Question

Cho đa thức: \(f\left(x\right)=x^5+x^2+1\) có 5 nghiệm là \(x_1,x_2,x_3,x_4,x_5\). Tính giá trị của biểu thức: \(A=q\left(x_1\right).q\left(x_2\right).q\left(x_3\right).q\left(x_4\right).q\left(x_5\right)\) với \(g\left(x\right)=x^2-4\)

Answer 1

Chắc là \(q\left(x\right)=x^2-4????\)

\(f\left(2\right)=2^5+2^2+1=37\) ; \(f\left(-2\right)=-27\)

Do \(f\left(x\right)\) có 5 nghiệm nên f(x) có dạng:

\(f\left(x\right)=\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)\left(x-x_4\right)\left(x-x_5\right)\)

\(\Rightarrow f\left(2\right)=\left(2-x_1\right)\left(2-x_2\right)\left(2-x_3\right)\left(2-x_4\right)\left(2-x_5\right)=37\)

\(f\left(-2\right)=\left(-2-x_1\right)\left(-2-x_2\right)\left(-2-x_3\right)\left(-2-x_4\right)\left(-2-x_5\right)=-27\)

\(\Rightarrow\left(2+x_1\right)\left(2+x_2\right)\left(2+x_3\right)\left(2+x_4\right)\left(2+x_5\right)=27\)

 

\(A=\left(x_1^2-4\right)\left(x^2_2-4\right)\left(x_3^2-4\right)\left(x_4^2-4\right)\left(x^2_5-4\right)\)

\(A=-\left(2-x_1\right)\left(2-x_2\right)\left(2-x_3\right)\left(2-x_4\right)\left(2-x_5\right)\left(2+x_1\right)\left(2+x_2\right)\left(2+x_3\right)\left(2+x_4\right)\left(2+x_5\right)\)

\(A=-37.27=-999\)