1) Chứng minh rằng: \(1+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{3\sqrt{3}}+...+\dfrac{1}{n\sqrt{n}}< 2\sqrt{2}\left(n\in N\right)\)
2) Chứng minh rằng: \(\dfrac{2}{3}+\sqrt{n+1}< 1+\sqrt{2}+\sqrt{3}+...+\sqrt{n}< \dfrac{2}{3}\left(n+1\right)\sqrt{n}\)
3) \(2\sqrt{n}-3< \dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}< 2\sqrt{n}-2\)
4) \(\dfrac{\sqrt{2}-\sqrt{1}}{2+1}+\dfrac{\sqrt{3}-\sqrt{2}}{3+2}+...+\dfrac{\sqrt{n+1}-\sqrt{n}}{n+1+n}< \dfrac{1}{2}\left(1-\dfrac{1}{\sqrt{n+1}}\right)\)
CMR:\(1+\dfrac{1}{2\sqrt{1}}+\dfrac{1}{3\sqrt{2}}+...+\dfrac{1}{\left(n+1\right)\sqrt{n}}< 2\)
Với số tự nhiên n, \(n\ge3\). Đặt \(S_n=\dfrac{1}{3\left(1+\sqrt{2}\right)}+\dfrac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+...+\dfrac{1}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}\). Chứng minh: \(S_n< \dfrac{1}{2}\)
Chứng minh rằng với mọi số nguyên dương n ta đều có \(\dfrac{1}{2\sqrt{1}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{4\sqrt{3}}+\dfrac{1}{5\sqrt{4}}+...+\dfrac{1}{\left(n+1\right)\sqrt{n}}< 2\)
CMR : \(2\left(\sqrt{n+1}-\sqrt{n}\right)< \dfrac{1}{\sqrt{n}}< 2\left(\sqrt{n}-\sqrt{n-1}\right)\) với n thuộc N*
Áp dụng cho : \(A=1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}\) . CMR : 18 < A < 19
@Akai Haruma
CMR, ∀n ≥ 1, n ∈ N : \(\dfrac{1}{2}\)+\(\dfrac{1}{3\sqrt{2}}\)+\(\dfrac{1}{4\sqrt{3}}\)+....+ \(\dfrac{1}{\left(n+1\right)\sqrt{n}}\)<2
cho 3 số thực a,b,c không âm thỏa mãn \(a+b+c=\sqrt{a}+\sqrt{b}+\sqrt{c}=2\)
CMR: \(\dfrac{\sqrt{a}}{1+a}+\dfrac{\sqrt{b}}{1+b}+\dfrac{\sqrt{c}}{1+c}=\dfrac{2}{\sqrt{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)
Cho \(P=\left(\dfrac{a-3\sqrt{a}+2}{3a-7\sqrt{a}+2}-\dfrac{\sqrt{a}-3}{3a-8\sqrt{a}-3}+\dfrac{8\sqrt{a}}{9a-1}\right):\left(1-\dfrac{2\sqrt{a}-a+1}{3\sqrt{a}+1}\right)\)
Tìm giá trị nguyên lớn nhất của a để \(P>\dfrac{3}{\left|1-3\sqrt{5}\right|}\)
1)tính
a)\(\left(\dfrac{1}{5}\sqrt{500}-3\sqrt{45}+5\sqrt{20}\right):\sqrt{5}\)
b)\(\left(\dfrac{\sqrt{3}+1}{\sqrt{3}-1}-\dfrac{\sqrt{3}-1}{\sqrt{3}+1}\right).\sqrt{\dfrac{1}{48}}\)
c)\(\left(\dfrac{2\sqrt{3}+3}{\sqrt{3}+2}+\dfrac{2\sqrt{2}}{\sqrt{2}+1}\right):\left(\sqrt{12}+\sqrt{18}\right)\)
