a: =2x^5-15x^3-x^2-2x^5-x^3=-16x^3-x^2

b: =x^3+3x^2-2x-3x^2-9x+6

=x^3-11x+6

c: \(=\dfrac{4x^3+2x^2-6x^2-3x-2x-1+5}{2x+1}\)

\(=2x^2-3x-1+\dfrac{5}{2x+1}\)

a) \(6x^3\left(\dfrac{1}{3}x^2-\dfrac{5}{2}-\dfrac{1}{6}\right)-2x^5-x^3\)

\(=6x^3\left(\dfrac{1}{3}x^2-\dfrac{16}{6}\right)-2x^5-x^3\)

\(=2x^5-16x^3-2x^5-x^3\)

\(=-17x^3\)

b) \(\left(x+3\right)\left(x^2+3x-2\right)\)

\(=x^3+3x^2-2x+3x^2+9x-6\)

\(=x^3+6x^2+7x-6\)

c) \(\left(4x^3-4x^2-5x+4\right):\left(2x+1\right)\)

\(=2x^2+4x^3-2x-4x^2-\dfrac{5}{2}-5x+\dfrac{2}{x}+4\)

\(=4x^3-2x^2-7x+\dfrac{2}{x}+\dfrac{3}{2}\)

a) Ta có: \(\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)

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

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

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

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

b) Ta có: \(\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{1-6x+9x^2}\)

\(=\dfrac{3x\left(3x+1\right)+2x\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}:\dfrac{2x\left(3x+5\right)}{\left(1-3x\right)^2}\)

\(=\dfrac{9x^2+3x+2x-6x^2}{\left(1-3x\right)\left(1+3x\right)}:\dfrac{2x\left(3x+5\right)}{\left(1-3x\right)^2}\)

\(=\dfrac{3x^2+5x}{\left(1-3x\right)\left(1+3x\right)}\cdot\dfrac{\left(1-3x\right)^2}{2x\left(3x+5\right)}\)

\(=\dfrac{x\left(3x+5\right)}{1+3x}\cdot\dfrac{1-3x}{2x\left(3x+5\right)}\)

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

c) Ta có: \(\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)

\(=\left(\dfrac{9}{x\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x\left(x+3\right)}-\dfrac{x}{3\left(x+3\right)}\right)\)

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

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

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

\(=\dfrac{-3}{x-3}\)

Bài 2:

a: \(=2x^4-x^3-10x^2-2x^3+x^2+10x=2x^3-3x^3-9x^2+10x\)

b: \(=\left(x^2-15x\right)\left(x^2-7x+3\right)\)

\(=x^4-7x^3+3x^2-15x^3+105x^2-45x\)

\(=x^4-22x^3+108x^2-45x\)

c: \(=12x^5-18x^4+30x^3-24x^2\)

d: \(=-3x^6+2.4x^5-1.2x^4+1.8x^2\)

B1: a)\(xy\left(3x-2y\right)-2xy^2=3x^2y-2y^2x-2xy^2=3x^2y-4xy^2\)

b) \(\left(x^2+4x+4\right):\left(x+2\right)=\left(x+2\right)^2:\left(x+2\right)=\left(x+2\right)\)

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

B2:

a)\(2x^2-4x+2=2\left(x^2-2x+1\right)=2\left(x-1\right)^2\)

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

Mấy bài này là mấy bài rất rất rất cơ bản, học sinh TB cũng phải tự làm được, mấy bài kiểu này đừng nên đăng lên hỏi nha:vv

`a)3x(2x^2-3x+4)`

`=6x^3-9x^2+12x`

______________________________________________

`b)(x+3)^2+(3x-2)(x+4)`

`=x^2+6x+9+3x^2+12x-2x-8`

`=4x^2+16x+1`

______________________________________________

`c)[2x-4]/[x-1]+[2x+2]/[x^2-1]`       `ĐK: x \ne +-1`

`=[(2x-4)(x+1)+2x+2]/[(x-1)(x+1)]`

`=[2x^2+2x-4x-4+2x+2]/[(x-1)(x+1)]`

`=[2x^2-2]/[x^2-1]`

`=2`

hếp

chào bê đê

a: \(=\dfrac{x^4-6x^3+12x^2-14x+3}{x^2-4x+1}\)

\(=\dfrac{x^4-4x^3+x^2-2x^3+8x^2-2x+3x^2-12x+3}{x^2-4x+1}\)

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

b: \(=\dfrac{x^5-3x^4+5x^3-x^2+3x-5}{x^2-3x+5}=x^2-1\)

c: \(=\dfrac{2x^4-5x^3+2x^2+2x-1}{x^2-x-1}\)

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

\(=2x^2-3x+1\)

Câu 1: =\(\frac{1}{4}x^2-\frac{1x}{2x}+\frac{1}{4x^2}\)

câu 2 dễ mà