So sánh:\(\frac{1}{\sqrt{1.2014}}+\frac{1}{\sqrt{2.2013}}+...+\frac{1}{\sqrt{2014.1}}\) và \(\frac{4028}{2015}\)
So sánh A và B
\(A=\dfrac{1}{\sqrt{1.2014}}+\dfrac{1}{\sqrt{2.2013}}+...+\dfrac{1}{\sqrt{2014.1}}\)
\(B=\dfrac{4028}{2015}\)
Áp dụng bất đẳng thức Cosi cho 2 số dương ta có:
\(\sqrt{1.2014} \leq \frac{1+2014}{2}=\frac{2015}{2} \\ \Rightarrow \frac{1}{\sqrt{1.2014}} \geq \frac{2}{2015}\)
Trong tổng A có 2014 phân thức, mỗi phân thức theo chứng minh tương tự, ta đều chỉ được nó lớn hơn hoặc bằng \( \frac{2}{2015}\)
Suy ra \(A\geq \frac{2.2014}{2015} = B\)
Dấu = xảy ra khi \(\Leftrightarrow\) \(1=2014\\ 2=2013\\ ...\\ 2014=1\) (vô lý)
Vậy A>B
Sử dụng BĐT: \(\dfrac{1}{\sqrt{ab}}>\dfrac{2}{a+b}\) (với \(a\ne b\)) ta được:
\(A>\dfrac{2}{1+2014}+\dfrac{2}{2+2013}+...+\dfrac{2}{2014+1}\) (2014 số hạng)
\(A>\dfrac{2}{2015}+\dfrac{2}{2015}+...+\dfrac{2}{2015}=\dfrac{2.2014}{2015}\)
\(A>\dfrac{4028}{2015}\)
Vậy \(A>B\)
So sánh \(A=\frac{1}{\sqrt{1.2014}}+\frac{1}{\sqrt{2.1013}}+...+\frac{1}{\sqrt{2014.1}}\)với \(B=\frac{4028}{2015}\)
Cho
S = \(\frac{1}{\sqrt{1.2014}}+\frac{1}{\sqrt{2.2013}}+......+\frac{1}{\sqrt{k.\left(2014-k+1\right)}}+.....+\frac{1}{\sqrt{2014.1}}\)
Hãy so sánh S với \(2.\frac{2014}{2015}\)
Theo BĐT \(AM-GM\) ta có : \(\sqrt{ab}< \frac{a+b}{2}\) với \(a;b>0;a\ne b\)\(\Rightarrow\frac{1}{\sqrt{ab}}>\frac{2}{a+b}\)
Áp dụng ta được :
\(S>\frac{2}{1+2014}+\frac{2}{2+2013}+...+\frac{2}{k+2014-k+1}+...+\frac{2}{2014+1}\)
\(=2\left(\frac{1}{2015}+\frac{1}{2015}+...+\frac{1}{2015}\right)=2.\frac{2014}{2015}\)
Vậy \(S>2.\frac{2014}{2015}\)
so sánh \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2015}}và\sqrt{2015}\)
Đặt \(A=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2015}}\)
Ta thấy: \(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{2015}}\)
\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{2015}}\)
\(\frac{1}{\sqrt{3}}>\frac{1}{\sqrt{2015}}\)
.........................
\(\frac{1}{\sqrt{2014}}>\frac{1}{\sqrt{2015}}\)
=>\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2014}}>\frac{1}{\sqrt{2015}}+\frac{1}{\sqrt{2015}}+\frac{1}{\sqrt{2015}}+...+\frac{1}{\sqrt{2015}}\)
=>\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2014}}+\frac{1}{\sqrt{2015}}>\frac{1}{\sqrt{2015}}+\frac{1}{\sqrt{2015}}+\frac{1}{\sqrt{2015}}+...+\frac{1}{\sqrt{2015}}+\frac{1}{\sqrt{2015}}\)
=>\(A>2015.\frac{1}{\sqrt{2015}}=\frac{2015}{\sqrt{2015}}=\sqrt{2015}\)
Vậy \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2015}}>\sqrt{2015}\)
Tính A= \(\frac{1.2014+2.2013+3.2012+...+2014.1}{\left(1+2+3+...+2014\right)+\left(1+2+3+...+2013\right)+...+\left(1+2\right)+1}\)
so sánh \(E=\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{2015\sqrt{2014}}\) và 2
Cho M=\(\frac{\sqrt{2}-\sqrt{1}}{1+1}+\frac{\sqrt{3}-\sqrt{2}}{2+3}+\frac{\sqrt{4}-\sqrt{3}}{3+4}+...+\frac{\sqrt{2015}-\sqrt{2014}}{2014+2015}\)
Hãy so sánh M với 1/2
Cho \(M=\frac{\sqrt{2}-\sqrt{1}}{1+2}+\frac{\sqrt{3}-\sqrt{2}}{2+3}+\frac{\sqrt{4}-\sqrt{3}}{3+4}+...+\frac{\sqrt{2015}-\sqrt{2014}}{2014+2015}\). Hãy so sánh M với \(\frac{1}{2}\)
\(\sqrt{17}+\sqrt{26}+1và\sqrt{99}\)
b)chứng minh:\(\frac{1}{\sqrt{ }1}+\frac{1}{\sqrt{ }2}+\frac{1}{\sqrt{ }3}+...+\frac{1}{\sqrt{ }99}+\frac{1}{\sqrt{ }100}>10\)
c)cho:S=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}+\frac{1}{2015}\)vàP=\(\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2014}+\frac{1}{2015}\)tính \(\left(S-P\right)^{2016}\)