Question

a/ Chứng minh rằng : 

\(x^7+\frac{1}{x^7}=\left(x^4+\frac{1}{x^4}\right)\left(x^3+\frac{1}{x^3}\right)-\left(x+\frac{1}{x}\right)\)

b. Cho x>0 thõa mãn \(x+\frac{1}{x}=7\)

Tính \(x^5+\frac{1}{x^5}\)

Help nhanh mình tick nhanh nhé!!!!

Answer 1

a/ \(\left(x^4+\frac{1}{x^4}\right)\left(x^3+\frac{1}{x^3}\right)-\left(x+\frac{1}{x}\right)\)

\(=x^7+x+\frac{1}{x}+\frac{1}{x^7}-\left(x+\frac{1}{x}\right)=x^7+\frac{1}{x^7}\)

b/ Ta có:

\(\left(x+\frac{1}{x}\right)^2=49\)

\(\Leftrightarrow x^2+\frac{1}{x^2}=49-2=47\)

\(\left(x+\frac{1}{x}\right)^3=343\)

\(\Leftrightarrow x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x}\right)=343\)

\(\Leftrightarrow x^3+\frac{1}{x^3}=343-3.7=322\)

\(\Rightarrow\left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right)=47.322=15134\)

\(\Leftrightarrow x^5+\frac{1}{x}+x+\frac{1}{x^5}=15134\)

\(\Leftrightarrow x^5+\frac{1}{x^5}=15134-7=15127\)

Answer 2

a)\(\left(x^4+\frac{1}{x^4}\right)\left(x^3+\frac{1}{x^3}\right)-\left(x+\frac{1}{x}\right)=x^7+x+\frac{1}{x}+\frac{1}{x^7}-x-\frac{1}{x}\)

=\(x^7+\frac{1}{x^7}\)

\(x+\frac{1}{x}=7\)

=>\(x\left(x+\frac{1}{x}\right)=7x\)

=>\(^{x^2-7x+1=0}\)

=>\(x=\frac{7+3\sqrt{5}}{2};x=\frac{7-3\sqrt{5}}{2}loại\)

=>\(x^5+\frac{1}{x^5}=15127\)