Question

cho 2 số a,b thỏa mãn a2+b2=1 , a4+b4=1/2, tính P=a2024+b2024

Answer 1

\(a^4+b^4=\left(a^2+b^2\right)^2-2a^2b^2=\dfrac{1}{2}\)

\(\Leftrightarrow1-2a^2b^2=\dfrac{1}{2}\)

\(\Leftrightarrow a^2b^2=\dfrac{1}{4}\)

mà \(a^2+b^2=1\)

Nên \(a^2;b^2\) là nghiệm của phương trình:

\(X^2-X+\dfrac{1}{4}=0\Leftrightarrow\left(X-\dfrac{1}{2}\right)^2=0\Leftrightarrow X=\dfrac{1}{2}\)

\(\Rightarrow a^2=b^2=\dfrac{1}{2}\)

\(P=a^{2024}+b^{2024}=\left(a^2\right)^{1012}+\left(b^2\right)^{1012}=\left(\dfrac{1}{2}\right)^{1012}+\left(\dfrac{1}{2}\right)^{1012}\)

\(\Rightarrow P=2.\left(\dfrac{1}{2}\right)^{1012}=\dfrac{1}{2^{1011}}\)