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a) \(4a^3b^3c^2x+12a^3b^4c^2-16a^4b^5cx\)
\(=4a^3b^3c\left(cx+3bc-4ab^2x\right)\)
b) \(\left(b-2c\right)\left(a-b\right)-\left(a+b\right)\left(2c-b\right)\)
\(=\left(b-2c\right)\left(a-b+a+b\right)=2a\left(b-2c\right)\)
c) \(3a\left(a+5\right)-2\left(5+a\right)=\left(a+5\right)\left(3a-2\right)\)
d) \(\left(x+1\right)^2-3\left(x+1\right)=\left(x+1\right)\left(x+1-3\right)\)
a)x^2.16-4xy+4y^2
<=>16.x^2-2x2y+(2y)^2
<=>16(x-2y)^2
b)x^5-x^4+x^3-x^2
<=>(x^5-x^4)+(x^3-x^2)
<=>x^4(x-1)+x^2(x-1)
<=>(x-1)(x^4+x^2)
c)x^5+x^3-x^2-1
<=>(x^5+x^3)-(x^2+1)
<=>x^3(x^2+1)-(x^2+1)
<=>(x^2+1)(x^3-1)
d)x^4-3x^3-x+3
<=>(x^4-3x^3)-(x-3)
<=>x^3(x-3)-(x_3)
<=>(x-3)(x^3-1)
\(a,x^2.16-4xy+4y^2\)
\(=16.x^2-4xy+4y^2\)
\(=16.\left[x^2-4xy+\left(2y\right)^2\right]\)
\(=16.\left(x-2y\right)^2\)
\(b,x^5-x^4+x^3-x^2\)
\(=x^4\left(x-1\right)+x^2\left(x-1\right)\)
\(=\left(x-1\right)\left(x^4+x^2\right)\)
\(=x^2\left(x-1\right)\left(x^2+1\right)\)
\(c,x^5+x^3-x^2-1\)
\(=x^3\left(x^2+1\right)-\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^3-1\right)\)
\(=\left(x^2+1\right)\left(x-1\right)\left(x^2+x+1\right)\)
\(d,x^4-3x^3-x+3\)
\(=x^3\left(x-3\right)-\left(x-3\right)\)
\(=\left(x-3\right)\left(x^3-1\right)\)
\(=\left(x-3\right)\left(x-1\right)\left(x^2+x+1\right)\)
\(a,\left(a^3-b^3\right)+\left(a-b\right)^2\)
\(=\left(a-b\right)\left(a^2+ab+b^2\right)+\left(a-b\right)^2\)
\(=\left(a-b\right)\left(a^2+ab+b^2+a-b\right)\)
\(b,\left(x^2+1\right)^2-4x^2\)
\(=x^4+2x^2+1-4x^2\)
\(=x^4-2x^2+1\)
\(\left(x^2-1\right)^2\)
\(c\left(y^3+8\right)+\left(y^2-4\right)\)
\(=\left(y+2\right)\left(y^2-8y+4\right)+\left(y-2\right)\left(y+2\right)\)
\(=\left(y+2\right)\left(y^2-8y+4+y-2\right)\)
\(=\left(y+2\right)\left(y^2-7y+2\right)\)
a) ( a3 - b3) + ( a - b)2
= (a-b) (a2 + ab + b2 ) + (a-b)2
= (a-b) (a2 + ab + b2 +a -b )
hok tốt
b) Ta có : x2 + 8x + 12
= x2 - 4 + 8x + 16
= (x - 2)(x + 2) + 8(x + 2)
= (x + 2)(x - 2 + 8)
= (x + 2)(x + 6)
ta có \(x^2+8x+12\)
=\(x^2-4+8x+16\)
=\(\left(x-2\right)\left(x+2\right)+8\left(x+2\right)\)
=\(\left(x+2\right)\left(x-2+8\right)\)
=\(\left(x+2\right)\left(x+6\right)\)
4) \(a^2-a-2012.2013=a^2-a-2012\left(2012+1\right)\)
\(=a^2-a-2012^2-2012=\left(a+2012\right)\left(a-2012\right)-\left(a+2012\right)\)
\(=\left(a+2012\right)\left(a-2013\right)\)
5) \(n^5-n=n\left(n^4-1\right)=n\left(n^2-1\right)\left(n^2+1\right)\)\(=n\left(n-1\right)\left(n+1\right)\left(n^2+1\right)\)
Đề sai nhé .Sửu lại
\(x^2-4x^2y^2+4+4x\)
\(=\left(x^2+4x+4\right)-4x^2y^2\)
\(=\left(x+2\right)^2-\left(2xy\right)^2\)
\(=\left(x+2+2xy\right)\left(x+2-2xy\right)\)
k ) \(125x^3-1\)
\(=\left(5x\right)^3-1\)
\(=\left(5x-1\right)\left[\left(5x\right)^2+5x.1+1^2\right]\)
\(=\left(5x-1\right)\left(25x^2+5x+1\right)\)
m ) \(x^6-y^3=\left(x^2\right)^3-y^3=\left(x^2-y\right).\left[\left(x^2\right)^2+x^2.y+y^2\right]=\left(x^2-y\right).\left(x^4+x^2y+y^2\right)\)
n ) \(a^4-2a^2+1\)
\(=\left(a^2\right)^2-2.a^2.1+1^2=\left(a^2-1\right)^2\)
i ) \(a^3+6a^2+12a+8\)
\(=\left(a+2\right)^3\)
k) \(125x^3-1=\left(5x\right)^3-1=\left(5x-1\right)\left(25x^2+5x+1\right)\)
m) \(x^6-y^3=\left(x^2\right)^3-y^3=\left(x^2-y\right)\left(x^4+x^2y+y^2\right)\)
n) \(a^4-2a^2+1=\left(a^2-1\right)^2=\left(a^2-1\right)\left(a^2-1\right)=\left(a-1\right)\left(a+1\right)\left(a-1\right)\left(a+1\right)\)
i) \(a^3+6a^2+12a+8=\left(a+2\right)^2\)