Lời giải:

a.

$=-2x^5+10x^4+2424x^3-x^3-3=-2x^5+10x^4+2423x^3-3$

b.

$=(x-5y)^2+2(x-5y)(x+y)+(x+y)^2$

$=[(x-5y)+(x+y)]^2=(2x-4y)^2=4x^2-16xy+16y^2$

Đáp án cần chọn là: B

a) x=1 hoặc x=0

b) x=5 hoặc x=6

c) x=64,8

d) x=1 hoặc x=0

a) 2016x = 2017x

=> 2016x - 2017x =0

=> x(2016 - 2017) =0

=> x(-1)=0

=>x=0:(-1)=0

b) (x-5)2015=(x-5)2014

=> (x-5)2015 - (x-5)2014=0

=> (x-5)(2015-2014)=0

=> x-5=0

=>x=5

c)5x + 5x +2 =650

=> 10x + 2 =650

=> 10x =648

=> x = \(\frac{648}{10}=64,8\)

d) 2017x =2x

=> 2017x -2x =0

=> 2015x=0

=>x=0

a: \(=xy\left(5x-1\right)\)

b: \(=\left(x-2\right)\left(3x-5\right)\)

\(a,=xy\left(5x-1\right)\\ b,=\left(x-2\right)\left(3x-5\right)\\ c,Sửa:x^2+2x+1-9y^2\\ =\left(x+1\right)^2-9y^2\\ =\left(x-3y+1\right)\left(x+3y+1\right)\\ d,=x\left(x-y\right)+5\left(x-y\right)=\left(x+5\right)\left(x-y\right)\)

â) \(=xy\left(5x-1\right)\)

b) \(=\left(3x-5\right)\left(x-2\right)\)

c) \(=\left(x+1\right)^2+\left(3y\right)^2=\left(x+1\right)^2+6y\left(x+1\right)+\left(3y\right)^2-6y\left(x+1\right)=\left(x+1+3y\right)^2-\sqrt{\left[6y\left(x+1\right)\right]^2}=\left(x+3y+1-\sqrt{6y\left(x+1\right)}\right)\left(x+3y+1+\sqrt{6y\left(x+1\right)}\right)\)

d) \(=\left(x^2+5x\right)-\left(5y+xy\right)=x\left(x+5\right)-y\left(5+x\right)=\left(x-y\right)\left(x+5\right)\)

\(a,=x^2-4-x^2+2x+4=2x\\ b,=\left(x-5y\right)^2:\left(5y-x\right)=\left(5y-x\right)^2:\left(5y-x\right)=5y-x\\ c,Sửa:\left(28x-9x^2+x^3-30\right):\left(x-3\right)\\ =\left(x^3-3x^2-6x^2+18x+10x-30\right):\left(x-3\right)\\ =\left(x-3\right)\left(x^2-6x+10\right)\left(x-3\right)=x^2-6x+10\)

b: (x-y)(x^2-2x+y)

\(=x^3-2x^2+xy-x^2y+2xy-y^2\)

\(=x^3-2x^2-x^2y+3xy-y^2\)

c: \(\left(x^2-y\right)\left(x+y^2\right)-\left(x-y\right)\left(x^2+xy+y^2\right)\)

\(=x^3+x^2y^2-xy-y^3-\left(x^3-y^3\right)\)

\(=x^2y^2-xy\)

d: \(3x\left(2xy-z\right)-5y\left(x^2-2\right)+3xz\)

\(=6x^2y-3xz-5x^2y+10y+3xz\)

\(=x^2y+10y\)