Question

[Lục Hoàng Phong]

Cho đa thức P(x) = \(x^5+ax^4\:+bx^3+cx^2+dx+e\)

Và cho biết: \(P\left(1\right)=1;P\left(2\right)=4;P\left(3\right)=9;P\left(4\right)=16;P\left(5\right)=25\)

Tính: \(P\left(6\right);P\left(7\right);P\left(8\right);P\left(9\right)\)

Answer 1

[Lightning Farron]

Ta có: \(P\left(x\right)=x^5+ax^4+bx^3+cx^2+dx+e\)

Suy ra \(P\left(1\right)=1^5+a\cdot1^4+b\cdot1^3+c\cdot1^2+d\cdot1+e=1\)

\(\Rightarrow a+b+c+d+e=0\)

\(P\left(2\right)=2^5+a\cdot2^4+b\cdot2^3+c\cdot2^2+d\cdot2+e=4\)

\(\Rightarrow16a+8b+4c+2d+e+28=0\)

\(P\left(3\right)=3^5+a\cdot3^4+b\cdot3^3+c\cdot3^2+d\cdot3+e=9\)

\(\Rightarrow81a+27b+9c+3d+e+234=0\)

\(P\left(4\right)=4^5+a\cdot4^4+b\cdot4^3+c\cdot4^2+d\cdot4+e=16\)

\(\Rightarrow256a+64b+16c+4d+e+1008=0\)

\(P\left(5\right)=5^5+a\cdot5^4+b\cdot5^3+c\cdot5^2+d\cdot5+e=25\)

\(\Rightarrow625a+125b+25c+5d+e+999=0\)

Thay lẫn lộn vào nhau đi nhé

Answer 2

[Aki Tsuki]

Cho phép lm tiếp....

\(\Rightarrow\left\{{}\begin{matrix}15a+7b+3c+d=-28\\80a+26b+8c+2d=-234\\255a+63b+15c+3d=-1008\\624a+124b+24c+4d=-3100\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}50a-12b+2c=-178\\210a+42b+6c=-924\\564a+96b+12c=-2988\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a=-15\\b=85\\c=-224\end{matrix}\right.\)

Thay bào pt \(15a+7b+3c+d=-28\) ta có: \(-225+595-672+d=-28\Rightarrow d=274\)

Thay vào pt \(a+b+c+d+e=0\) ta có:

\(-15+85-224+274+e=0\Rightarrow e=-120\)

Thay a,b,c,d,e vào r` tính là ra!

p/s: cho a,b,c bấm casio nhé!