Ôn tập phép nhân và phép chia đa thức

Bài 6:

a: =(3x)^3-(2y)^3

=27x^3-8y^3

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

`(8x^3y^4z) : (4x^2y^4)`

`= (8:4) . (x^3 : x^2) . (y^4 : y^4) . z`

`= 2xz.`

=2xz

\(\left(x+3\right)^2+\left(x-2\right)\left(x+2\right)-2\left(x-1\right)^2=23\)

\(x^2+6x+9+x^2-4-2\left(x^2-2x+1\right)=23\)

\(x^2+6x+9+x^2-4-2x^2+4x-2=23\)

\(10x+3=23\)

\(10x=20\)

\(x=2\)

=>x^2+6x+9+x^2-4-2x^2+4x-2=23

=>10x+5=23

=>10x=18

=>x=1,8

a: \(=\dfrac{15x^4-x^3-x^2+41x-70}{3x^2-2x+7}\)

\(=\dfrac{15x^4-10x^3+35x^2+9x^3-6x^2+21x-30x^2+20x-70}{3x^2-2x+7}\)

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

b: \(=\dfrac{x^4-10x^2-9}{x^2-2x-3}\)

\(=\dfrac{x^4-2x^3-3x^2+2x^3-4x^2-6x-3x^2+6x+9-18}{x^2-2x-3}\)

\(=x^2+2x-3+\dfrac{-18}{x^2-2x-3}\)

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

( x^4 − 2 x^2 + 1 ) : ( 1 − x^2 ) = ( x^2 − 1 ) 2 : ( 1 − x^2 ) = ( 1 − x^2 ) 2 : ( 1 − x^2 ) = 1 − x^2

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

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

\(=\dfrac{x+1}{x-1}\)

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

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

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

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

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

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

|x+2|=3

=>x+2=3 hoặc x+2=-3

=>x=-5(loại) hoặc x=1(nhận)

Khi x=1 thì \(A=\dfrac{3}{1+5}=\dfrac{3}{6}=\dfrac{1}{2}\)

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

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

c: \(P=\dfrac{A}{B}=\dfrac{3}{x+5}:\dfrac{-3}{x-3}=\dfrac{-x+3}{x+5}\)

Để P nguyên thì \(-x-5+8⋮x+5\)

=>\(x+5\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\)

=>\(x\in\left\{-4;-6;-7;-1;-9;-13\right\}\)

Bài 2:

a. 

$(2x-y)(4x^2+2xy+3y^2)=4x^2(2x-y)+2xy(2x-y)+3y^2(2x-y)$

$=8x^3-4x^2y+4x^2y-2xy^2+6xy^2-3y^3$
$=8x^3+4xy^2-3y^3$
b.

$=\frac{9x^5y^8}{2x^3}-\frac{4x^4y^2}{2x^3}+\frac{7x^3}{2x^3}$

$=\frac{9}{2}x^2y^8-2xy^2+\frac{7}{2}$

Bài 3:

a.

$(2x+4)(3x-5x^2y+2)=3x(2x+4)-5x^2y(2x+4)+2(2x+4)$
$=6x^2+12x-10x^3y-20x^2y+4x+8$

$=-10x^3y-20x^2y+6x^2+16x+8$

b.

$=\frac{15x^4}{-5x^2}-\frac{20x^3}{-5x^2}+\frac{10x^2}{-5x^2}$

$=-3x^2+4x-2$

Bài 1:

a.

$(2x+3)(x^2+5x-4)=x^2(2x+3)+5x(2x+3)-4(2x+3)$
$=2x^3+3x^2+10x^2+15x-8x-12$

$=2x^3+13x^2+7x-12$
b.

$(6x^5y^8-4x^4y^2+7x^3):(2x^3)=\frac{6x^5y^8}{2x^3}-\frac{4x^4y^2}{2x^3}+\frac{7x^3}{2x^3}$

$=3x^2y^8-2xy^2+\frac{7}{2}$

\(=\dfrac{x^2+2x-x-2}{x^2-3x-2x-6}\cdot\dfrac{x^2+4x-3x-12}{x^2+4x-x-4}\cdot\dfrac{y^2-5y+3y-15}{y^2-5y+2y-10}\cdot\dfrac{\left(y-2\right)\left(y+2\right)}{y^2+3y-2y-6}\)

\(=\dfrac{x\left(x+2\right)-\left(x+2\right)}{x\left(x-3\right)-2\left(x-3\right)}\cdot\dfrac{x\left(x+4\right)-3\left(x+4\right)}{x\left(x+4\right)-\left(x+4\right)}\cdot\dfrac{y\left(y-5\right)+3\left(y-5\right)}{y\left(y-5\right)+2\left(y-5\right)}\cdot\dfrac{\left(y-2\right)\left(y+2\right)}{y\left(y+3\right)-2\left(y+3\right)}\)

\(=\dfrac{\left(x+2\right)\left(x-1\right)}{\left(x-3\right)\left(x-2\right)}\cdot\dfrac{\left(x+4\right)\left(x-3\right)}{\left(x+4\right)\left(x-1\right)}\cdot\dfrac{\left(y-5\right)\left(y+3\right)}{\left(y-5\right)\left(y+2\right)}\cdot\dfrac{\left(y-2\right)\left(y+2\right)}{\left(y+3\right)\left(y-2\right)}\)

\(=\dfrac{\left(x+2\right)\left(x-1\right)\left(x+4\right)\left(x-3\right)\left(y-5\right)\left(y+3\right)\left(y-2\right)\left(y+2\right)}{\left(x-3\right)\left(x-2\right)\left(x+4\right)\left(x-1\right)\left(y-5\right)\left(y+2\right)\left(y+3\right)\left(y-2\right)}\)

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