a: =12x^4-6x^3+3x+4x^3-2x^2+1

=12x^4-2x^3-2x^2+3x+1

b: =14x^4+28x^2+6x^2+12x

=14x^4+34x^2+12x

`@` `\text {Ans}`

`\downarrow`

\((3x + 1)(4x³ - 2x² + 1)\)

`= 3x(4x^3-2x^2+1) + 4x^3 - 2x^2 + 1`

`= 12x^4 - 6x^3 + 3x + 4x^3 - 2x^2 + 1`

`= 12x^4 + (-6x^3 + 4x^3) - 2x^2 + 3x + 1`

`= 12x^4 - 2x^3 - 2x^2 + 3x + 1`

\((7x² + 3x)(2x + 4)\)

`= 7x^2(2x+4) + 3x(2x+4)`

`= 14x^3 + 28x^2 + 6x^2 + 12x`

`= 14x^3 + (28x^2 + 6x^2)+12x`

`= 14x^3 + 34x^2 + 12x`

`@` `\text {Kaizuu lv uuu}`

a: \(A=4x-3x^2+20-15x-9x^2-12x-4+\left(2x+1\right)^3-\left(8x^3-1\right)\)

\(=-12x^2-23x+16+8x^3+12x^2+6x+1-8x^3+1\)

\(=-17x+18\)

Cho bất phương trình  - 4x + 12 > 0 . Phép biến đổi nào dưới đây đúng ?

- 4x + 12 < 0 

<=> -4x < - 12 

<=> 4x > 12 

C 

\(\left|-4x\right|-3x+13=-4x-3x+13=-7x+13\)

=> D

\(B=\dfrac{x^2-x}{x^2-3x}-\dfrac{7x-9}{x^2-9}\)

\(B=\dfrac{x\left(x-1\right)}{x\left(x-3\right)}-\dfrac{7x-9}{x^2-3^2}\)

\(B=\dfrac{x-1}{x-3}-\dfrac{7x-9}{\left(x+3\right)\left(x-3\right)}\)

\(B=\dfrac{\left(x-1\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{7x-9}{\left(x+3\right)\left(x-3\right)}\)

\(B=\dfrac{\left(x-1\right)\left(x+3\right)-7x-9}{\left(x+3\right)\left(x-3\right)}\)

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

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

\(B=\dfrac{x\left(x-5\right)+6}{\left(x+3\right)\left(x-3\right)}\)

\(=7x^2+28x-14x^3+7x^2-28x-6x^2+3x-12\)

\(=-14x^3+8x^2+3x-12\)

\(7x\left(x+4\right)-\left(7x+3\right)\left(2x^2-x+4\right)\)

\(=7x^2+28x-14x^3+7x^2-28x-6x^2+3x-12\)

\(=14x^3+8x^2+3x-12\)

a) \(C=\left(\dfrac{x}{x^2-x-6}-\dfrac{x-1}{3x^2-4x-15}\right):\dfrac{x^4-2x^2+1}{3x^2+11x+10}\cdot\left(x^2-2x+1\right)\) (ĐK: \(x\ne-\dfrac{5}{3};x\ne3;x\ne-2;x\ne1\))

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

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

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

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

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

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

b) Thay x = 2003 ta có: 

\(C=\dfrac{2}{\left(2003-1\right)^4\left(2003-3\right)}=\dfrac{2}{2002^4\cdot2000}=\dfrac{1}{2002^4\cdot1000}\)

c) \(C>0\) khi: 

\(\dfrac{2}{\left(x-1\right)^4\left(x-3\right)}>0\) mà: \(\left\{{}\begin{matrix}2>0\\\left(x-1\right)^4>0\end{matrix}\right.\)

\(\Leftrightarrow x-3>0\)

\(\Leftrightarrow x>3\) (đpcm) 

a) ĐKXĐ: 3x + 6 khác 0

x khác -2

b) A = (x² + 4x + 4)/(3x + 6)

= (x + 2)²/[3(x + 2)]

= (x + 2)/3

c) Khi x = 1/4, ta có:

A = (1/4 + 2)/3

= (9/4)/3

= 3/4