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

\(=4x^2-12x+9-x^3+5x^2-4x+3x^2-15x+12\)

\(=-x^3+12x^2-31x+21\)

\(a,3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)

\(=3x^2-6x-5x+5x^2-8x^2+24\)

\(=\left(3x^2+5x^2-8x^2\right)+\left(-6x-5x\right)+24\)

\(=0-11x+24\)

\(=-11x+24\)

\(b,\left(7x-3\right)\left(2x+1\right)-\left(5x-2\right)\left(x+4\right)-9x^2+17x\)

\(=14x^2+7x-6x-3-5x^2-20x+2x+8-9x^2+17x\)

\(=\left(14x^2-5x^2-9x^2\right)+\left(7x-6x-20x+2x+17x\right)+\left(-3+8\right)\)

\(=0+0+5\)

\(=5\)

hello

3x(x – 2) – 5x(1 – x) – 8( x 2  – 3)

= 3x.x + 3x .( -2) – [5x.1 + 5x. (- x)] – [8 x 2  + 8.(- 3)]

= (3 x 2  – 6x) – (5x – 5 x 2 ) – (8 x 2  – 24)

= 3 x 2  – 6x – 5x + 5 x 2  – 8 x 2 + 24

= ( 3 x 2  +5 x 2  – 8 x 2 )- ( 6x + 5x) + 24

= - 11x + 24

\(a,x\left(2x^2-3\right)-x^2\left(5x+1\right)+x^2\)

\(=2x^3-3x-5x^3-x^2+x^2\)

\(=-3x^3-3x\)

\(b,3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)

\(=3x^2-6x-5x+5x^2-8x^2+24\)

\(=-11x+24\)

=.= hok tốt!!

a)

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

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

b)

Nhân ra:

\(3x^2-6x-5x+5x^2-8x^2+24=-11x+24\)

Chúc bạn học tốt!

a)

ĐKXĐ: \(x\notin\left\{0;3;-3\right\}\)

Ta có: \(A=\left(\dfrac{1}{3}+\dfrac{3}{x^2-3x}\right):\left(\dfrac{x^2}{27-3x^2}+\dfrac{1}{x+3}\right)\)

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

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

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

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

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

b) Để A nguyên thì \(-x-3⋮x\)

mà \(-x⋮x\)

nên \(-3⋮x\)

\(\Leftrightarrow x\inƯ\left(-3\right)\)

\(\Leftrightarrow x\in\left\{1;-1;3;-3\right\}\)

Kết hợp ĐKXĐ, ta được: \(x\in\left\{1;-1\right\}\)

Vậy: Để A nguyên thì \(x\in\left\{1;-1\right\}\)

a: \(=2x^3-3x-5x^3-x^2+x^2=-3x^3-3x\)

b: \(=3x^2-6x-5x+5x^2-8x^2+24\)

=-11x+24