a: \(\frac{4\left(x+3\right)}{3x^2-x}:\frac{x^2+3x}{1-3x}\)

\(=\frac{4\left(x+3\right)}{x\left(3x-1\right)}\cdot\frac{-\left(3x-1\right)}{x\left(x+3\right)}=\frac{-4}{x^2}\)

b: \(\frac{x+1}{x^2-2x-8}\cdot\frac{4-x}{x^2+x}\)

\(=\frac{x+1}{\left(x-4\right)\left(x+2\right)}\cdot\frac{-\left(x-4\right)}{x\left(x+1\right)}=\frac{-1}{x\left(x+2\right)}\)

c: \(\frac{9x+5}{2\left(x-1\right)\left(x+3\right)^2}-\frac{5x-7}{2\left(x-1\right)\left(x+3\right)^2}\)

\(=\frac{9x+5-5x+7}{2\left(x-1\right)\left(x+3\right)^2}=\frac{4x+12}{2\left(x-1\right)\left(x+3\right)^2}\)

\(=\frac{4\left(x+3\right)}{2\left(x-1\right)\left(x+3\right)^2}=\frac{2}{\left(x-1\right)\left(x+3\right)}\)

d: \(\frac{18}{\left(x-3\right)\left(x^2-9\right)}-\frac{3}{x^2-6x+9}-\frac{x}{x^2-9}\)

\(=\frac{18}{\left(x+3\right)\left(x-3\right)^2}-\frac{3}{\left(x-3\right)^2}-\frac{x}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{18-3\left(x+3\right)-x\left(x-3\right)}{\left(x-3\right)^2\cdot\left(x+3\right)}=\frac{18-3x-9-x^2+3x}{\left(x-3\right)^2\cdot\left(x+3\right)}=\frac{-x^2+9}{\left(x-3\right)^2\left(x+3\right)}=\frac{-1}{x-3}\)

e: \(\frac{1}{x^2-x+1}+\frac{1}{1-x^2}+\frac{2}{x^3+1}\)

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

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

\(=\frac{3x-4}{\left(x+1\right)\left(x-1\right)\left(x^2-x+1\right)}\)

Dễ

 Thế

Mà

Cũnhoir

Dc

Ạ

Chịu

Chắc

Phải

Ngu 

Lamqs

Mới

Hỏi

Câu

Này

\(a,=\left[x^2\left(x^2-x-1\right)+x^3+x^2-3x-1\right]:\left(x^2-x-1\right)\\ =\left[x^2\left(x^2-x-1\right)+x\left(x^2-x-1\right)+2x^2-2x-1\right]\\ =\left[x^2\left(x^2-x-1\right)+x\left(x^2-x-1\right)+2\left(x^2-x-1\right)+1\right]:\left(x^2-x-1\right)\\ =\left[\left(x^2+x+2\right)\left(x^2-x-1\right)+1\right]:\left(x^2-x-1\right)=x^2+x+2R1\)

a: \(x^2\left(x-2x^3\right)=x^2\cdot x-x^2\cdot2x^3=x^3-2x^5\)

b: \(\left(x^2+1\right)\cdot5x=x^2\cdot5x+1\cdot5x=5x^3+5x\)

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

\(=x^3+2x^2+4x-2x^2-4x-8=x^3-8\)

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

\(=x^3+2x^2+4x-2x^2-4x-8=x^3-8\)

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

f: (2x-1)(3x+2)(3-x)

\(=\left(6x^2+4x-3x-2\right)\left(3-x\right)\)

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

\(=-6x^3+18x^2-x^2+3x+2x-6=-6x^3+17x^2+5x-6\)

\(a,\left(x^3+5x^2-2x+1\right)\left(x-7\right)\\ =x^4-7x^3+5x^3-35x^2-2x^2+14x+x-7\\ =x^4-2x^3-37x^2+15x-7\\ b,\left(2x^2-3xy+y^2\right)\left(x+y\right)\\ =2x^3+2x^2y-3x^2y-3xy^2+xy^2+y^3\\ =2x^3-x^2y-2xy^2+y^3\\ c,\left(x-2\right)\left(x^2-5x+1\right)-x\left(x^2+11\right)\\ =x^3-5x^2+x-2x^2+10x--x^3-11x\\ =x^3-7x^2\\ d,x\left(1-3x\right)\left(4-3x\right)-\left(x-4\right)\left(3x+5\right)\\ =x\left(4-15x+9x^2\right)-\left(3x^2-7x-20\right)\\ =4x-15x^2+9x^3-3x^2+7x+20\\ =9x^3-18x^2+11x+20\)

c: \(=\dfrac{x^3+2x^2+x^2+2x-10x-20}{x+2}\)

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

Bài 5: 

a. 1 - 2y + y²

= (1 - y)²

b. (x + 1)² - 25

= (x + 1)² - 5²

= (x + 1 - 5)(x + 1 + 5)

= (x - 4)(x + 6)

c. 1 - 4x²

= 1² - (2x)²

= (1 - 2x)(1 + 2x)

d. 8 - 27x³

= 2³ - (3x)³

= (2 - 3x)(4 + 6x + 9x²)

e. (đề hơi khó hiểu ''x3'' !?)

g. x³ + 8y³

= (x + 2y)(x² - 2xy + y²)

$=x^3-2x^5$

b) $(x^2+1)(5-x)$

$=5x^2-x^3+5-x$

$=-x^3+5x^2-x+5$

$=x^3+3x^2-4x-2x^2-6x+8$

$=x^3+x^2-10x+8$

$=x^2-x^3+4x-2x+2x^2-8$

$=-x^3+3x^2+2x-8$

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

Trước hết:

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

$=-3x^2+7x+6$

Do đó:

$(2x-1)(-3x^2+7x+6)$

$=-6x^3+14x^2+12x+3x^2-7x-6$

$=-6x^3+17x^2+5x-6$

g) $(x+3)(x^2+3x-5)$

$=x^3+3x^2-5x+3x^2+9x-15$

$=x^3+6x^2+4x-15$

$=x^4y-2x^2y-6xy-2x^3+4x+12$

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

$=20x^5-9x^4+19x^3-16x^2+7x-6$

a: \(\dfrac{4x^4y-7x^2y+3y}{-3x^2+2y}\)

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

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

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

a) `(4x^4y-7x^2y+3y).(2y-3x^2y)`

`=8x^4y^2-14x^2y^2+6y^2-12x^6y^2+21x^4y^2-9x^2y^2`

`=29x^4y^2-12x^6y^2-23x^2y^2+6y^2`

b) `(x^2+3x-3/2 x^3):2x - x/2 . (1-3/2 x)`

`=(x+3-3/2 x^2):2 - (x/2 - 3/4 x^2)`

`=x/2 + 3/2 - 3/4 x^2 -x/2 +3/4 x^2`

`=3/2`

c) `(-2x^3-x-3+5x^2):(3-2x)`

`=(3-2x)(x^2-x-1) : (3-2x)`

`=x^2-x-1`

1) \(\left(x^3-8\right):\left(x-2\right)=\left[\left(x-2\right)\left(x^2+2x+4\right)\right]:\left(x-2\right)=x^2+2x+4\)

2) \(\left(x^3-1\right):\left(x^2+x+1\right)=\left[\left(x-1\right)\left(x^2+x+1\right)\right]:\left(x^2+x+1\right)=x-1\)

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

4) \(\left(25x^2-4y^2\right):\left(5x-2y\right)=\left[\left(5x-2y\right)\left(5x+2y\right)\right]:\left(5x-2y\right)=5x+2y\)