Bài 1: Căn bậc hai

a, \(A=\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)

\(\dfrac{A}{\sqrt{2}}=\dfrac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\dfrac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\)

\(\dfrac{A}{\sqrt{2}}=\dfrac{2+\sqrt{3}}{3+\sqrt{3}}+\dfrac{2-\sqrt{3}}{3-\sqrt{3}}\)

\(\dfrac{A}{\sqrt{2}}=\dfrac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{6}=1\)

\(\Rightarrow A=\sqrt{2}\)

nếu k có ai làm chiều tớ làm cho h lười quá

\(B=\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{5}}\)

\(=\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}-\sqrt{23-4\sqrt{5}}\)

\(=\sqrt{5}-\sqrt{3}-\sqrt{23-4\sqrt{5}}\)

a) \(\sqrt{\dfrac{2-\sqrt{5}}{\sqrt{5}-3}}:\sqrt{\left(\sqrt{5}-3\right)\left(2-\sqrt{5}\right)}\)

\(=\sqrt{\dfrac{2-\sqrt{5}}{\sqrt{5}-3}}:\left(\left(\sqrt{5}-3\right)\cdot\left(2-\sqrt{5}\right)\right)\)

\(=\sqrt{\dfrac{2-\sqrt{5}}{\sqrt{5}-3}:\left(2\sqrt{5}-5-6+3\sqrt{5}\right)}\)

\(=\sqrt{\dfrac{2-\sqrt{5}}{\sqrt{5}-3}:\left(5\sqrt{5}-11\right)}\)

\(=\sqrt{\dfrac{2-\sqrt{5}}{\sqrt{5}-3}\cdot\dfrac{1}{5\sqrt{5}-11}}\)

\(=\sqrt{\dfrac{2-\sqrt{5}}{\left(\sqrt{5}-3\right)\cdot\left(5\sqrt{5}-1\right)}}\)

\(=\sqrt{\dfrac{\left(2-\sqrt{5}\right)\cdot\left(\sqrt{5}+3\right)}{-4\left(5\sqrt{5}-1\right)}}\)

\(=\sqrt{\dfrac{2\sqrt{5}+6-5-3\sqrt{5}}{-4\left(5\sqrt{5}-11\right)}}\)

\(=\sqrt{\dfrac{-\sqrt{5}+1}{-4\left(5\sqrt{5}-11\right)}}\)

\(=\sqrt{-\dfrac{\left(-\sqrt{5}+1\right)\cdot\left(5\sqrt{5}+11\right)}{16}}\)

\(=\sqrt{-\dfrac{-25-11\sqrt{5}+5\sqrt{5}+11}{16}}\)

\(=\sqrt{-\dfrac{-14-6\sqrt{5}}{16}}\)

\(=\sqrt{-\dfrac{2\left(-7-3\sqrt{5}\right)}{16}}\)

\(=\sqrt{-\dfrac{-7-3\sqrt{5}}{8}}\)

\(=\dfrac{\sqrt{-\left(-7-3\sqrt{5}\right)}}{\sqrt{8}}\)

\(=\dfrac{\sqrt{7+3\sqrt{5}}}{2\sqrt{2}}\)

\(=\dfrac{\sqrt{\left(7+3\sqrt{5}\right)\cdot2}}{4}\)

\(=\dfrac{\sqrt{14+6\sqrt{5}}}{4}\)

\(=\dfrac{\sqrt{\left(3+\sqrt{5}\right)^2}}{4}\)

\(=\dfrac{3+\sqrt{5}}{4}\)

b) \(\dfrac{2+3\sqrt{5}}{\sqrt{5}-2}-\dfrac{\sqrt{5}+1}{\sqrt{5}+2}\)

\(=\left(2+3\sqrt{5}\right)\cdot\left(\sqrt{5}+2\right)-\left(\sqrt{5}+1\right)\cdot\left(\sqrt{5}-2\right)\)

\(=2\sqrt{5}+4+15+6\sqrt{5}-\left(5-2\sqrt{5}+\sqrt{5}-2\right)\)

\(=2\sqrt{5}+4+15+6\sqrt{5}-\left(3-\sqrt{5}\right)\)

\(=2\sqrt{5}+4+15+6\sqrt{5}-3+\sqrt{5}\)

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

c) \(\dfrac{1+\sqrt{2}}{\sqrt{4-2\sqrt{3}}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)

\(=\dfrac{1+\sqrt{2}}{\sqrt{\left(1-\sqrt{3}\right)^2}}\cdot\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)

\(=\dfrac{1+\sqrt{2}}{\sqrt{3}-1}\cdot\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)

\(=\dfrac{\left(1+\sqrt{2}\right)\cdot\left(\sqrt{2}-1\right)}{\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+1\right)}\)

\(=\dfrac{\left(\sqrt{2}+1\right)\cdot\left(\sqrt{2}-1\right)}{3-1}\)

\(=\dfrac{2-1}{2}\)

