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

\(\Leftrightarrow x=\sqrt{5+\sqrt{13+x}}\) (\(x\ge0\))

\(\Leftrightarrow x^2=5+\sqrt{13+x}\)

\(\Leftrightarrow x^2-9=\sqrt{13+x}-4\)

\(\Leftrightarrow\left(x-3\right).\left(x+3\right)=\dfrac{x-3}{\sqrt{13+x}+4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x+3=\dfrac{1}{\sqrt{x+13}+4}\left(∗\right)\end{matrix}\right.\)

Xét (*) ta có VT \(\ge3\) (1)

mà \(VP=\dfrac{1}{\sqrt{x+13}+4}\le\dfrac{1}{4}\) (2)

Từ (1) và (2) dễ thấy (*) vô nghiệm 

Hay x = 3

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

\(\Rightarrow x^2-5=\sqrt{13+\sqrt{5+\sqrt{13...}}}\)

\(\Rightarrow x^4-10x^2+25-13=x\)

\(\Leftrightarrow x^4-10x^2-x+12=0\)

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

Dễ thấy \(x=\sqrt{5+\sqrt{13+\sqrt{5+\sqrt{13...}}}}>\sqrt{4}=2\)nên \(\left(x+3\right)\left(x+1\right)\left(x-1\right)-1>5\cdot3\cdot1-1=14>0\)nên x = 3

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

\(\Leftrightarrow x^2-5=\sqrt{13+\sqrt{5+\sqrt{13+...}}}\)

\(\Leftrightarrow\left(x^2-5\right)^2=13+x\)

\(\Leftrightarrow x^4-10x^2-x+12=0\)

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

do x>2 nen x=3

online ngu 

Bạn Tuyển Trần Thị cho mình hỏi là x > 2 ở đâu vậy?

dk  \(x>2\)

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

    \(\left(x^2-5\right)^2=13+x\)

\(\Leftrightarrow x^4-10x^2-x+12=0\)

\(\Leftrightarrow\left(x^4-9x^2\right)-\left(x^2-9\right)-\left(x-3\right)=0\)

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

tiếp :  vì \(x>2\Rightarrow\left(x+3\right)\left(x+1\right)\left(x-1\right)-1>0\)

do đó \(x-3=0\Leftrightarrow x=3\)

x= √5+√13+√5+√13=√5+√13+√5+√16=

 = √5+√13+√5+4=√5+√13+√9=√5+√13+3

x= \(\sqrt{5+\sqrt{13+\sqrt{5+\sqrt{13}}}}=\sqrt{5+\sqrt{13+\sqrt{5+\sqrt{16}}}}=\)

 = \(\sqrt{5+\sqrt{13+\sqrt{5+4}}}=\sqrt{5+\sqrt{13+\sqrt{9}}}=\)\(\sqrt{5+\sqrt{13+3}}\)

= \(\sqrt{5+\sqrt{16}}=\sqrt{5+4}=\sqrt{9}=3\)

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

Nhận xét : x > 0

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

\(\Leftrightarrow\left(x^2-5\right)^2-13=\sqrt{5+\sqrt{13+\sqrt{5+...}}}=x\)

\(\Rightarrow x^4-10x^2-x+12=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^3+3x^2-x-4\right)=0\)

Suy ra x = 3

Chú ý : Ta có \(x>\sqrt{5}>\sqrt{4}=2\)

Do đó , các nghiệm của pt \(x^3+3x^2-x-4=0\) không thỏa mãn

x xấp xỉ bằng 3

Dễ dàng nhận thấy \(x>0\)

a/ \(x^2=6+\sqrt{6+\sqrt{6+...}}\)

\(\Leftrightarrow x^2=6+x\)

\(\Leftrightarrow x^2-x-6=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)

\(\Leftrightarrow x=3\)

b/ \(x^2=5+\sqrt{13+\sqrt{5+\sqrt{13+...}}}\)

\(\Leftrightarrow x^2=5+\sqrt{13+x}\)

\(\Leftrightarrow x^2-5=\sqrt{x+13}\) (\(x\ge\sqrt{5}\))

\(\Leftrightarrow\left(x^2-5\right)^2=x+13\)

\(\Leftrightarrow x^4-10x^2-x+12=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^3+3x^2-x-4\right)=0\)

Do \(x\ge\sqrt{5}\Rightarrow\left\{{}\begin{matrix}x-1>0\Rightarrow x^3-x=x^2\left(x-1\right)>0\\x^2\ge5\Rightarrow3x^2-4>0\end{matrix}\right.\)

\(\Rightarrow x^3+3x^2-x-4>0\)

\(\Rightarrow x=3\)

bài 2 là bài 21 trong nâng cao phát triển toán 9, chắc bạn có chứ

Bài 1: Ta có:

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

\(\Rightarrow x^2-5=\sqrt{13+x}\Rightarrow x^4-10x^2+25=13+x\Leftrightarrow x^4-10x^2-x+12=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^3+3x^2-x-4\right)=0\)

Pt này có 1 nghiệm x = 3 và 3 nghiệm nhỏ hơn 2.

Vì \(x>\sqrt{4}=2\)

Vậy x = 3.

b2

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

\(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}=1+\frac{1}{2}-\frac{1}{3}\)

.....................................................................

\(\sqrt{1+\frac{1}{2013^2}+\frac{1}{2014^2}}=1+\frac{1}{2013}-\frac{1}{2014}\)

BT = 2012-1/2014