Cho `P(x) = 3x^2-7x+7` `Q(x) = 4x^2-5x+3` `a)` Tính `2P(-1)-|Q(1/2)|` `b)` Tính `M(x) = Q(x) -2P(x)+H(x)` 12/08/2021 Bởi Adalyn Cho `P(x) = 3x^2-7x+7` `Q(x) = 4x^2-5x+3` `a)` Tính `2P(-1)-|Q(1/2)|` `b)` Tính `M(x) = Q(x) -2P(x)+H(x)`
Ta có : $P(-1)=3.(-1)²-7.(-1)+7=3+7+7=17$ $Q(\frac{1}{2})=4.(\frac{1}{2})²-5.\frac{1}{2}+3=1-\frac{5}{2}+3=4-\frac{5}{2}=\frac{3}{2}$ Thay $P(-1)=17$ và $Q(\frac{1}{2})=\frac{3}{2}$ vào $2P(-1)$-l$Q(\frac{1}{2})$l ta đc $2.17$-l$\frac{3}{2}$l=$34-\frac{3}{2}$=$\frac{68}{2}-\frac{3}{2}$=$\frac{65}{2}$ b)$M(x)=Q(x)-2P(x)+H(x)$ =$(4x²-5x+3)-2.(3x²-7x+7)+H(x)$ =$4x²-5x+3-6x²+14x-14+H(x)$ =$(4x²-6x²)+(14x-5x)+(3-14)+H(x)$ =$-2x²+11x-11+H(x)$ Bình luận
Ta có : $P(-1)=3.(-1)²-7.(-1)+7=3+7+7=17$
$Q(\frac{1}{2})=4.(\frac{1}{2})²-5.\frac{1}{2}+3=1-\frac{5}{2}+3=4-\frac{5}{2}=\frac{3}{2}$
Thay $P(-1)=17$ và $Q(\frac{1}{2})=\frac{3}{2}$ vào $2P(-1)$-l$Q(\frac{1}{2})$l ta đc
$2.17$-l$\frac{3}{2}$l=$34-\frac{3}{2}$=$\frac{68}{2}-\frac{3}{2}$=$\frac{65}{2}$
b)$M(x)=Q(x)-2P(x)+H(x)$
=$(4x²-5x+3)-2.(3x²-7x+7)+H(x)$
=$4x²-5x+3-6x²+14x-14+H(x)$
=$(4x²-6x²)+(14x-5x)+(3-14)+H(x)$
=$-2x²+11x-11+H(x)$