\(a,=4x^2-4x+1-4x^2+4-x^2-x+6=-x^2-5x+11\\ b,=8x^3+27-8x^3+72x=72x+27\)

a) \(=4x^2-4x+1-4\left(x^2-1\right)-\left(x^2-2x+3x-6\right)=4x^2-4x+1-4x^2+4-x^2-x+6=-x^2-5x+11\)

b) \(=8x^3+27-8x\left(x^2-9\right)=8x^3+27-8x^3+72x=72x+27\)

\(Bài1:\\ a,\left(4x-1\right)\left(2x^2-x-1\right)=4x\left(2x^2-x-1\right)-\left(2x^2-x-1\right)=8x^3-4x^2-4x-2x^2+x+1=8x^3-6x^2-3x+1\\ b,\left(4x^3+8x^2-2x\right):2x\\ =2x\left(2x^2+4x-1\right):2x\\ =2x^2+4x-1\)

\(Bài2:\\ a,2x^3-8x^2+8x=2x\left(x^2-4x+4\right)=2x\left(x-2\right)^2\\ b,2xy+2x+yz+z=2x\left(y+1\right)+z\left(y+1\right)=\left(y+1\right)\left(2x+z\right)\\ c,x^2+2x+1-y^2=\left(x+1\right)^2-y^2=\left(x-y+1\right)\left(x+y+1\right)\)

a,

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

b,

$12x^2y^3z^4+(-7x^2y^3z^4)=12x^2y^3z^4-7x^2y^3z^4=5x^2y^3z^4$

c,

$-6xy^3-(-6xy^3)+6x^3=-6xy^3+6xy^3+6x^3=0+6x^3=6x^3$

d,

$\frac{-x^2}{2}+\frac{7}{2}x^2+x=(\frac{7}{2}-\frac{1}{2})x^2+x$

$=3x^2+x$

e,

$2x^3+3x^3-\frac{1}{3}x^3=(2+3-\frac{1}{3})x^3=\frac{14}{3}x^3$

f,

$5xy^2+\frac{1}{2}xy^2+\frac{1}{4}xy^2=(5+\frac{1}{2}+\frac{1}{4})xy^2$

$=\frac{23}{4}xy^2$

\(2x\left(x^2-7x-3\right)=2x^3-14x-6x\)

\(4xy^2\left(-2x^3+y^2-7xy\right)=-8x^4y^2+4xy^5-28x^2y^3\)

a: \(=\dfrac{2x-2x+y}{2\left(2x-y\right)}=\dfrac{y}{2\left(2x-y\right)}\)

b: \(=\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{x}{2\left(x-1\right)}\)

\(=\dfrac{6x+2-x^2-x}{2\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{-x^2+5x+2}{2\left(x-1\right)\left(x+1\right)}\)

c: \(=\dfrac{1}{x+2}+\dfrac{x+8}{3x\left(x+2\right)}\)

\(=\dfrac{3x+x+8}{3x\left(x+2\right)}=\dfrac{4x+8}{3x\left(x+2\right)}=\dfrac{4}{3x}\)

d: \(=\dfrac{4x+6-2x^2+3x+2x+1}{\left(2x-3\right)\left(2x+3\right)}\)

\(=\dfrac{-2x^2+9x+7}{\left(2x-3\right)\left(2x+3\right)}\)

Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người dễ theo dõi hơn.

a: =1/3x^3-2x^2-7/3x

b: =5/4xyz^2+1/4xyz

c: =9/8x^2y^3-3x^2y