a: \(x^2-5x+1=0\)
=>\(x^2+1=5x\)
\(A=x+\frac{1}{x}=\frac{x^2+1}{x}=\frac{5x}{x}=5\)
b: \(B=x^2+\frac{1}{x^2}=\left(x+\frac{1}{x}\right)^2-2\cdot x\cdot\frac{1}{x}\)
\(=5^2-2=25-2=23\)
c: \(C=x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)^3-3\cdot x\cdot\frac{1}{x}\left(x+\frac{1}{x}\right)\)
\(=\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)\)
\(=5^3-3\cdot5=125-15=110\)
d: \(D=x^4+\frac{1}{x^4}=\left(x^2+\frac{1}{x^2}\right)^2-2\cdot x^2\cdot\frac{1}{x^2}\)
\(=23^2-2=527\)
e: \(\left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right)=x^5+\frac{1}{x}+x+\frac{1}{x^5}\)
=>\(x^5+\frac{1}{x^5}+5=110\cdot23=2530\)
=>E+5=2530
=>E=2530-5=2525
f: \(F=x^6+\frac{1}{x^6}=\left(x^2+\frac{1}{x^2}\right)^3-3\cdot x^2\cdot\frac{1}{x^2}\left(x^2+\frac{1}{x^2}\right)\)
\(=23^3-3\cdot23=12098\)
giúp mình bài này với ạ,ghi từng bước giúp mình lun ạ
