\(a,2x^3-8x^2+8x\)

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

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

\(=2x\left(x-2\right)-4x\left(x-2\right)\)

\(=\left(2x-4x\right)\left(x-2\right)\)

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

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

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

\(c,x^2y-x^3-9y+9x\)

\(=\left(x^2y-x^3\right)-\left(9y-9x\right)\)

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

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

\(\left(12x^2y^3-15x^4y^3+9x^2y^2\right):3xy^2\)

\(=4xy-5x^3y+3x\)

Độ dài của đường kính EF là 20cm

a)  4x²-y²+2y-1

=4x² -(y²-2y+1)

=(2X)² -(y -1)²

=(2x-y+1)(2x+y-1)

b) 5x(x-2)-(2-x)

=5x(x-2)+(x-2)

=(x-2)(5x+1)

\(x^2-10x+25\)

\(=x^2-2\cdot x\cdot5+5^2\)

\(=\left(x-5\right)^2\)

x^2 - 10x + 25

= x^2 - 5x - 5x + 5^2

= x(x - 5) - 5(x - 5)

= (x - 5)(x - 5)

\(x+y=\frac{1}{3}\Leftrightarrow\left(x+y\right)^3=\frac{1}{27}\)

\(\Leftrightarrow x^3+y^3+3xy\left(x+y\right)=\frac{1}{27}\)

\(\Leftrightarrow x^3+y^3+xy.3.\frac{1}{3}=\frac{1}{27}\)

\(\Leftrightarrow x^3+y^3+xy=\frac{1}{27}\)

Do đó \(B=\frac{1}{27}\)

Có: x³ + y³ = (x + y)³ - 3xy (x + y)

=> B = x³ + y³ + xy 

         = (x + y)³ - 3xy (x + y) + xy

         = (1/3)³ - 3xy . 1/3 + xy (do x + y =1/3)

         = 1/9 - xy + xy

         = 1/9

\(Q=\frac{10x^2+6x+3}{x^2+2}\)

\(Q=\frac{x^2+2+9x^2+6x+1}{x^2+2}\)

\(Q=\frac{x^2+2}{x^2+2}+\frac{9x^2+6x+1}{x^2+2}\)

\(Q=1+\frac{\left(3x+1\right)^2}{x^2+2}\)

Vì \(\left(3x+1\right)^2\ge0\forall x\)

\(\Rightarrow Q\ge1+0=1\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow3x+1=0\Leftrightarrow x=\frac{-1}{3}\)

Vậy \(Q_{min}=1\Leftrightarrow x=\frac{-1}{3}\)

ĐỂ x⁴ - x³ + 6x² -x \(⋮x^2-x+5\)

\(\Rightarrow x-5=0\Rightarrow x=5\)

b , ta có : \(3x^3+10x^2-5⋮3x+1\)

\(\Rightarrow3x^3+x^2+9x^2+3x-3x-1-4⋮3x+1\)

\(\Rightarrow x\left(3x+1\right)+3x\left(3x+1\right)-\left(3x+1\right)-4⋮3x+1\)

mà : \(\left(3x+1\right)\left(4x-1\right)⋮3x+1\)

\(\Rightarrow4⋮3x+1\Rightarrow3x+1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

Nếu : 3x + 1 = 1 => x = 0 ( TM ) 

    3x + 1 = -1 => x = -2/3 ( loại ) 

    3x + 1 = 2 => x = 1/3 ( loại ) 

  3x + 1 = -2 => x = -1 ( TM ) 

 3x + 1 = 4 => x = 1 ( TM ) 

3x + 1 = -1 => x = -5/3 ( loại ) 

\(\Rightarrow x\in\left\{0;\pm1\right\}\)