...+19+(-20)
=[1+(-2)]+[3+(-4)]+...+[19+(-20)]
=(-1)+(-1)+..+(-1)
=(-1).10
=-10
2) 1-2+3-4+...+99-100
=(1-2)+(3-4)+...+(99-100)
=(-1)+(-1)+..+(-1)
=(-1).50
=-50
3) 2-4+6-8+..+48-50
=(2-4)+(6-8)+..+(48-50)
=(-2)+(-2)+..+(-2)
=(-2).25
=-50
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