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2) b)
Do \(a+b+c=9\Rightarrow\left(a+b+c\right)^2=81\)
\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=81\)
\(\Rightarrow2\left(ab+bc+ac\right)=81-141=-60\)
\(ab+bc+ac=-60:2=-30\)
a, B=x^3 + 3xy +y^3 = x^3 +3xy(x+y)+y^3 (vì x+y=1)
= (x+y)^3
= 1^3 =1
b, (a+b+c)^2 =a^2 +b^2 +c^2 +2ab +2bc +2ac
9^2 = 141 +2(ab+bc+ac)
-60 = 2(ab+bc+ac)
ab+ac+bc=-30
Vậy M=-30
c, N =(x+y)^3 -3(x+y)(x^2+y^2) +2(x^3+y^3)
= x^3 + 3x^2 .y + 3xy^2 + -3(x^3+xy^2 +x^2 .y+y^3)+ 2x^3 +2y^3
= x^3 +3x^2 .y + 3xy^2 - 3x^3 -3xy^2 -3x^2 .y -3y^3 +2x^3 +2y^3
= 0
Vậy N=0 .Chúc bạn học tốt.
2a) \(4x^2-1=\left(2x\right)^2-1^2=\left(2x+1\right)\left(2x-1\right)\)
b) \(x^2+16x+64=\left(x+8\right)^2\)
c) \(x^3-8y^3=x^3-\left(2y\right)^3\)
\(=\left(x-2y\right)\left(x^2+2xy+4y^2\right)\)
d) \(9x^2-12xy+4y^2=\left(3x-2y\right)^2\)
a)4x3y-6xy2
=2xy(2x2-3y)
b)4x2-4x+1
=(2x)2-2*2x*1+12
=(2x-1)2
c)x2-2xy-3x+6y
=x(x-2y)-3(x-2y)
=(x-3)(x-2y)
d)x3-2x2+x-xy2
=x(x2-2x+1-y2)
=x[(x-1)2-y2]
=x(x-y-1)(x+y-1)
e)x2-x+y2-y-x2y2+xy
=xy2-x+y2-y-x2y2+x2-xy2+xy
=(xy2-x+y2-y)-x(xy2-x+y2-y)
=(1-x)(xy2-x+y2-y)
=(1-x)[xy2+xy+y2-(xy+y+x)]
=(1-x)[y(xy+y+x)-(xy+y+x)]
=(1-x)(y-1)(xy+y+x)
Bài 2:
a)x(x-y)+y(y-x)
=x2-xy+y2-xy
=(x-y)2.Tại x=53 và y=3 ta có:
N=(53-3)2=502=2500
b) x2013-53x2012+103x2011-51x2010
=x2010(x3-53x2+103x-51)
=x2010[x3-2x2+x-51x2+102x-51]
=x2010[x(x2-2x+1)-51(x2-2x+1)]
=x2010(x-51)(x2-2x+1).Tại x=51 ta có:
M=512010(51-51)(512-2*51+1)=0
\(x^{2010}+y^{2010}=x^{2011}+y^{2011}=x^{2012}+y^{2012}\)
\(\Leftrightarrow\left(x^{2012}+x^{2010}-2x^{2011}\right)+\left(y^{2012}+y^{2010}-2y^{2011}\right)=9\)\(\rightarrow x^{2010}\left(x^2-2x+1\right)+y^{2010}\left(y^2-y+1\right)=0\)
\(\Leftrightarrow x^{2010}\left(x-1\right)^2+y^{2010}\left(y-1\right)^2=0\)
Do x;y dương => x=y=1
121^2 - 21^2 = (121 - 21)(121 + 21) = 100.142 = 14200
2011^2 - 2010.2012 = 2011^2 - (2011^2 - 1) = 1