Question

Cho x² - 4x + 1 = 0. Tính giá trị biểu thức A = \(\frac{x^4+x^2+1}{x^2}\)

Answer 1

\(x^2-4x+1=0\Rightarrow x\ne0\)

Khi đó: \(x^2+1=4x\Leftrightarrow x+\frac{1}{x}=4\Rightarrow\left(x+\frac{1}{x}\right)^2=16\)

\(\Rightarrow x^2+\frac{1}{x^2}+2=16\Rightarrow x^2+\frac{1}{x^2}=14\)

\(A=x^2+\frac{1}{x^2}+1=14+1=15\)

Answer 2

x^2-4x+1=0=>x khác0

khi đó:x^2+1=4x<=>x+1/x=4=>(x+1/x)^2=16

=>x^2+1/x^2+2=16=>x^2+1/x^2=14

A=x^2+1/x^2+1=14+1=15