Question

Tính:

\(A=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+...}}}}}\)

\(B=\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+...}}}}}\)

\(C=\sqrt{5+\sqrt{13+\sqrt{5+\sqrt{13+\sqrt{5+\sqrt{13+...}}}}}}\)

Answer 1

Ta có  A > 0 

Từ đó  \(A^2=2+\sqrt{2+\sqrt{2+...}}\Leftrightarrow A^2=2+A\Leftrightarrow A^2-A-2=0\)

\(\Leftrightarrow\left(A+1\right)\left(A-2\right)=0\Leftrightarrow\orbr{\begin{cases}A+1=0\\A-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}A=-1\\A=2\end{cases}}\)

Do A > 0 nên A= 2

Answer 2

b, tương tự

c,\(C>2\)

Xét \(C^2=5+\sqrt{13+\sqrt{5+\sqrt{13...}}}\)

\(\left(C^2-5\right)^2=13+C\Leftrightarrow C^4-10C^2-C+12=0\Leftrightarrow\left(C^4-9C^2\right)-\left(C^2-9\right)-\left(C-3\right)=0\)

\(\Leftrightarrow\left(C-3\right)\left[\left(C+3\right)\left(C-1\right)\left(C+1\right)-1\right]=0\)

VÌ C> 2 =>  C-3 = 0 => C=3