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TH1: x=0 suy ra 2011x=2012x(cùng bằng 0)
Th2: x>0 do 2011<2012 nên 2011x<2012x
Th3: x<0 do 2011<2012 nên 2011x>2012x
ban luong ledu oi!ban co cach khac chi tiet hon khong
\(x+2x+3x+4x+.....+2011x=2012.2013\)
\(1x+2x+3x+4x+.....+2011x=2012.2013\)
\(1x+2x+3x+4x+.....+2011x=4050156\)
\(x(1+2+3+4+.....+2011)=4050156\)
\(x.2011.(2011+1):2=4050156\)
\(x.2023066=4050156\)
\(x=4050156:2023066\)
\(x=2...............\)
Ta có
x + 2x + 3x + 4x + 5x + ... + 2011x = 2012.2013
x + 2x + 3x + 4x + 5x + ... + 2011x = 4050156
x(2 + 3 + 4 + ... + 2011) = 4050156
x.2023066 = 40501156
x = 40501156 : 2023066
x = 20,...
1 (3y - 0,8 ) : y + 14,5 = 15
( 3y - 0,8 ) : y = 0,5
3y : y - 0,8 : y = 0,5
3 - 0,8 : y = 0,5
0,8 : y = 2,5
y = 0,8 : 2,5
y = 0,32
Ta có :
Tử số = 2012 x 14 + 1997 + 2010 x 2011
= ( 2011 + 1 ) x 14 + 1997 + 2010 x 2011
= 2011 x 14 + 1 x 14 + 1997 + 2010 x 2011
= 2011 x 14 + 14 + 1997 + 2010 x 2011
= ( 2011 x 14 ) + ( 14 + 1997 ) + ( 2010 x 2011 )
= 2011 x 14 + 2011 + 2010 x 2011
= 2011 x ( 14 + 1 + 2010 )
= 2011 x 2025
Mẫu số = 2011 x 5 + 2011 x 1008 + 1012 x 2011
= 2011 x ( 5 + 1008 + 1012 )
= 2011 x 2025
=> \(A=\frac{2011\times2025}{2011\times2025}=1\)
\(\dfrac{1}{2011}.x=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\)
\(\dfrac{1}{2011}.x=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}\)
\(\dfrac{1}{2011}.x=\dfrac{1}{4}\)
=> \(x=\dfrac{1}{4}:\dfrac{1}{2011}\)
=> \(x=\dfrac{2011}{4}\)
\(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}
\)
\(=\frac{210}{840}+\frac{70}{840}+\frac{35}{840}+\frac{21}{840}+\frac{14}{840}+\frac{10}{840}\)
\(=\frac{210+70+35+21+14+10}{840}\)
\(=\frac{360}{840}\)
\(=\frac{3}{7}\)
a ) \(2011^2.2011^x=2011^6\)
\(2011^{x+2}=2011^6\)
=> x + 2 = 6
=> x = 4
b) \(5^x+1=126\)
\(5^x=125\)
\(5^x=5^3\)
=> x = 3