Tính giá trị biểu thức: \(A=\dfrac{1}{1\sqrt{3}+3\sqrt{1}}+\dfrac{1}{3\sqrt{5}+5\sqrt{3}}+\dfrac{1}{5\sqrt{7}+7\sqrt{5}}+...+\dfrac{1}{79\sqrt{81}+81\sqrt{79}}\)
Tính giá trị biểu thức: \(A=\dfrac{1}{1\sqrt{3}+3\sqrt{1}}+\dfrac{1}{3\sqrt{5}+5\sqrt{3}}+\dfrac{1}{5\sqrt{7}+7\sqrt{5}}+...+\dfrac{1}{79\sqrt{81}+81\sqrt{79}}\)
Với mọi \(n\in\text{ℕ*}\), ta có:
\(\dfrac{2}{n\sqrt{n+2}+\left(n+2\right)\sqrt{n}}\)\(=\dfrac{2\left(n\sqrt{n+2}-\left(n+2\right)\sqrt{n}\right)}{\left(n+2\right)^2n-n^2\left(n+2\right)}\)\(=\dfrac{2\left[\left(n+2\right)\sqrt{n}-n\sqrt{n+2}\right]}{n\left(n+2\right)}\)\(=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+2}}\)
Vậy ta có:
\(2A=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{3}}-\dfrac{1}{\sqrt{5}}+...-\dfrac{1}{\sqrt{81}}\)
\(=1-\dfrac{1}{\sqrt{81}}\)
\(A=\dfrac{1-\dfrac{1}{\sqrt{81}}}{2}\)
CMR A = \(\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{2\sqrt{3}+2\sqrt{5}}+....+\frac{1}{40\sqrt{79}+40\sqrt{81}}\) <\(\frac{8}{9}\)
Chứng tỏ P < \(\frac{8}{9}\)
P=\(\frac{\sqrt{3}-\sqrt{1}}{2}+\frac{\sqrt{5}-\sqrt{3}}{4}+\frac{\sqrt{7}-\sqrt{5}}{6}+...+\frac{\sqrt{81}-\sqrt{79}}{80}\)
Tính A=\(\frac{\sqrt{3}-\sqrt{1}}{2}+\frac{\sqrt{5}-\sqrt{3}}{4}+...+\frac{\sqrt{81}-\sqrt{79}}{80}\)
tính H =\(\frac{1}{1+\sqrt{3}}\)+\(\frac{1}{\sqrt{3}+\sqrt{5}}\)+...+\(\frac{1}{\sqrt{79}+\sqrt{81}}\)
Dạng tổng quát :
\(\frac{1}{\sqrt{x}+\sqrt{x+2}}=\frac{\sqrt{x}-\sqrt{x+2}}{\left(\sqrt{x}+\sqrt{x+2}\right)\left(\sqrt{x}-\sqrt{x+2}\right)}\)
\(=\frac{\sqrt{x}-\sqrt{x+2}}{x-x-2}=\frac{\sqrt{x}-\sqrt{x+2}}{-2}=\frac{\sqrt{x+2}}{2}-\frac{\sqrt{2}}{2}\)
Từ đó :
\(H=\frac{\sqrt{3}}{2}-\frac{1}{2}+\frac{\sqrt{5}}{2}-\frac{\sqrt{3}}{2}+...+\frac{\sqrt{81}}{2}-\frac{\sqrt{79}}{2}\)
\(H=\frac{\sqrt{81}}{2}-\frac{1}{2}\)
\(H=\frac{9}{2}-\frac{1}{2}=4\)
Tìm |-79|; |10,7|; \(\left| {\sqrt {11} } \right|;\left| {\frac{{ - 5}}{9}} \right|\)
\(\left| { - 79} \right| = 79;{\rm{ }}\left| {10,7} \right| = 10,7;\)\(\left| {\sqrt {11} } \right| = \sqrt {11} ;\left| {\frac{{ - 5}}{9}} \right| = \frac{5}{9}\)
|-79|=79
|10,7|=10,7
\(\left|\sqrt{11}\right|=\sqrt{11};\left|-\dfrac{5}{9}\right|=\dfrac{5}{9}\)
Ta có: \(|-79|=79\)
\(\left|10,7\right|=10,7\)
\(|\sqrt{11}|=11\)
\(\left|\dfrac{-5}{9}\right|=\dfrac{5}{9}\)
Cho A=\(\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{2\left(\sqrt{3}+\sqrt{5}\right)}+\frac{1}{3\left(\sqrt{5}+\sqrt{7}\right)}+...+\frac{1}{40\left(\sqrt{79}+\sqrt{81}\right)}\)
Chứng minh rằng A<\(\frac{8}{9}\)
Giúp mình với, mình đang rối quá
So sánh
\(A=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{79}+\sqrt{80}}\) và \(B=\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{4}+\sqrt{5}}+...+\dfrac{1}{\sqrt{80}+\sqrt{81}}\)
Lời giải:
Ta thấy:
\(A-B=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+..+\frac{1}{\sqrt{79}+\sqrt{80}}-\left(\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{4}+\sqrt{5}}+...+\frac{1}{\sqrt{80}+\sqrt{81}}\right)\)
\(=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{5}+\sqrt{6}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}-\frac{1}{\sqrt{80}+\sqrt{81}}\)
\(> \frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{77}+\sqrt{78}}>0\)
\(\Rightarrow A>B\)
Tính A=\(\frac{\sqrt{3}}{2}+\frac{\sqrt{5}}{4}+\frac{\sqrt{7}}{6}+...+\frac{\sqrt{81}}{80}\)
Tính B=\(\frac{\sqrt{1}}{2}+\frac{\sqrt{3}}{4}+\frac{\sqrt{5}}{6}+...+\frac{\sqrt{79}}{80}\)
1) Chứng minh rằng : \(\dfrac{1}{\sqrt{1}+\sqrt{2}}\) +\(\dfrac{1}{\sqrt{3}+\sqrt{4}}\)+....+\(\dfrac{1}{\sqrt{79}+\sqrt{80}}\) >4
\(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}...+\dfrac{1}{\sqrt{79}+\sqrt{80}}\)
\(=\left(\sqrt{2}-\sqrt{1}\right)+\left(\sqrt{3}-\sqrt{2}\right)+...+\left(\sqrt{80}-\sqrt{79}\right)\)
\(=\sqrt{80}-\sqrt{2}\)
Đến đây bấm máy rồi đối chiếu kết quả cho nhanh, hoặc nếu em thik "màu mè" hơn thì giả sử lớn hơn rồi biến đổi tương đương thôi :)