Chứng minh đẳng thức sau:
\(\dfrac{1}{1+\sqrt{2}}\)+\(\dfrac{1}{\sqrt{2}+\sqrt{3}}\)+...+\(\dfrac{1}{\sqrt{99}+\sqrt{100}}\)=9
Những câu hỏi liên quan
Bài 40: Chứng minh rằng:
a) \(A=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}=9\)
b) \(B=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}=\dfrac{9}{10}\)
Chứng minh đẳng thức sau \(\sqrt{n+1}-\sqrt{n}=\dfrac{1}{\sqrt{n+1}+\sqrt{n}}\) với n là một số tự nhiên tùy ý. Từ đó tính giá trị của biểu thức
\(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}.\)
Ta có : \(\dfrac{1}{\sqrt{n+1}+\sqrt{n}}\)
\(=\dfrac{\sqrt{n+1}-\sqrt{n}}{\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\)
\(=\dfrac{\sqrt{n+1}-\sqrt{n}}{n+1-n}\)
\(=\dfrac{\sqrt{n+1}-\sqrt{n}}{1}\)
\(=\sqrt{n+1}-\sqrt{n}\)
Vậy đẳng thức đã được chứng minh .
Áp dụng :
\(\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+....+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)
\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+.....+\sqrt{100}-\sqrt{99}\)
\(=-1+\sqrt{100}\)
\(=-1+10=9\)
Đúng 0
Bình luận (0)
chứng minh bất đẳng thức:
\(\dfrac{1}{\sqrt{1}+\sqrt{3}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+\dfrac{1}{\sqrt{9}+\sqrt{11}}+...+\dfrac{1}{\sqrt{97}+\sqrt{99}}\)<\(\dfrac{9}{4}\)
Cho \(x=\dfrac{\sqrt{2}-\sqrt{1}}{1+\sqrt{2}}+\dfrac{\sqrt{3}-\sqrt{2}}{2+\sqrt{3}}+\dfrac{\sqrt{4}-\sqrt{3}}{3+4}+...+\dfrac{\sqrt{100}-\sqrt{99}}{99+100}\). Chứng minh \(x< \dfrac{1}{2}\)
Chứng minh đẳng thức :\(\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\dfrac{1}{9}}-\sqrt[3]{\dfrac{2}{9}}+\sqrt[3]{\dfrac{4}{9}}\)
Chứng minh đẳng thức
\(\left(4-\sqrt{7}\right)^2=23-8\sqrt{7}\)
\(\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2\)
\(\dfrac{\sqrt{4-2\sqrt{3}}}{1+\sqrt{2}}:\dfrac{\sqrt{2}-1}{\sqrt{3}+1}=2\)
\(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right).\dfrac{1}{\sqrt{6}}=-1,5\)
\(\left(4-\sqrt{7}\right)^2=4^2-2\cdot4\cdot\sqrt{7}+7\)
\(=16-8\sqrt{7}+7=23-8\sqrt{7}\)
\(\sqrt{9-4\sqrt{5}}-\sqrt{5}\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot2+4}-\sqrt{5}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}\)
\(=\left|\sqrt{5}-2\right|-\sqrt{5}\)
\(=\sqrt{5}-2-\sqrt{5}=-2\)
\(\dfrac{\sqrt{4-2\sqrt{3}}}{1+\sqrt{2}}:\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)
\(=\dfrac{\sqrt{3-2\cdot\sqrt{3}\cdot1+1}}{\sqrt{2}+1}\cdot\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)
\(=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}+1}\cdot\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)
\(=\dfrac{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}=\dfrac{3-1}{2-1}=2\)
\(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-\dfrac{6\sqrt{6}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{1}{2}\sqrt{6}-2\sqrt{6}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\dfrac{1}{2}-2=-\dfrac{3}{2}=-1,5\)
Đúng 1
Bình luận (0)
Chứng minh đẳng thức
\(\left(4-\sqrt{7}\right)^2=23-8\sqrt{7}\)
\(\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2\)
\(\dfrac{\sqrt{4-2\sqrt{3}}}{1+\sqrt{2}}:\dfrac{\sqrt{2}-1}{\sqrt{3}-1}=2\)
\(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right).\dfrac{1}{\sqrt{6}}=-1,5\)
Tính các tổng sau:
\(T=\dfrac{1}{1+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{9}}+\dfrac{1}{\sqrt{9}+\sqrt{13}+......+\dfrac{1}{\sqrt{2013}+\sqrt{2017}}}\)
\(S=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+.....+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}\)
Chứng minh đẳng thức:
\(\dfrac{1+\dfrac{\sqrt{3}}{2}}{1+\sqrt{1+\dfrac{\sqrt{3}}{2}}}+\dfrac{1-\dfrac{\sqrt{3}}{2}}{1-\sqrt{1-\dfrac{\sqrt{3}}{2}}}=1\)
\(=\left[\dfrac{2+\sqrt{3}}{2}:\left(1+\sqrt{\dfrac{4+2\sqrt{3}}{4}}\right)\right]+\left[\dfrac{2-\sqrt{3}}{2}:\left(1-\sqrt{\dfrac{4-2\sqrt{3}}{4}}\right)\right]\)
\(=\left(\dfrac{2+\sqrt{3}}{2}:\dfrac{2+\sqrt{3}+1}{2}\right)+\left(\dfrac{2-\sqrt{3}}{2}:\dfrac{2-\sqrt{3}+1}{2}\right)\)
\(=\dfrac{2+\sqrt{3}}{3+\sqrt{3}}+\dfrac{2-\sqrt{3}}{3-\sqrt{3}}\)
\(=1\)
Đúng 0
Bình luận (0)