Câu 1: C

Câu 2: B

a) \(\frac{5\left(a-b\right)^3+2\left(a-b\right)^2}{\left(b-a\right)^2}=\frac{\left(a-b\right)^2\cdot\left[5\cdot\left(a-b\right)+2\right]}{\left(a-b\right)^2}=5\cdot\left(a-b\right)+2\)

b) \(\frac{5\left(x-2y\right)^3}{5x-10y}=\frac{5\left(x-2y\right)^3}{5\left(x-2y\right)}=\left(x-2y\right)^2\)

c) \(\frac{x^3+8y^3}{x+2y}=\frac{\left(x+2y\right)\left(x^2-2xy+4y^2\right)}{x+2y}=x^2-2xy+4y^2\)

a)=\(3x^3-15x^2+21x\)

b)\(=-2x^4y-10x^2y+2xy\)

c)\(=-x^3+6x^2+5x-4x^2+24x+20=-x^3+2x^2+29x+20\)

d)\(=2x^4-3x^3+4x^2-2x^2+3x-4=2x^4-3x^32x^2+3x-4\)

e)\(=x^2-4y^2\)

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

g)\(=3xy^4-\dfrac{1}{2}y^2+2x^2y\)

h)\(=9x^2-6x+1-7x^2-14=2x^2-6x-13\)

i)\(=x^2-x-3\)

j)\(=\left(x+2y\right)\left(x^2-2y+4y^2\right):\left(x+2y\right)=x^2-2y+4y^2\)

Tại sao ý b có dấu - trước ngoặc đâu mà đổi dấu mong bn giải đáp

a) \(\dfrac{\left(x+y\right)^2}{2}\)

b)\(\dfrac{5a-5b}{2}\)

c)\(\left(x-2y\right)^2\)

a:

1: Sai vì \(-3\left(2x+7\right)=-3\cdot2x-3\cdot7=-3x-21\)

2: Sai vì ac+3bc không thể bằng 3abc^2 được

3: Sai vì 3x+4y<>9x^2+4xy

b:

1: 8(x+5)-3(2x+7)=8x+40-6x-21

=2x+19

2: a(2b+c)+b(3c-2a)

\(=2ab+ac+3bc-2ab=ac+3bc\)

3: 2y(y+5x)+x(3x+4y)

\(=2y^2+10xy+3x^2+4xy=3x^2+14xy+2y^2\)

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\(^2\)