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\(a,\frac{6}{4+\sqrt{4-2\sqrt{3}}}=\frac{6}{4+\sqrt{\sqrt{3}^2-2\sqrt{3}+\sqrt{1}^2}}\)
\(=\frac{6}{4+\sqrt{\left(\sqrt{3}-\sqrt{1}\right)^2}}=\frac{6}{4+|\sqrt{3}-1|}=\frac{6}{3+\sqrt{3}}\)
\(=\frac{6}{\sqrt{3}\left(\sqrt{3}+1\right)}=\frac{\sqrt{36}}{\sqrt{3}\left(\sqrt{3}+1\right)}=\frac{\sqrt{3}.\sqrt{12}}{\sqrt{3}\left(\sqrt{3}+1\right)}=\frac{\sqrt{12}}{\sqrt{3}+1}\)
\(d,\frac{1}{\sqrt{7-2\sqrt{10}}}+\frac{1}{\sqrt{7+2\sqrt{10}}}\)
\(=\frac{1}{\sqrt{\sqrt{5}^2-2.\sqrt{2}.\sqrt{5}+\sqrt{2}^2}}+\frac{1}{\sqrt{\sqrt{5}^2+2.\sqrt{2}.\sqrt{5}+\sqrt{2}^2}}\)
\(=\frac{1}{\sqrt{\left(\sqrt{5}-\sqrt{2}\right)}}+\frac{1}{\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}}\)
\(=\frac{1}{\sqrt{5}-\sqrt{2}}+\frac{1}{\sqrt{5}+\sqrt{2}}=\frac{\sqrt{5}+\sqrt{2}+\sqrt{5}-\sqrt{2}}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}\)
\(=\frac{2\sqrt{5}}{\sqrt{5}^2-\sqrt{2}^2}=\frac{\sqrt{5.4}}{5-2}=\frac{\sqrt{20}}{3}\)
Câu 1,2,3 Ez quá rồi :3
Câu 4:
Tổng quát:
\(\frac{1}{\sqrt{a}+\sqrt{a+1}}=\frac{\sqrt{a}-\sqrt{a+1}}{a-a-1}=\sqrt{a+1}-\sqrt{a}.\) Game là dễ :v
Câu 5 ko khác câu 4 lắm :v
Câu 5:
Tổng quát:
\(\frac{1}{\sqrt{a}-\sqrt{a+1}}=\frac{\sqrt{a}+\sqrt{a+1}}{a-a-1}=-\sqrt{a}-\sqrt{a+1}.\) Game là dễ :v
A.\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}\) \(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)\left(n+1-n\right)}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}\)
=\(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
b. ap dungtinh B =\(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}=1-\frac{1}{10}=\frac{9}{10}\)
Với mọi \(n\inℕ^∗\)ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}\)
\(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
đến đây bạn áp dụng đẳng thức trên để tính gtbt nhé: kết quả: 9/10
b/ Ta có: \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}.\sqrt{n+1}.\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}.\sqrt{n}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Áp dụng vào bài toán ta được
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{99}-\frac{1}{\sqrt{100}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{100}}=1-\frac{1}{10}=\frac{9}{10}\)
Cả 2 câu là n tự nhiên khác 0 hết nhé
a/ Ta có: \(\frac{1}{\sqrt{n}+\sqrt{n+1}}=\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\sqrt{n+1}-\sqrt{n}\)
Áp đụng vào bài toán được
\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{1680}+\sqrt{1681}}\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{1681}-\sqrt{1680}\)
\(=\sqrt{1681}-\sqrt{1}=41-1=40\)
\(a,A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+..+\frac{1}{\sqrt{99}+\sqrt{100}}\)
\(=\frac{\sqrt{1}-\sqrt{2}}{1-2}+\frac{\sqrt{2}-\sqrt{3}}{2-3}+...+\frac{\sqrt{99}-\sqrt{100}}{99-100}\)
\(=\frac{1-\sqrt{2}+\sqrt{2}-\sqrt{3}+...+\sqrt{99}-\sqrt{100}}{-1}\)
\(=\frac{1-\sqrt{100}}{-1}=9\)
\(b,B=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+..+\frac{1}{\sqrt{99}}\)
\(=\frac{2}{\sqrt{1}+\sqrt{1}}+\frac{2}{\sqrt{2}+\sqrt{2}}+\frac{2}{\sqrt{3}+\sqrt{3}}+...+\frac{2}{\sqrt{99}+\sqrt{99}}>\frac{2}{\sqrt{1}+\sqrt{2}}+\frac{2}{\sqrt{2}+\sqrt{3}}+...+\frac{2}{\sqrt{99}+\sqrt{100}}\)\(\Rightarrow B>2\left(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+..+\frac{1}{\sqrt{99}+\sqrt{100}}\right)\)
\(\Rightarrow B>2\left(\frac{\sqrt{1}-\sqrt{2}+\sqrt{2}-\sqrt{3}+...+\sqrt{99}-\sqrt{100}}{-1}\right)\)
\(\Rightarrow B>2\left(\frac{1-\sqrt{100}}{-1}\right)\)
\(\Rightarrow B>2.9=18\left(ĐPCM\right)\)