a: Ta có: \(\left(7x+4\right)^2-\left(7x-4\right)\left(7x+4\right)\)

\(=\left(7x+4\right)\left(7x+4-7x+4\right)\)

\(=8\left(7x+4\right)\)

=56x+32

b: Ta có: \(8\left(x-2\right)^2-3\left(x^2-4x-5\right)-5x^2\)

\(=8x^2-32x+32-3x^2+12x+15-5x^2\)

\(=-20x+47\)

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

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

=2

a: \(=49x^2-64-10\left(4x^2+12x+9\right)+5x\left(9x^2-12x+4\right)+4x\left(x^2-10x+25\right)\)

\(=49x^2-64-40x^2-120x-90+45x^3-60x^2+20x+4x^3-40x^2+100x\)

\(=49x^3-91x^2-154\)

b: \(=27x^3+189x^2+441x+343-125x^3+y^3+x^3+6x^2y+12xy^2+8y^3\)

\(=-97x^3+189x^2+441x+6x^2y+12xy^2+9y^3+343\)

N = ( 3x² - 7x ) ( x - 2 )

N = 3x² . x - 3x² . 2 - 7x . x - 7x . ( -2 )

N = 3x³ - 6x² - 7x² + 14x

N = 3x³ - 13x² + 14x

N = ( 3x² - 7x ) ( x - 2 )

 = 3x² . x - 3x² . 2 - 7x . x - 7x . ( -2 )

 = 3x³ - 6x² - 7x² + 14x

 = 3x³ - 13x² + 14x

P/s tham khảo nha

a: \(=-2x^2\cdot3x+2x^2\cdot4X^3-2x^2\cdot7+2x^2\cdot x^2\)

\(=8x^5+2x^4-6x^3-14x^2\)

b: \(=2x^3-3x^2-5x+6x^2-9x-15\)

\(=2x^3+3x^2-14x-15\)

c: \(=\dfrac{-6x^5}{3x^3}+\dfrac{7x^4}{3x^3}-\dfrac{6x^3}{3x^3}=-2x^2+\dfrac{7}{3}x-2\)

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

e: \(=\dfrac{2x^4-8x^3-6x^2-5x^3+20x^2+15x+x^2-4x-3}{x^2-4x-3}\)

=2x^2-5x+1

a: \(\Leftrightarrow3x^3-2x^2+6x^2-4x-3x+2+a-2⋮3x-2\)

=>a-2=0

=>a=2

b: \(\Leftrightarrow3x^3-2x^2+6x^2-4x-3x+2+3⋮3x-2\)

=>\(3x-2\in\left\{1;-1;3;-3\right\}\)

mà x là số nguyên

nên x=1

c: \(\Leftrightarrow x^2+x-3x-3-a+3⋮x+1\)

=>3-a=0

=>a=3

a) \(35x^9y^n=5.\left(7x^9y^n\right)\)

Để \(35x^9y^n⋮\left(-7x^7y^2\right)\)

\(\Rightarrow n\in\left\{0;1;2\right\}\)

b) \(5x^3-7x^2+x=3x\left(\dfrac{5}{3}x^2-\dfrac{7}{3}x+\dfrac{1}{3}\right)\)

Để \(\left(5x^3-7x^2+x\right)⋮3x^n\)

\(\Rightarrow3x\left(\dfrac{5}{3}x^2-\dfrac{7}{3}x+\dfrac{1}{3}\right)⋮3x^n\)

\(\Rightarrow n\in\left\{0;1\right\}\)

A=(x+1)(x+2)(x+3)(x+4)-24

=(x²+5x+4)(x²+5x+6)-24

Đặt t=(x²+5x+4) ta có:

t(t+2)-24=t²+6t-2t-24

=t(t+6)-4(t+6)

=(t-4)(t+6).Thay vào ta đc:

(x²+5x+4-4)(x²+5x+4+6)=(x²+5x)(x²+5x+10) 

=x(x+5)(x²+5x+10)

B=(x²+3x+2)(x²+7x+120-24)

=(x²+3x+2)(x²+7x+96)

=(x²+2x+x+2)(x²+7x+96)

=[x(x+2)+(x+2)](x²+7x+96)

=(x+1)(x+2)(x²+7x+96)

C và D bn cx lm tương tự

ĐK: ` x \ne 2/7`

`(2x+3)((3x+8)/(2-7x)+1)=(x-5)((3x+8)/(2-7x)+1)`

`<=> ((3x+8)(2-7x)+1)(2x+3-x+5)=0`

`<=> ((3x+8)/(2-7x)+1)(x+8)=0`

 \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{3x+8}{2-7x}=-1\\x+8=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-8\end{matrix}\right.\)

Vậy `S={5/2 ; -8}`.