\(=\dfrac{1}{2}\)

a) \(\sqrt{\dfrac{2-\sqrt{5}}{\sqrt{5}-3}}:\sqrt{\left(\sqrt{5}-3\right)\left(2-\sqrt{5}\right)}\)= \(\dfrac{\sqrt{2-\sqrt{5}}}{\sqrt{\sqrt{5}-3}}.\dfrac{1}{\sqrt{\sqrt{5}-3}\sqrt{2-\sqrt{5}}}\)

= \(\dfrac{1}{\sqrt{\sqrt{5}-3}}.\dfrac{1}{\sqrt{\sqrt{5}-3}}\) = \(\dfrac{1}{\sqrt{\sqrt{5}-3}^2}\) = \(\dfrac{1}{3-\sqrt{5}}\)

b) \(\dfrac{2+3\sqrt{5}}{\sqrt{5}-2}-\dfrac{\sqrt{5}+1}{\sqrt{5}+2}\) = \(\dfrac{\left(2+3\sqrt{5}\right)\left(\sqrt{5}+2\right)-\left(\sqrt{5}+1\right)\left(\sqrt{5}-2\right)}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\)

= \(\dfrac{2\sqrt{5}+4+15+6\sqrt{5}-\left(5-2\sqrt{5}+\sqrt{5}-2\right)}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\)

= \(\dfrac{8\sqrt{5}+19-5+2\sqrt{5}-\sqrt{5}+2}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\) = \(\dfrac{9\sqrt{5}+16}{5-4}\) = \(9\sqrt{5}+16\)

c) \(\dfrac{1+\sqrt{2}}{\sqrt{4-2\sqrt{3}}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\) = \(\dfrac{1+\sqrt{2}}{\sqrt{\left(\sqrt{3}-1\right)^2}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)

= \(\dfrac{1+\sqrt{2}}{\sqrt{3}-1}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\) = \(\dfrac{\left(1+\sqrt{2}\right)\left(\sqrt{2}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\) = \(\dfrac{\sqrt{2}-1+2-\sqrt{2}}{3-1}\)

= \(\dfrac{1}{2}\)

\(M=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{2012}+\sqrt{2013}}\)

\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{2013}-\sqrt{2012}\)

\(=\sqrt{2013}-1\)

\(\sqrt{x-2013}+x^3=\sqrt{y-2013}+y^3\)

\(\Leftrightarrow\sqrt{x-2013}-\sqrt{y-2013}+x^3-y^3=0\)

\(\Leftrightarrow\dfrac{x-y}{\sqrt{x-2013}+\sqrt{y-2013}}+\left(x-y\right)\left(x^2+xy+y^2\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(\dfrac{1}{\sqrt{x-2013}+\sqrt{y-2013}}+\left(x^2+xy+y^2\right)\right)=0\)

\(\Leftrightarrow x=y\)

\(\Rightarrow B=\dfrac{2013x+2014y}{2013y+2014x}=1\)

Tìm câu hỏi tt trc khi đăng: Câu hỏi của Lee Je Yoon - Toán lớp 9 | Học trực tuyến

Ta có

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

\(\Leftrightarrow A^2=2+A\)

\(\Leftrightarrow\left[{}\begin{matrix}A=-1\left(l\right)\\A=2\end{matrix}\right.\)

ko có yêu cầu à

Tìm trước khi hỏi Câu hỏi của Uchiha Sasuke - Toán lớp 9 | Học trực tuyến

Đặt \(A=\sqrt{2+\sqrt{2+\sqrt{2+...+\sqrt{2}}}}>0\)

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

\(\Rightarrow A^2-A=2\)

\(\Rightarrow A^2-A-2=0\)

\(\Rightarrow\left(A-2\right)\left(A+1\right)=0\)

Do \(A>0\) nên \(A=-1< 0\Leftrightarrow A+1< 0\)(loại)

Tức là \(A-2=0\Rightarrow A=2\)(đpcm)

\(\sqrt{313^2-312^2}\cdot\left(\sqrt{\dfrac{49}{8}}:\sqrt{3\dfrac{1}{8}}\right)\)

\(=\sqrt{\left(313-312\right)\left(313+312\right)}\cdot\left(\sqrt{\dfrac{49}{8}}:\sqrt{\dfrac{25}{8}}\right)\)

\(=\sqrt{1\cdot625}\cdot\left(\sqrt{\dfrac{49}{8}:\dfrac{25}{8}}\right)\)

\(=\sqrt{625}\cdot\left(\sqrt{\dfrac{49}{8}\cdot\dfrac{8}{25}}\right)\)

\(=25\cdot\left(49\cdot\dfrac{1}{25}\right)\)

\(=25\cdot\left(7\cdot\dfrac{1}{5}\right)\)

\(=25\cdot\dfrac{7}{5}\)

\(=35\)

\(\sqrt{313^2-312^2}\)

\(=\sqrt{\left(313-312\right)\left(313+312\right)}\)

\(=\sqrt{1\cdot625}\)

\(=\sqrt{625}\)

\(=25\)

Vế phải bằng bao nhiêu bạn?