tìm \(x=9-\frac{1}{\sqrt{\frac{9}{4}-\sqrt{5}}}+\frac{1}{\sqrt{\frac{9}{4}+\sqrt{5}}}\)
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)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x=9-\frac{1}{\sqrt{\frac{9}{4}-\sqrt{5}}}+\frac{1}{\sqrt{\frac{9}{4}+\sqrt{5}}}\)
\(=9-\frac{2}{\sqrt{9-4\sqrt{5}}}+\frac{2}{\sqrt{9+4\sqrt{5}}}\)
\(=9-\frac{2}{\sqrt{\left(\sqrt{5}-2\right)^2}}+\frac{2}{\sqrt{\left(\sqrt{5}+2\right)^2}}\)
\(=9-\frac{2}{\sqrt{5}-2}+\frac{2}{\sqrt{5}+2}\)
\(=9-\frac{4+2\sqrt{5}-2\sqrt{5}+4}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\)
\(=9-\frac{8}{5-4}\)
= 1
\(f\left(x\right)=\left(1^4-3+1\right)^{2016}=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(\left(\frac{1}{\sqrt{\frac{9}{4}+\sqrt{5}}}-\frac{1}{\sqrt{\frac{9}{4}-\sqrt{5}}}\right)^2\)
\(=\frac{1}{\frac{9}{4}+\sqrt{5}}+\frac{1}{\frac{9}{4}-\sqrt{5}}-2.\frac{1}{\sqrt{\frac{9}{4}+\sqrt{5}}}.\frac{1}{\sqrt{\frac{9}{4}-\sqrt{5}}}\)
\(=\frac{\frac{9}{4}-\sqrt{5}+\frac{9}{4}+\sqrt{5}}{\frac{1}{16}}-2.\frac{1}{\frac{1}{4}}\)
\(=72-8=64\)
Mà \(\frac{1}{\sqrt{\frac{9}{4}+\sqrt{5}}}< \frac{1}{\sqrt{\frac{9}{4}-\sqrt{5}}}\)
\(\Rightarrow\frac{1}{\sqrt{\frac{9}{4}+\sqrt{5}}}-\frac{1}{\sqrt{\frac{9}{4}-\sqrt{5}}}< 0\)
Do đó : \(\frac{1}{\sqrt{\frac{9}{4}+\sqrt{5}}}-\frac{1}{\sqrt{\frac{9}{4}-\sqrt{5}}}=-8\)
Khi đó : \(x=9-8=1\)
Với x =1 ta có ;
\(f\left(1\right)=\left(1^4-3.1+1\right)^{2016}=\left(-1\right)^{2016}=1\)
Chúc bạn học tốt !!!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}.\)
\(\Rightarrow A^2=4+\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{2}}\right)\left(4-\sqrt{10+2\sqrt{2}}\right)}+4-\sqrt{10+2\sqrt{5}}\)
\(=8+2\sqrt{16-\left(10+2\sqrt{5}\right)}\)
\(=8+2\sqrt{6-2\sqrt{5}}\)
\(=8+2\sqrt{5-2\sqrt{5.1}+1}=8+2\left(\sqrt{5}-1\right)\)
\(=8+2\sqrt{5}-2=6+2\sqrt{5}\)
\(=\left(\sqrt{5}+1\right)^2\)
\(\Rightarrow A=\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}+1\)
\(B=\frac{1}{1+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{13}}+...+\frac{1}{\sqrt{2001}+\sqrt{2005}}\)
\(=\frac{1-\sqrt{5}}{\left(1+\sqrt{5}\right)\left(1-\sqrt{5}\right)}+\frac{\sqrt{5}-\sqrt{9}}{\left(\sqrt{5}+\sqrt{9}\right)\left(\sqrt{5}-\sqrt{9}\right)}+...+\frac{\sqrt{2001}-\sqrt{2005}}{\left(\sqrt{2001}+\sqrt{2005}\right)\left(\sqrt{2001}-\sqrt{2005}\right)}\)
\(=\frac{1-\sqrt{5}}{1-5}+\frac{\sqrt{5}-\sqrt{9}}{5-9}+...+\frac{\sqrt{2001}-\sqrt{2005}}{2001-2005}\)
\(=-\frac{1}{4}\left(1-\sqrt{5}+\sqrt{5}-\sqrt{9}+....+\sqrt{2001}-\sqrt{2005}\right)\)
\(=-\frac{1}{4}\left(1-\sqrt{2005}\right)\)
\(=10,94430659\)
\(\text{Lm hơi vắn tắt thông cảm nha!!}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1.a) \(\sqrt{x^2-4}-\sqrt{x-2}=0\)
\(\Leftrightarrow\sqrt{\left(x-2\right)\left(x+2\right)}-\sqrt{x-2}=0\)
\(\Leftrightarrow\sqrt{x-2}.\sqrt{x+2}-\sqrt{x-2}=0\)
\(\Leftrightarrow\sqrt{x-2}.\left(\sqrt{x+2}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x-2}=0\\\sqrt{x+2}-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\\sqrt{x+2}=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x+2=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)
Vậy x=2 hoặc x=-1
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{1}{4}+\frac{1}{3}:2x=-5\)
\(\frac{1}{3}:2x=\frac{-21}{4}\)
\(2x=\frac{-4}{63}\)
\(x=\frac{2}{63}\)
b) \(\left(3x-\frac{1}{4}\right)\left(x+\frac{1}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-\frac{1}{4}=0\\x+\frac{1}{2}=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{12}\\x=\frac{-1}{2}\end{cases}}\)
Vậy.........