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a. 1⋅2⋅3+2⋅4⋅6+3⋅6⋅9+4⋅8⋅12
= 6+2⋅4⋅6+3⋅6⋅9+4⋅8⋅12
= 6+48+3⋅6⋅9+4⋅8⋅12
= 6+48+162+4⋅8⋅12
= 6+48+162+384
= 600
b . Ta có \(A=\frac{2010+2011}{2011+2012}=\frac{2010}{2011+2012}+\frac{2011}{2011+2012}.\)
Ta có : \(\frac{2010}{2011+2012}< \frac{2010}{2011}\) và \(\frac{2011}{2011+2012}< \frac{2011}{2012}\)
=> \(\frac{2010+2011}{2011+2012}< \frac{2010}{2011}+\frac{2011}{2012}\)
=> A < B
`x xx 2/3 xx 3/4 xx 4/5 xx ... xx 2010/2011 = 2/2012`
`<=> x/2011 = 1/1006`
`=> x = 2011/1006`
=\(\frac{1}{2}\)x\(\frac{2}{3}\)x........x \(\frac{2010}{2011}\)
=\(\frac{1x2x......x2010}{2x3x....x2011}\)
=\(\frac{1}{2011}\)
a)(3/2-0,5)/x=7/2+1/4
(3/2-1/2)/x=14/4+1/4
1/x=15/4
x=1:15/4
x=4/15
b)(x*0,25+2010)*2011=(53+2010)*(2012-1)
(x*0,25+2010)*2011=2063*2011
=>0,25x+2010=2063
0,25x=2063-2010
0,25x=53
x=53/0,25
x=212
\(A=\frac{1}{\left(2\times2\right)}+\frac{1}{\left(3\times3\right)}+....+\frac{1}{\left(2011\times2011\right)}\)
\(A=1+\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+.....+\frac{1}{2011}-\frac{1}{2011}\)
\(A=1+\frac{1}{2}\)
\(A=\frac{3}{2}\)
\(A>\frac{3}{4}\)