a) 4x²(x² - 5x + 2)

= 4x².x² - 4x².5x + 4x².2

= 4x⁴ - 20x³ + 8x²

b) (2x²  - 5x + 3) : (2x - 3)

= (2x² - 3x - 2x + 3) : (2x - 3)

= [(2x² - 3x) - (2x - 3)] : (2x - 3)

= [x(2x - 3) - (2x - 3)] : (2x - 3)

= (2x - 3)(x - 1) : (2x - 3)

= x - 1

a, \(4x^2\left(x^2-5x+2\right)\\ =4x^4-20x^3+8x^2\)

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

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

=-4/3x^2+8/3-10/3

=-4/3x^2-2/3

d: \(\dfrac{3x^3-5x+2}{x-3}=\dfrac{3x^3-9x^2+9x^2-27x+22x-66+68}{x-3}\)

\(=3x^2+9x+22+\dfrac{68}{x-3}\)

a) kết quả là x^2-2x+3

b) CM NÈ:

X^2-2X+3=(X^2-2X+1)+2=(X-1)^2+2

VÌ (X-1)^2>=0 VỚI MỌI X=>(X-1)^2+2>0 VỚI MỌI x=> GIÁ TRỊ BIỂU THỨC LUÔN DƯƠNG

Lời giải:
a.

$5x-[2x+1-(2x-3)-(4x+1)]=5x-(2x+1-2x+3-4x-1)$

$=5x-(-4x+3)=5x+4x-3=9x-3$

b.

$(-3x^2+2x-1)+(4x^2-2x+3)$

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

cô giải đúng qué

a: \(x^3-x^2-14x+24\)

\(=x^3+x-12-x^2-15x+36\)

=>\(\left(x^3-x^2-14x+24\right):\left(x^3+x-12\right)=1+\frac{-x^2-15x+36}{x^3+x-12}\)

Để dư là 0 thì \(-x^2-15x+36=0\)

=>\(x^2+15x-36=0\) (1)

\(\Delta=15^2-4\cdot1\cdot\left(-36\right)=225+144=369>0\)

Do đó: (1) có hai nghiệm phân biệt là:

\(\left[\begin{array}{l}x=\frac{-15-\sqrt{369}}{2\cdot1}=\frac{-15-3\sqrt{41}}{2}\\ x=\frac{-15+3\sqrt{41}}{2}\end{array}\right.\)

b: \(x^5+4x^3+3x^2-5x+15\)

\(=x^5-x^3+3x^2+5x^3-5x+15=\left(x^3-x+3\right)\left(x^2+5\right)\)

=>\(\frac{x^5+4x^3+3x^2-5x+15}{x^3-x+3}=x^2+5\)

=>Đây là phép chia hết

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

\(=2x^4+2x^3+8x^2-5x^2-5x-20=\left(x^2+x+4\right)\left(2x^2-5\right)\)

=>\(\frac{2x^4+2x^3+3x^2-5x-20}{x^2+x+4}=2x^2-5\)

d: \(2x^4-14x^3+19x^2-20x+9\)

\(=2x^4-8x^3+2x^2-6x^3+24x^2-6x-7x^2+28x-7-42x+16\)

\(=\left(x^2-4x+1\right)\left(2x^2-6x-7\right)-42x+16\)

=>\(\frac{2x^4-14x^3+19x^2-20x+9}{x^2-4x+1}=2x^2-6x-7\) dư -42x+16

để dư bằng 0 thì -42x+16=0

=>-42x=-16

=>\(x=\frac{16}{42}=\frac{8}{21}\)

a) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x-6}{9x^2-4}\)

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

\(=\dfrac{-3x+10}{\left(3x-2\right)\left(3x+2\right)}\)

b) \(\dfrac{x+25}{2x^2-50}-\dfrac{x+5}{x^2-5x}-\dfrac{5-x}{2x^2+10x}\)

\(=\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}+\dfrac{x-5}{2x\left(x+5\right)}\)

\(=\dfrac{x^2+25x-2\left(x+5\right)^2+\left(x-5\right)^2}{2x\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{x^2+25x-2x^2-20x-50+x^2-10x+25}{2x\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{-5x-25}{2x\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{-5\left(x+5\right)}{2x\left(x-5\right)\left(x+5\right)}=\dfrac{-5}{2x\left(x-5\right)}\)

c) Ta có: \(\dfrac{1-2x}{2x}-\dfrac{4x}{2x-1}-\dfrac{3}{2x-4x^2}\)

\(=\dfrac{-\left(2x-1\right)^2-8x^2+3}{2x\left(2x-1\right)}\)

\(=\dfrac{-\left(4x^2-4x+1\right)-8x^2+3}{2x\left(2x-1\right)}\)

\(=\dfrac{-4x^2+4x-1-8x^2+3}{2x\left(2x-1\right)}\)

\(=\dfrac{-12x^2+4x+2}{2x\left(2x-1\right)}\)

Bài 3:

3: \(6x\left(x-y\right)-9y^2+9xy\)

\(=6x\left(x-y\right)+9xy-9y^2\)

\(=6x\left(x-y\right)+9y\left(x-y\right)\)

\(=\left(x-y\right)\left(6x+9y\right)\)

\(=3\left(2x+3y\right)\left(x-y\right)\)

Bài 4: