\(\left(x-10\right)^2-2\left(x-10\right)\left(5x-3\right)+\left(5x-3\right)^2\)

\(=\left(x-10-5x+3\right)^2=\left(-4x-7\right)^2=\left(4x+7\right)^2\)

\(=16x^2+56x+49\)

Ta có: \(\left(x-10\right)^2-2\left(x-10\right)\left(5x-3\right)+\left(5x-3\right)^2\)

\(=\left(x-10-5x+3\right)^2\)

\(=\left(-4x+6\right)^2\)

\(=16x^2-48x+36\)

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

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

(5x+1)(x+3)-(x-2)(5x-4)

=5x²+15x+x+3-(5x²-4x-10x+8)

=5x²+15x+x+3-5x²+4x+10x-8

=30x-5

(x^2+x-3)(x^2-x+3)

=[x²+(x-3)][x²-(x-3)]

=x²-(x-3)²

=x²-(x²-6x+9)

=x²-x²+6x-9

=6x-9

\(4x\left(x^2-5x+3\right)=4x^3-20x^2+12x\)

=> Chọn A

   `x^2-5x-2x^3+x^4+1 + (-5x^3) - 3+8x^4+x^2`

`= ( x^4 + 8x^4 ) - ( 2x^3 + 5x^3 ) + ( x^2 + x^2 ) - 5x + ( 1 - 3 )`

`= 9x^4 - 7x^3 + 2x^2 - 5x - 2`

= x²-5x-2x³+x⁴+1+(-5x³)-3+8x⁴+x²

=(x²+x²)+(-2x³-5x³)+(x⁴+8x⁴)-5x+(1-3)

=2x²+(-7x³)+9x⁴-5x+(-2)

\(Câu8\)

\(a,A=\dfrac{1}{2}x^3\times\dfrac{8}{5}x^2=\left(\dfrac{1}{2}\times\dfrac{8}{5}\right)x^{3+2}=\dfrac{4}{5}x^5\)

b, \(P\left(0\right)=0^2-5.0+6=6\\ P\left(2\right)=2^2-5.2+6=0\)

Câu 9

\(a,A\left(x\right)+B\left(x\right)=5x^3+x^2-3x+5+5x^3+x^2+2x-3\\ =\left(5x^3+5x^3\right)+\left(x^2+x^2\right)+\left(-3x+2x\right)+\left(5-3\right)\\ =10x^3+2x^2-x+2\)

\(b,H\left(x\right)=A\left(x\right)-B\left(x\right)=5x^3+x^2-3x+5-\left(5x^3+x^2+2x-3\right)\\ =5x^3+x^2-3x+5-5x^3-x^2-2x+3\\ =\left(5x^3-5x^3\right)+\left(x^2-x^2\right) +\left(-3x-2x\right)+\left(5+3\right)\\ =-5x+8\)

\(H\left(x\right)=0\\ \Rightarrow-5x+8=0\\ \Rightarrow x=\dfrac{8}{5}\)

vậy nghiệm của đa thức là \(x=\dfrac{8}{5}\)

b: \(=\dfrac{x^3+6x^2-25}{x\left(x+5\right)\left(x-2\right)}+\dfrac{x+5}{x\left(x-2\right)}\)

\(=\dfrac{x^3+6x^2-25+x^2+10x+25}{x\left(x+5\right)\left(x-2\right)}=\dfrac{x^3+7x^2+10x}{x\left(x+5\right)\left(x-2\right)}=\dfrac{x+2}{x-2}\)