Question

Cho hai đa thức P(x)= x⁵-5x³+4x+1, Q(x)=2x²+x-1. Gọi x1,x2,x3,x4,x5 là các ng của P(x)

Tính Q(x1).Q(x2).Q(x3).Q(x4).Q(x5)

Answer 1

Vì P(x) có hệ số bậc cao nhất là 1

Nên P(x) có thể được viết dưới dạng: \(P\left(x\right)=\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)\left(x-x_4\right)\left(x-x_5\right)\)

Và \(P\left(-1\right)=\left(-1\right)^5-5\left(-1\right)^3+4\left(-1\right)+1=1\)

\(P\left(\frac{1}{2}\right)=\frac{77}{32}\)

Ta có: \(Q\left(x\right)=2x^2+x-1=2x^2+2x-x-1=2x\left(x+1\right)-\left(x+1\right)=\left(x+1\right)\left(2x-1\right)\)

=> \(Q\left(x_1\right).\text{​​}\text{​​}Q\left(x_2\right).\text{​​}\text{​​}Q\left(x_3\right).\text{​​}\text{​​}Q\left(x_4\right).\text{​​}\text{​​}Q\left(x_5\right)\text{​​}\text{​​}\)

\(=\left(x_1+1\right)\left(2x_1-1\right)\left(x_2+1\right)\left(2x_2-1\right)\left(x_3+1\right)\left(2x_3-1\right)\left(x_4+1\right)\left(2x_4-1\right)\left(x_5+1\right)\left(2x_5-1\right)\)

\(=32\left(-1-x_1\right)\left(\frac{1}{2}-x_1\right)\left(-1-x_2\right)\left(\frac{1}{2}-x_2\right)\left(-1-x_3\right)\left(\frac{1}{2}-x_3\right)\left(-1-x_4\right)\left(\frac{1}{2}-x_4\right)\left(-1-x_5\right)\left(\frac{1}{2}-x_5\right)\)\(=32.P\left(-1\right).P\left(\frac{1}{2}\right)=32.1.\frac{77}{32}=77\)

Answer 2

\(p\left(x\right)=x^5-5x^3+4x+1=\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)\left(x-x_4\right)\left(x-x_5\right)\)

\(Q\left(x\right)=2\left(\frac{1}{2}-x\right)\left(-1-x\right)\)

Do đó \(Q\left(x_1\right)\cdot Q\left(x_2\right)\cdot Q\left(x_3\right)\cdot Q\left(x_4\right)\cdot Q\left(x_5\right)\)

\(=2^5\left[\left(\frac{1}{2}-x_1\right)\left(\frac{1}{2}-x_2\right)\left(\frac{1}{2}-x_3\right)\left(\frac{1}{2}-x_4\right)\left(\frac{1}{2}-x_5\right)\right]\)

\(=\left(-1-x_1\right)\left(-1-x_2\right)\left(-1-x_3\right)\left(-1-x_4\right)\left(-1-x_5\right)\)

\(=32P\left(\frac{1}{2}\right)\cdot\left[P\left(-1\right)\right]\)

\(=32\cdot\left(\frac{1}{32}-\frac{5}{8}+\frac{4}{2}+1\right)\left(-1+5-4+1\right)\)

\(=4300\)

*Mình không chắc*