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Lời giải :
\(x^2-2014xy-2016xz+\left(2015^2-1\right)yz\)
\(=x^2-2014xy-2016xz+\left(2015-1\right)\left(2015+1\right)yz\)
\(=x^2-2014xy-2016xz+2014\cdot2016\cdot yz\)
\(=x\left(x-2014y\right)-2016z\left(x-2014y\right)\)
\(=\left(x-2014y\right)\left(x-2016z\right)\)
x2 - 2014xy - 2016xz + (20152 - 1)yz
= x2 - 2014xy - 2016xz + (2015 - 1)(2015 + 1)yz
= x2 - 2014xy - 2016xz + 2014.2016.yz
= (x2 - 2014xy) - (2016xz - 2014.2016.yz)
= x(x - 2014y) - 2016z(x - 2014y)
= (x - 2014y)(x - 2016z)
#TT
bÀI LÀM
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
a)\(2a^2-3ab+b^2\)
=\(a^2+a^2-2ab-ab+b^2\)
=\(\left(a-b\right)^2+a\left(a-b\right)\)
=\(\left(a-b\right)\left(2a-b\right)\)
b)\(x^2-7x-30\)
=\(x^2-10x+3x-30\)
=\(x\left(x-10\right)+3\left(x-10\right)\)
=\(\left(x-10\right)\left(x+3\right)\)
c)\(6a^2-5ab-6b^2\)
=\(6a^2-9ab+4ab-6b^2\)
=\(3a\left(2a-3b\right)+2b\left(2a-3b\right)\)
=\(\left(2a-3b\right)\left(3a+2b\right)\)
d)\(a^4+a^2+1\)
=\(a^4+2a^2-a^2+1\)
=\(\left(a^2+1\right)^2-a^2\)
=\(\left(a^2+1-a\right)\left(a^2+1+a\right)\)
e)\(x^3+6x^2+11x+6\)
=\(x\left(x^2+6x+9+2\right)+6\)
\(=x\left(\left(x+3\right)^2+2\right)+6\)
=\(x\left(x+3\right)^2+2x+6\)
=\(x\left(x+3\right)^2+2\left(x+3\right)\)
=\(\left(x+3\right)\left(x^2+3x+2\right)\)
a) \(a^3-b^3-3ab\left(a-b\right)\)
\(=a^3-3a^2b+3ab^2-b^3\)
\(=\left(a-b\right)^3\)
b) \(2x^3+x^2-4x-12\)
\(=2x^3-4x^2+5x^2-10x+6x-12\)
\(=2x^2\left(x-2\right)+5x\left(x-2\right)+6\left(x-2\right)\)
\(=\left(2x^2+5x+6\right)\left(x-2\right)\)
c) \(x^3-3x^2+2\)
\(=x^3-x^2-2x^2+2x-2x+2\)
\(=x^2\left(x-1\right)-2x\left(x-1\right)-2\left(x-1\right)\)
\(=\left(x^2-2x-2\right)\left(x-1\right)\)
a) \(a^3-b^3-3ab\left(a-b\right)=\left(a-b\right)^3\)
b) \(2x^3+x^2-4x-12=2x^3-4x^2+5x^2-10x+6x-12\)
\(=2x^2\left(x-2\right)+5x\left(x-2\right)+6\left(x-2\right)\)
\(=\left(x-2\right)\left(2x^2+5x+6\right)\)
c) \(x^3-3x^2+2=x^3-x^2-2x^2+2\)
\(=x^2\left(x-1\right)-2\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x^2+2x+2\right)\)
1/ \(\left(a-b\right)\left(a^2+3ab+b^2\right)+\left(a+b\right)^3+ab\left(b-a\right)=\left(a^2+2ab+b^2+ab\right)\left(a-b\right)+\left(a+b\right)^3+ab\left(b-a\right)\)= \(\left(a^2+2ab+b^2\right)\left(a-b\right)+\left(a+b\right)ab+\left(a-b\right)^3-ab\left(a-b\right)\)
= \(\left(a+b\right)^2\left(a-b\right)+\left(a+b\right)^3\)
= \(\left(a+b\right)^2\left(a-b+a+b\right)=2a\left(a+b\right)^2\)
k mình nhé!