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![](https://rs.olm.vn/images/avt/0.png?1311)
a)\([\)5(a-b)\(^3\)+2(a-b)\(^2]\):(b-a)\(^2\)
=\([\)5(a-b)\(^3\)+2(a-b)\(^2]\):(a-b)\(^2\)
=5(a-b)+2
b)5(x-2y)\(^3\):(5x-10y)
=5(x-2y)\(^3\):5(x-2y)
=(x-2y)\(^2\)
c)(x\(^3\)+8y\(^3\)):(x+2y)
=\([\)x\(^3\)+(2y)\(^3]\):(x+2y)
=(x+2y)(x\(^2\)-2xy+4y\(^2\)):(x+2y)
=x\(^2\)-2xy+4y\(^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b: Ta có: \(\left(4x^4-3x^3\right):\left(-x^3\right)+\left(15x^2+6x\right):3x=0\)
\(\Leftrightarrow-4x+3+5x+2=0\)
\(\Leftrightarrow x=-5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (x + 2)(x2 + 3x + 1)
= x.x2 + x.3x + x.1 + 2.x2 + 2.3x + 2.1
= x3 + 3x2 + x + 2x2 + 6x + 2
= x3 + 5x2 + 7x + 2
b) (2x3 + 10x2 + 9x + 4) : (x + 4)
= (2x3 + 8x2 + 2x2 + 8x + x + 4) : (x + 4)
= [(2x3 + 8x2) + (2x2 + 8x) + (x + 4)] : (x + 4)
= [2x2(x + 4) + 2x(x + 4) + (x + 4)] : (x + 4)
= (x + 4)(2x2 + 2x + 1) : (x + 4)
= 2x2 + 2x + 1
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(5x^2y^4:10x^2y=\dfrac{1}{2}y^3\)
c: \(\left(-xy\right)^{10}:\left(-xy\right)^5=-x^5y^5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(=\dfrac{6x^2+15x-2x-5}{2x+5}=3x-1\)
b: \(=\dfrac{x^2\left(x+3\right)+\left(x-3\right)}{x-3}=x^2+1\)
c: \(=\dfrac{2x^4-6x^2+x^3-3x+x^2-3}{x^2-3}=2x^2+x+1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(5x+5\right)^2\)
\(A=\left[\left(3x+1\right)-\left(5x+5\right)\right]^2\)
\(A=\left(-2x-4\right)^2\)
A = (3x + 1)2 - 2(3x + 1)(5x + 5) + (5x + 5)2
= [(3x + 1)-(5x + 5)]2
= (3x + 1 - 5x - 5)2
= [(-2x) - 4]2
B = (3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 +1)(332 + 1)
=> (3 - 1)B = (3 - 1)(3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 +1)(332 + 1)
=>2B = (32 - 1)(32 + 1)(34 + 1)(38 + 1)(316 +1)(332 + 1)
= (34 - 1)(34 + 1)(38 + 1)(316 +1)(332 + 1)
= (38 - 1)(38 + 1)(316 +1)(332 + 1)
= (316 - 1)316 +1)(332 + 1)
= (332 - 1)(332 + 1)
= 364 - 1
vì 2B = 364 - 1
=> B = \(\dfrac{3^{64}-1}{2}\)
C = a2 + b2 + c2 + 2ab - 2ac - 2bc + a2 + b2 + c2 - 2ab + 2ac - 2bc - 2( b2 - 2bc + c2)
= 2a2 + 2b2 + 2c2 - 4bc - 2b2 + 4bc - 2c2
= 2a2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(f\left(-1\right)=-4\Rightarrow-1+a-b+c=-4\)
\(\Rightarrow a-b+c=-3\)
\(f\left(2\right)=5\Rightarrow8+4a+2b+c=5\Rightarrow4a+2b+c=-3\)
\(\Rightarrow3a+3b=0\Rightarrow a=-b\)
\(\Rightarrow a^{2019}=-b^{2019}\Rightarrow a^{2019}+b^{2019}=0\)
\(\Rightarrow A=0\)