Question

Cho đa thức \(P\left(x\right)=x^4+ãx^3+bx^2+cx+d\)có P(1)=7, P(2)=28,P(3)=63 TÍnh\(A=\frac{P\left(100\right)+P\left(-96\right)}{8}\)

Answer 1

Ta có:

\(P\left(1\right)=7=7.1^2\); \(P\left(2\right)=28=7.2^2\); \(P\left(3\right)=63=7.3^2\)

\(\Rightarrow\)Đặt \(g\left(x\right)=7x^2\).

Đặt \(Q\left(x\right)=P\left(x\right)-g\left(x\right)\).

Ta có:

\(Q\left(1\right)=Q\left(2\right)=Q\left(3\right)=0\)

\(\Rightarrow x=1;x=2;x=3\)là các nghiệm của đa thức Q(x)

\(\Rightarrow Q\left(x\right)⋮\left(x-1\right);\left(x-2\right);\left(x-3\right)\)

Do Q(x) là đa thức bậc 4 có hệ số cao nhất bằng 1 nên

\(Q\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-m\right).\)

\(\Rightarrow P\left(x\right)=Q\left(x\right)+g\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-m\right)+7x^2\)

Ta có:

\(P\left(100\right)=\left(100-1\right)\left(100-2\right)\left(100-3\right)\left(100-m\right)+7.100^2\)

\(=99.98.97\left(100-m\right)+7.100^2==97.98.99.100-97.98.99m+7.100^2\)

\(P\left(-96\right)=\left(-96-1\right)\left(-96-2\right)\left(-96-3\right)\left(-96-m\right)+7.\left(-96\right)^2\)

\(=\left(-97\right).\left(-98\right).\left(-99\right).\left(-96-m\right)+7.96^2\)

\(=\left(-96\right).\left(-97\right).\left(-98\right).\left(-99\right)-\left(-97\right).\left(-98\right).\left(-99\right).m+7.96^2\)

\(=96.97.98.99+97.98.99m+7.96^2\)

\(A=\frac{P\left(100\right)+P\left(-96\right)}{8}\)

\(=\frac{97.98.99.100-97.98.99m+7.100^2+96.97.98.99+97.98.99m+7.96^2}{8}\)

\(=\frac{97.98.99\left(100+96\right)+7.\left(100^2+96^2\right)}{8}=112244867\)

Answer 2

Ta có:

\(P\left(1\right)=7=7.1^2\); \(P\left(2\right)=28=7.2^2\); \(P\left(3\right)=63=7.3^2\)

\(\Rightarrow\)Đặt \(g\left(x\right)=7x^2\).

Đặt \(Q\left(x\right)=P\left(x\right)-g\left(x\right)\).

Ta có:

\(Q\left(1\right)=Q\left(2\right)=Q\left(3\right)=0\)

\(\Rightarrow x=1;x=2;x=3\)là các nghiệm của đa thức Q(x)

\(\Rightarrow Q\left(x\right)⋮\left(x-1\right);\left(x-2\right);\left(x-3\right)\)

Do Q(x) là đa thức bậc 4 có hệ số cao nhất bằng 1 nên

\(Q\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-m\right).\)

\(\Rightarrow P\left(x\right)=Q\left(x\right)+g\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-m\right)+7x^2\)

Ta có:

\(P\left(100\right)=\left(100-1\right)\left(100-2\right)\left(100-3\right)\left(100-m\right)+7.100^2\)

\(=99.98.97\left(100-m\right)+7.100^2==97.98.99.100-97.98.99m+7.100^2\)

\(P\left(-96\right)=\left(-96-1\right)\left(-96-2\right)\left(-96-3\right)\left(-96-m\right)+7.\left(-96\right)^2\)

\(=\left(-97\right).\left(-98\right).\left(-99\right).\left(-96-m\right)+7.96^2\)

\(=\left(-96\right).\left(-97\right).\left(-98\right).\left(-99\right)-\left(-97\right).\left(-98\right).\left(-99\right).m+7.96^2\)

\(=96.97.98.99+97.98.99m+7.96^2\)

\(A=\frac{P\left(100\right)+P\left(-96\right)}{8}\)

\(=\frac{97.98.99.100-97.98.99m+7.100^2+96.97.98.99+97.98.99m+7.96^2}{8}\)

\(=\frac{97.98.99\left(100+96\right)+7.\left(100^2+96^2\right)}{8}=112244867\)