a) \(\dfrac{9x^2-6x+1}{9x^2-1}\)

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

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

\(=\dfrac{3\cdot\left(-3\right)-1}{3\cdot\left(-3\right)+1}=\dfrac{-9-1}{-9+1}=\dfrac{-10}{-8}=\dfrac{5}{4}\)

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

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

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

\(=\dfrac{-\dfrac{1}{3}-3}{3\cdot\dfrac{-1}{3}}=\dfrac{-\dfrac{10}{3}}{-1}=\dfrac{10}{3}\)

c) Ta có: \(\dfrac{x^2-4x+4}{2x^2-4x}\)

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

\(=\dfrac{x-2}{2x}\)

\(=\dfrac{\dfrac{-1}{2}-2}{2\cdot\dfrac{-1}{2}}=\dfrac{-\dfrac{5}{2}}{-1}=\dfrac{5}{2}\)

a) Ta có: \(\left(y+3\right)\left(y^2-3y+9\right)-\left(60-y^3\right)\)

\(=y^3+27-60+y^3\)

\(=2y^3-33\)

b) Ta có: \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)

\(=8x^3+y^3-8x^3+y^3\)

\(=2y^3\)

`@` `\text {Ans}`

`\downarrow`

`A= (2x - 3)^2 - (2x + 3)^2`

`= [(2x - 3) - (2x + 3)]*[(2x - 3) + (2x + 3)]`

`= (2x - 3 - 2x - 3) * (2x - 3 + 2x + 3)`

`= -6 * 4x`

`= -24x`

`A=(2x-3)^2-(2x+3)^2`

`A=(2x-3-2x-3)(2x-3+2x+3)`

`A=-6.4x=-24x`

b: B=(x+1)^2-2(2x-1)(x+1)+4x^2-4x+1

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

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

=(-x+2)^2=x^2-4x+4

\(a,\dfrac{\left(x-1\right)^2}{x^2-1}=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{x+1}\\ b,\dfrac{x^2-16}{4x-x^2}=\dfrac{\left(x-4\right)\left(x+4\right)}{x\left(4-x\right)}=\dfrac{-\left(4-x\right)\left(x+4\right)}{x\left(4-x\right)}=\dfrac{-\left(x+4\right)}{x}\\ c,\dfrac{x^2+6x+9}{2x+6}=\dfrac{\left(x+3\right)^2}{2\left(x+3\right)}=\dfrac{x+3}{2}\)

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

\(e,\dfrac{x^2-x+1}{x^3+1}=\dfrac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{1}{x+1}\\ f,\dfrac{\left(x+y\right)^2-z^2}{x+y+z}=\dfrac{\left(x+y-z\right)\left(x+y+z\right)}{x+y+z}=x+y-z\)

\(A=\left(2x+3\right)\left(4x^2-6x+9\right)-2\left(4x^2-1\right)\)

\(=8x^3-12x^2+18x+12x^2-18x+27-8x^2+2\)

\(=8x^3-8x^2+29\)

ta có 

\(A=\left(2x+3\right)\left(4x^2-6x+9\right)-2\left(4x^2-1\right)\)

\(A=8x^3-12x^2+18x+12x^2-18x+27-8x^2+2\)

\(A=8x^3-8x^2+29\)

Trả lời:

A = ( 2x + 3 ) ( 4x² - 6x + 9 ) - 2 ( 4x² - 1 )

= 8x³ - 12x² + 18x + 12x² - 18x + 27 - 8x² + 2

= 8x³ - 8x² + 29

a: \(=2x^2-6x+x-3-20x+8x^2\)

\(=10x^2-25x-3\)

b: \(=x^2+4x+4-2\left(x^2-9\right)+10\)

\(=x^2+4x+14-2x^2+18\)

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

\(=8x^3-27-8x^3+50=23\)

mình chỉ biết làm một nửa k biết có đứng k bạn có chắc đề bài đúng k

5x^2 - 1^2 - (2x^3-3^3)= (5x^2-1x^2)-(2x^3-3^3) hdt số 3 và số 7

\(\left(2x+3\right)\left(4x^2-6x+9\right)-8x\left(x^2-2\right)\)

\(=\left(2x+3\right)\left[\left(2x^2\right)-2x.3^2\right]-8x\left(x^2-2\right)\)

\(=\left(2x\right)^3+3^3-8x^3+16x\)

\(=18x^3+27-8x^3+16x\)

\(=16x+27\)

(2x + 3)(4x² - 6x + 9) - 8x(x² - 2)

= (2x)³ + 3³ - 8x(x² - 2)

= 8x³ + 9  - 8x³ + 16x

= 9 + 16x

 Chúc bạn học tốt

\(\left(2x+3\right)\left(4x^2-6x+9\right)-8x\left(x^2-2\right)\)

\(=8x^3+27-8x^3+16x\)

=16x+27