Question

Biết \(x^2-y^2=1.\) Gía trị của biểu thức :

\(A=2\left(x^6-y^6\right)-3\left(x^4+y^4\right)+1\) là A =

Answer 1

Ta có :

+ ) \(x^2-y^2=1\)

\(\Rightarrow\left(x^2-y^2\right)^3=1^3\)

\(\Rightarrow x^2-y^6-3x^2y^2\left(x^2-y^2\right)=1\)

\(\Rightarrow x^6-y^6=1+3x^2y^2\left(x^2-y^2\right)\)

\(\Rightarrow x^6-y^6=1+3x^2y^2\)

+ ) \(x^2-y^2=1\)

\(\Rightarrow\left(x^2-y^2\right)^2=1^2\)

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

\(\Rightarrow x^4+y^4=1+2x^2y^2\)

Khi đó :

\(A=2\left(x^6-y^6\right)-3\left(x^4+y^4\right)+1\)

\(=2\left(1+3x^2y^2\right)-3\left(1+2x^2y^2\right)+1\)

\(=0\)

Vậy \(A=0\).