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\(A=9\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)
\(A=9\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=9\left(1-\frac{1}{100}\right)\)
\(A=9\times\frac{99}{100}\)
\(A=\frac{891}{100}\) hoặc 8,91
A=9/1.2+9/2.3+9/3.4+.....+9/98.99+9/99.100
=9.(1/1.2+1/2.3+1/3.4+....+1/98.99+1/99.100
=9.(1/1-1/2+1/2-1/3+1/3-1/4+...+1/98-1/99+1/99-1/100)
=9.(1/1-1/100)
=9.99/100
=891/100
CHÚC BẠN HỌC TỐT!
\(A=9.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)
\(=9.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=9.\left(1-\frac{1}{100}\right)\)
\(=9.\frac{99}{100}\)
\(=\frac{891}{100}\)
Answer:
`B=1+3-5-7+9+11-....-397-399`
`=(1+3-5-7)+(9+11-13-15)+...+(393+395-397-399)`
`=(-8)+(-8)+...+(-8)`
`=(-8).100`
`=-800`
`C=1-2-3+4+5-6-7+...+97-98-99+100`
`=(1-2-3+4)+(5-6-7+8)+...+(97-98-99+100)`
`=0+0+...+0`
`=0`
\(D=2^{100}-2^{99}-2^{98}-...-2^2-2-1\)
\(\Rightarrow2D=2^{101}-2^{100}-2^{99}-...-2^3-2^2-2\)
\(\Rightarrow2D-D=\left(2^{101}-2^{100}-...-2^2-2\right)-\left(2^{100}-2^{99}-...-2-1\right)\)
\(\Rightarrow D=2^{101}-2^{100}-2^{100}+1\)
\(\Rightarrow D=2^{101}-2^{101}+1\)
\(\Rightarrow D=1\)
Giải:
\(A=\dfrac{9}{1.2}+\dfrac{9}{2.3}+\dfrac{9}{3.4}+...+\dfrac{9}{98.99}+\dfrac{9}{99.100}\)
\(\Leftrightarrow A=9\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)
\(\Leftrightarrow A=9\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(\Leftrightarrow A=9\left(\dfrac{1}{1}-\dfrac{1}{100}\right)\)
\(\Leftrightarrow A=9.\dfrac{99}{100}\)
\(\Leftrightarrow A=\dfrac{891}{100}\)
Vậy ...
\(\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{99.100}\)
=\(9.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
=\(9.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
=\(9.\left(\frac{1}{1}-\frac{1}{100}\right)\)
=\(9.\frac{99}{100}\)
=\(\frac{891}{100}\)
1. 1-2+3-4+5-6-.....+99-100
=(1-2)+(3-4)+(5-6)+...+(99-100) (50 cặp)
=(-1)+(-1)+(-1)+...+(-1) (50 số -1)
=(-1).50
=-50
2.1+3-5-7+9+11-.....-397-399
=(1+3-5-7)+(9+11-13-15)+....+(387+389-391-393)+395-397-399 (99 cặp)
=(-8)+(-8)+(-8)+...+(-8)+(-401)(có 99 có -8)
=(-8).99+(-401)
=(-792)+(-401)
=-1193
3. 1-2-3+4+5-6-7+...+96+97-98-99+100
=(1-2-3+4)+(5-6-7+8)+...+(93-94-95+96)+(97-98-99+100) (25 cặp)
=0+0+0+...+0
=0
4. A=2100-299-298-.....-22-2-1
2A=2101-2100-299-....-23-22-2
2A-A=A=2101-2100-2100+1
A=2101-2.2100+1
A=2101-2101+1
A=1
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
Đặt \(A=2+2^2+2^3+...+2^{99}\)
nên \(2A=2^2+2^3+...+2^{99}+2^{100}=\left(2+2^2+..+2^{99}\right)+2^{100}-2\)
hay \(2A=A+2^{100}-2\)
vậy \(A=2^{100}-2\text{ hay }2^{100}-2^{99}-..-2=2\)