Yêu cầu đề là gì vậy bạn?

a: \(=2x^3-14x^2-6x\)

c: \(=-10x^5-15x^4+25x^3\)

a) 2x. (x2 – 7x -3)

= 2x³- 14x²- 6x

b) ( -2x3 + y2 -7xy). 4xy2 

= -8x⁴y²+ 4xy⁴- 28x²y³

c)(-5x3).(2x2+3x-5)

= -10x⁵-15x⁴+25x³

d) (2x2 - xy+ y2).(-3x3)

=-6x⁵+ 3x⁴y -3x³y²

e)(x2 -2x+3). (x-4) 

=x³-2x²+3x -4x²+8x-12

=x³-6x²+11x-12

f) ( 2x3 -3x -1). (5x+2)

=10x⁴-15x²-5x +4x³-6x-2

=10x⁴+4x³-15x²-11x-2

A) -2x(3x+2)(3x-2)+5(x+2)² - (x-1)(2x+1)(2x+1)

= -2x(9x²-4)+5(x²+4x+4) - (x-1)(4x²-1)

= -18x³+8x+5x²+20x+20-(4x³-x-4x²+1)

= -18x³+5x²+28x+20-4x³+x+4x²+1

= -22x³+9x²+29x+21

B) (7x-8)(7x+8)-10(2x+3)²+5x(3x-2)²-4x(x-5)²

= 49x² - 64 -10(4x²+ 12x + 3) + 5x(9x² - 12x +4) - 4x(x² - 10x +25)

= 49x² - 64 -40x² - 120x - 30 + 45x³ - 60x² - 20x - 4x³ + 40x² -100x

= 41x³ -11x² -240x -94

C) \(\left(x^2-3\right)\left(x^2+3\right)-5x^2\left(x+1\right)^2-\left(x^2-3x\right)\left(x^2-2x\right)+4x\left(x+2\right)^2\)

\(\left(x^4-9\right)-5x^2\left(x^2+2x+1\right)-\left(x^4-2x^3-3x^3+6x^2\right)+4x\left(x^2+4x+4\right)\)

\(x^4-9-5x^4-10x^3-5x^2-x^4+5x^3-6x^2+4x^3+16x^2+16x\)

\(-5x^4-x^3+5x^2+20x-9\)

D) \(-6x^2\left(x+5\right)^2-\left(x-3\right)^2+\left(x^2-2\right)\left(2x^2+1\right)-4x^2\left(3x-4\right)^2\)

\(-6x^2\left(x^2+10x+25\right)-\left(x^2-6x+9\right)+2x^4-3x^2-2-4x^2\left(9x^2-24x+16\right)\)

\(-6x^4-60x^3+150x^2-x^2+6x-9+2x^4-3x^2-2-36x^4+96x^3-64x^2\)

\(-40x^4+36x^3+82x^2+6x-11\)

\(a\\ -5x^2+3x.\left(x+2\right)=-5x^2+3x^2+6x=-2x^2+6x\\ b\\ -2x.\left(1-x^2\right)-2x^3=-2x+2x^3-2x^3=-2x\\ c\\ 4x.\left(x-1\right)-4.\left(x^2+2x-1\right)\\ =4x^2-4x-4x^2-8x+4=-12x+4\)

\(d\\ 6x^3-2x^2.\left(-x^2-3x\right)=6x^3+2x^4+6x^3=2x^4+12x^3\\ e\\ 3x.\left(x-1\right)-\left(1+2x\right).5x\\ =3x^2-3x-5x-10x^2=-7x^2-8x\\ f\\ -5x^2-\left(x-6\right).\left(-2x^2\right)=-5x^2+2x^3-12x^2=2x^3-17x^2\)

a) 2x(x+3) – 3x²(x+2) + x(3x² + 4x – 6)

= (2x . x + 2x . 3) – (3x² . x + 3x² . 2) + (x . 3x² + x . 4x – x . 6)

= 2x² + 6x – (3x³ + 6x²) + (3x³ + 4x² - 6x)

= 2x² + 6x – 3x³ – 6x² + 3x³ + 4x² - 6x

= (– 3x³ + 3x³ ) + (2x²  - 6x² + 4x² ) + (6x – 6x)

= 0 + 0 + 0

= 0

b) 3x(2x² – x) – 2x²(3x+1) + 5(x² – 1)

= [3x . 2x² + 3x . (-x)] – (2x² . 3x + 2x² . 1) + [5x² + 5 . (-1)]

= 6x³ – 3x² – (6x³ +2x²) + 5x² – 5

= 6x³ – 3x² – 6x³ - 2x² + 5x² – 5

= (6x³ – 6x³ ) + (-3x² – 2x² + 5x²) – 5

= 0 + 0 – 5

= - 5

lớp 8 có pt bậc 2 ak??

\(m,x^2+6x-16=0\)

\(\Leftrightarrow x^2-2x+8x-16=0\)

\(\Leftrightarrow x\left(x-2\right)+8\left(x-2\right)=0\)

\(\Leftrightarrow\left(x+8\right)\left(x-2\right)=0\)

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

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

\(n,2x^2+5x-3=0\)

\(\Leftrightarrow2x^2-x+6x-3=0\)

\(\Leftrightarrow x\left(2x-1\right)+3\left(2x-1\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(2x-1\right)=0\)

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

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

\(k,x\left(2x-7\right)-4x+14=0\)

\(\Leftrightarrow2x^2-4x-7x+14=0\)

\(\Leftrightarrow2x\left(x-2\right)-7\left(x-2\right)=0\)

\(\Leftrightarrow\left(2x-7\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-7=0\\x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\end{matrix}\right.\)

`@` `\text {Ans}`

`\downarrow`

^{*Máy tớ cam hơi mờ, cậu thông cảm ._.*}

Cậu viết lại rõ đề câu c, nhé.