Question

1)Tính GT biểu thức 3x⁴-5x³y+6x²-10xy+2 với x,y thỏa mãn 3x-5y=0

2)Tính x⁵-2010x⁴+ 2010x³ -2010x²⁺2010x-2020 với x = 2009

 

Answer 1

1)3x⁴-5x³y+6x²-10xy+2

=(3x⁴-5x³y)+(6x²-10xy)+2

=x³(3x-5y)+2x(3x-5y)+2

=x³.0+2x.0+2

=0+0+2

=2

2) x⁵-2010x⁴+2010x³-2010x²+2010x-2020

=x⁵-(2009+1)x⁴+(2009+1)x³-(2009+1)x²+(2009+1)x-2009-11

=x⁵-(x+1)x⁴+(x+1)x³-(x+1)x²+(x+1)x-x-11

=x⁵-x⁵-x⁴+x⁴+x³-x³-x²+x²+x-x-11

=-11

Answer 2

2, Với x= 2009 => 2010=x+1

=> \(x^5-2010\text{x}^4+2010\text{x}^3-2010\text{x}^2+2010\text{x}-2020=x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-2020\)

\(=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-2020\)

\(=x-2020\)

\(=2009-2020\\ =-11\)