Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x=\sqrt{5+\sqrt{13+\sqrt{5}+\sqrt{13+..............}}}\)
\(\Rightarrow x^2=5+\sqrt{13+\sqrt{5+\sqrt{13+.......}}}\)
\(\Rightarrow x^2-5=\sqrt{13+\sqrt{5+\sqrt{13+..........}}}\)
\(\Rightarrow x^2-5=\sqrt{13+x}\)
\(\Rightarrow x^4-10x^2+25-13-x=0\)
\(\Rightarrow x^4-10x^2-x+12=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^3+3x^2-x-4\right)=0\)
Hình như trong ngoặc có 2 nghiệm dạng lượng giác :v xài lượng giác hóa thử bạn nhé :) ko thì Cardano :))))))
![](https://rs.olm.vn/images/avt/0.png?1311)
Nhận xét x > 0
Ta có : \(x^2=5+\sqrt{5+\sqrt{13+\sqrt{5+\sqrt{13+...}}}}\)
\(\Leftrightarrow x^2-5=\sqrt{13+\sqrt{5+\sqrt{13+....}}}\)
\(\Leftrightarrow\left(x^2-5\right)^2=13+\sqrt{5+\sqrt{13+...}}\)
\(\Leftrightarrow\left(x^2-5\right)^2-13=x\)
\(\Leftrightarrow x^4-10x^2-x+12=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^3+3x^2-x-4\right)=0\)
Vì pt \(x^3+3x^2-x-4=0\) luôn có nghiệm \(x< 2\) mà \(x>\sqrt{5}>\sqrt{4}=2\)
Vậy x = 3
![](https://rs.olm.vn/images/avt/0.png?1311)
b: \(=\sqrt{5}-1-\sqrt{5}-1=-2\)
c: \(=\dfrac{\left(2\sqrt{2}+\sqrt{3}-2\sqrt{2}+\sqrt{3}\right)}{2\sqrt{3}}=1\)
d: \(=\dfrac{\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}-1-\sqrt{5}-1}{\sqrt{2}}=-\sqrt{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(H=2\sqrt{27}+\sqrt{243}-6\sqrt{12}\\ =2\cdot\sqrt{9}\cdot\sqrt{3}+\sqrt{81}\cdot\sqrt{3}-6\cdot\sqrt{4}\cdot\sqrt{3}\\ =2\cdot3\cdot\sqrt{3}+9\cdot\sqrt{3}-6\cdot2\cdot\sqrt{3}\\ =6\sqrt{3}+9\sqrt{3}-12\sqrt{3}\\ =3\sqrt{3}=\sqrt{9}\cdot\sqrt{3}=\sqrt{27}\)
\(I=\sqrt{14-2\sqrt{13}}+\sqrt{14+2\sqrt{13}}\\ =\sqrt{13-2\cdot\sqrt{13}\cdot1+1}+\sqrt{13+2\cdot\sqrt{13}\cdot1+1}\\ =\sqrt{\sqrt{13}^2-2\cdot\sqrt{13}\cdot1+1^2}+\sqrt{\sqrt{13}^2+2\cdot\sqrt{13}\cdot1+1^2}\\ =\sqrt{\left(\sqrt{13}-1\right)^2}+\sqrt{\left(\sqrt{13}+1\right)^2}\\ =\left|\sqrt{13}-1\right|+\left|\sqrt{13}+1\right|\\ =\sqrt{13}-1+\sqrt{13}+1\\ =2\sqrt{13}=\sqrt{4}\cdot\sqrt{13}=\sqrt{52}\)
\(I=\sqrt{10-4\sqrt{6}}+\sqrt{10+4\sqrt{6}}\\ =\sqrt{6-2\cdot\sqrt{6}\cdot2+4}+\sqrt{6+2\cdot\sqrt{6}\cdot2+4}\\ =\sqrt{\sqrt{6}^2-2\cdot\sqrt{6}\cdot2+2^2}+\sqrt{\sqrt{6}^2+2\cdot\sqrt{6}\cdot2+2^2}\\ =\sqrt{\left(\sqrt{6}-2\right)^2}+\sqrt{\left(\sqrt{6}+2\right)^2}\\ =\left|\sqrt{6}-2\right|+\left|\sqrt{6}+2\right|\\ =\sqrt{6}-2+\sqrt{6}+2\\ =2\sqrt{6}=\sqrt{4}\cdot\sqrt{6}=\sqrt{24}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
câu đầu có \(3-12\sqrt{6}< 0\) nên không căn được nên đề bạn sai
\(\sqrt{31-8\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)
\(=\sqrt{4^2-2.4.\sqrt{15}+\left(\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}\right)^2-2.\sqrt{15}.3+3^2}\)
\(=\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}-3\right)^2}=\left|4-\sqrt{15}\right|+\left|\sqrt{15}-3\right|\)
\(=4-\sqrt{15}+\sqrt{15}-3=1\)
\(\sqrt{49-5\sqrt{96}}-\sqrt{49+5\sqrt{96}}=\sqrt{49-20\sqrt{6}}-\sqrt{49+20\sqrt{6}}\)
\(=\sqrt{5^2-2.5.2\sqrt{6}+\left(2\sqrt{6}\right)^2}-\sqrt{5^2+2.5.4\sqrt{6}+\left(2\sqrt{6}\right)^2}\)
\(=\sqrt{\left(5-2\sqrt{6}\right)^2}-\sqrt{\left(5+2\sqrt{6}\right)^2}=\left|5-2\sqrt{6}\right|-\left|5+2\sqrt{6}\right|\)
\(=5-2\sqrt{6}-5-2\sqrt{6}=-4\sqrt{6}\)
\(\sqrt{31-8\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)
\(=4-\sqrt{15}+\sqrt{15}-3\)
=1
![](https://rs.olm.vn/images/avt/0.png?1311)
b: \(=\dfrac{\sqrt{5}+1}{\sqrt{5}-1}+\dfrac{\sqrt{5}-1}{\sqrt{5}+1}\)
\(=\dfrac{6+2\sqrt{5}+6-2\sqrt{5}}{4}=\dfrac{12}{4}=3\)
c: \(=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)
\(=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{43+30\sqrt{2}}\)
e: \(=\dfrac{2\sqrt{3+\sqrt{5-2\sqrt{3}-1}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{\sqrt{2}\cdot\sqrt{3+\sqrt{3}-1}}{\sqrt{3}-1}=\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}-1}=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\)
\(=\dfrac{4-2\sqrt{3}}{2}=2-\sqrt{3}\)
Mih chỉ lm đc câu R thôi:
\(R=\sqrt{5+\sqrt{13+\sqrt{5+\sqrt{13+\sqrt{5...}}}}}\)
\(\Rightarrow R^2=5+\sqrt{13+\sqrt{5+\sqrt{13+\sqrt{5...}}}}\)
\(\Rightarrow\left(R^2-5\right)^2=13+\sqrt{5+\sqrt{13+\sqrt{5...}}}\)
\(\Rightarrow R^4-10R^2+12=R\) (Vì R là lặp lại vô hạn cách viết nên nếu mũ chẵn lên thì R vẫn là R)
\(\Rightarrow\left(R-3\right)\left(R^3+3R^2-R-4\right)=0\)
Mà \(R^3+3R^2-R-4=\left(R+3\right)\left(R-1\right)\left(R+1\right)-1>0\forall R>\sqrt{5}\)
Nên ta dễ dàng suy ra đc R-3=0 => R=3
câu R có trên đienantoanhoc òi