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\(A=1.2+2.3+3.4+...+2018.2019\)
\(3A=1.2.3+2.3.3+3.4.3+...+2018.2019.3\)
\(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+2018.2019.\left(2020-2017\right)\)
\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+2018.2019.2020-2017.2018.2019\)
\(3A=2018.2019.2020\)
\(A=2018.673.2020\)
\(A=2743390280\)
Chúc bạn học tốt ~
Đặt A = 1x2 + 2x3 + 3x4 + ... + 99 x 100
3A = 1x2x3 + 2x3x3 + ... + 99x100x3
3A = 1x2x( 3 - 0 ) + 2x3x(4 - 1 ) +...+ 99x100x(101 - 98)
3A = ( 1x2x3 + 2x3x4 + ... + 99x100x101 ) - ( 0x1x2 + 1x2x3 +...+ 98x99x100)
3A = 99x100x101 - 0x1x2
3A = 99x100x101 - 0
A = 99x100x101 : 3
A = 333300
Vậy A = 3333000
A=1.2+2.3+3.4+4.5+...+2014.2015
=>3A=1.2.3+2.3.3+3.4.3+4.5.3+...+2014.2015.3
=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+4.5.(6-3)+...+2014.2015.(2016-2013)
=1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+2014.2015.2016-2013.2014.2015
=(1.2.3-1.2.3)+(2.3.4-2.3.4)+(3.4.5-3.4.5)+(4.5.6-4.5.6)+...+(2013.2014.2015-2013.2014.2015)+0.1.2+2014.2015.2016
=0+2014.2015.2016
=>A=\(\frac{2014.2015.2016}{3}\)
\(\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...............+\dfrac{2}{2008.2009}\)
\(=2\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+................+\dfrac{1}{2008.2009}\right)\)
\(=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.................+\dfrac{1}{2008}-\dfrac{1}{2009}\right)\)
\(=2\left(1-\dfrac{1}{2009}\right)\)
\(=2.\dfrac{2008}{2009}=\dfrac{4016}{2009}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018+2019}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)
=\(1-\frac{1}{2019}< 1\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)
\(A=\frac{1}{1}-\frac{1}{2019}< 1\)
Vậy \(A< 1\)
1) (x-3)(x-5) = 0
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=5\end{cases}}\)
(x+7).35 = 2.35
\(\Rightarrow x+7=2\)
\(\Rightarrow x=2-7=-5\)
Vậy x = -5
2) 1.2 + 2.3 + 3.4 + .... + 99.100
Đặt A = 1.2 + 2.3 + .... + 99.100
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
Vậy A = 333300
ta có \(3S=1\cdot2\cdot3+2\cdot3\cdot3+.....+99\cdot100\cdot3\)
\(3S=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)....+99\cdot100\cdot\left(101-98\right)\)
\(3S=1\cdot2\cdot3-1\cdot2\cdot3+2\cdot3\cdot4-......-98\cdot99\cdot100+99\cdot100\cdot101\)
\(3S=99.100.101\)
\(S=\frac{99\cdot100\cdot101}{3}\)
S=...
3S=1.2.3+2.3.3+3.4.3+4.5.3+...+99.100.3
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
S=33.100.101
S=333300
Vậy S=333300