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\(A=2011.2013-2012^2\)
Gọi 2012 là a ta có:
\(2011=a-1;2013=a+1\)
\(\Rightarrow A=\left(a+1\right).\left(a-1\right)-a^2\)
\(\Rightarrow A=a^2-a+a-1-a^2\)
\(\Rightarrow A=a^2-1-a^2\)
\(\Rightarrow A=-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 2x² - xy + 4x - 2y
<=> (2x² + 4x)-(xy + 2y)
<=> 2x(x + 2) - y(x + 2)
<=> (x + 2)(2x - y)
b) (a²−a+2012)(a²−a+2014)−3
Đặt a²−a+2012 là x , ta có :
x(x + 2) - 3
<=> x² + 2x - 3
<=> x² + 3x - x - 3
<=> x(x + 3) - (x + 3)
<=> (x +3)(x - 1)
Thay x = a²−a+2012 , ta được :
(a²−a+2015)(a²−a+2011)
![](https://rs.olm.vn/images/avt/0.png?1311)
Phân tích các đa thức sau thành nhân tử
a) (x+y+z)^3 - x^3 - y^3 - z^3
b) x^4 + 2012x^2 + 2011x + 2012
![](https://rs.olm.vn/images/avt/0.png?1311)
= x3 + y3 + z3 + 3x2yz + 3xy2z + 3xyz2 - x3 -y3 - z3
=3x2yz + 3xy2z + 3xyz2
= 3xyz( x + y + z)
b.
x^4+2012x^2+2012x-x+2012=
(x^4-x)+2012(x^2+x+1)=
x(x-1)(x^2+x+1)+2012(x^2+x+1)=
(x+2012)(x^2+x+1)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,3x^2-11x+6=3x^2-9x-2x+6=3x\left(x-3\right)-2\left(x-3\right)=\left(3x-2\right)\left(x-3\right)\)
\(b,8x^2+10x-3=8x^2+12x-2x-3=4x\left(2x+3\right)-\left(2x+3\right)=\left(4x-1\right)\left(2x+3\right)\)
\(c,8x^2-2x-1=9x^2-x^2-2x-1=9x^2-\left(x+1\right)^2=\left(3x-x-1\right)\left(3x+x+1\right)\)
\(=\left(2x-1\right)\left(4x+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a^4+a^3+a^2+a\)
\(=a^3\left(a+1\right)+a\left(a+1\right)\)
\(=\left(a+1\right)\left(a^3+a\right)\)
nha !!!
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![](https://rs.olm.vn/images/avt/0.png?1311)
giúp mik vs
ta có: (a^2-a+2012)(a^2-a+2014)-3
=(a^2-a+2013-1)(a^2-a+2013+1)-3
=(a^2-a+2013)^2-1-3
=(a^2-a+2013)^2-4
=(a^2-a+2013-2)(a^2-a+2013+2)
=(a^2-a+2011)(a^2-a+2015)
chúc bn học tốt