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Theo bđt Cauchy ta có \(\frac{a+b}{2}>\sqrt{ab}\) \(\left(a,b\ge0;a\ne b\right)\)
\(\Leftrightarrow\)\(\frac{2}{a+b}< \frac{1}{\sqrt{ab}}\)
Đặt \(A=\frac{1}{\sqrt{1.2014}}+\frac{1}{\sqrt{2.2013}}+...+\frac{1}{\sqrt{2014.1}}\)
\(A=\frac{2}{1+2014}+\frac{2}{2+2013}+...+\frac{2}{2014+1}\)
\(A=2\left(\frac{1}{1+2014}+\frac{1}{2+2013}+...+\frac{1}{2014+1}\right)\)
\(A=2\left(\frac{1}{2015}+\frac{1}{2015}+...+\frac{1}{2015}\right)\)
\(A=2.\frac{2014}{2015}\)
\(A=\frac{4028}{2015}\)
Vậy \(A=\frac{4028}{2015}\)
Chúc bạn học tốt ~
sorry mk nhầm
Sửa lại các dấu "=" thành dấu ">" nha bn
Chúc bạn học tốt ~
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 \(S>2.\frac{2014}{2015}\)
(bài 1) a) \(\dfrac{1}{5+2\sqrt{6}}-\dfrac{1}{5-2\sqrt{6}}\) = \(\dfrac{5-2\sqrt{6}-5-2\sqrt{6}}{25-24}\)
= \(\dfrac{-4\sqrt{6}}{1}\) = \(-4\sqrt{6}\)
b) \(\sqrt{6+2\sqrt{5}}-\dfrac{\sqrt{15}-\sqrt{3}}{\sqrt{3}}\) = \(\sqrt{\left(\sqrt{5}+1\right)^2}-\dfrac{\sqrt{3}\left(\sqrt{5}-1\right)}{\sqrt{3}}\)
= \(\left(\sqrt{5}+1\right)-\left(\sqrt{5}-1\right)\) = \(\sqrt{5}+1-\sqrt{5}+1\) = \(2\)
c) \(\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}:\dfrac{1}{\sqrt{16}}\) = \(\dfrac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}:\dfrac{1}{\sqrt{16}}\)
= \(\sqrt{6}.\sqrt{16}\) = \(4\sqrt{6}\)
d) \(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{1+\sqrt{2}}-\dfrac{1}{2-\sqrt{3}}\)
= \(\dfrac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{1+\sqrt{2}}-\dfrac{1}{2-\sqrt{3}}\)
= \(\sqrt{3}+2+\sqrt{2}-\dfrac{1}{2-\sqrt{3}}\) = \(\dfrac{\left(\sqrt{3}+2+\sqrt{2}\right)\left(2-\sqrt{3}\right)-1}{2-\sqrt{3}}\)
= \(\dfrac{2\sqrt{3}-3+4-2\sqrt{3}+2\sqrt{2}-\sqrt{6}-1}{2-\sqrt{3}}\)
= \(\dfrac{2\sqrt{2}-\sqrt{6}}{2-\sqrt{3}}\) = \(\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{2}}\) = \(\sqrt{2}\)
e) \(\dfrac{4}{1+\sqrt{3}}-\dfrac{\sqrt{15}+\sqrt{3}}{1+\sqrt{5}}\) = \(\dfrac{4}{1+\sqrt{3}}-\dfrac{\sqrt{3}\left(\sqrt{5}+1\right)}{1+\sqrt{5}}\)
= \(\dfrac{4}{1+\sqrt{3}}-\sqrt{3}\) = \(\dfrac{4-\sqrt{3}-3}{1+\sqrt{3}}\) = \(\dfrac{1-\sqrt{3}}{1+\sqrt{3}}\)
= \(\dfrac{\left(1-\sqrt{3}\right)\left(1-\sqrt{3}\right)}{1-3}\) = \(\dfrac{1-2\sqrt{3}+3}{-2}\) = \(\dfrac{4-2\sqrt{3}}{-2}\)
= \(\dfrac{-2\left(-2+\sqrt{3}\right)}{-2}\) = \(\sqrt{3}-2\)
bài 2)
a)\(\dfrac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\dfrac{1}{\sqrt{a}+\sqrt{b}}=\dfrac{\left(a+b-2\sqrt{ab}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\)
= \(\dfrac{a\sqrt{a}+a\sqrt{b}+b\sqrt{a}+b\sqrt{b}-2a\sqrt{b}-2b\sqrt{a}}{\sqrt{a}-\sqrt{b}}\)
= \(\dfrac{a\sqrt{a}+-a\sqrt{b}+b\sqrt{b}-b\sqrt{a}}{\sqrt{a}-\sqrt{b}}\) = \(\dfrac{a\left(\sqrt{a}-\sqrt{b}\right)-b\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\)
= \(\dfrac{\left(a-b\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\) = \(a-b\)
b) \(\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right).\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)
= \(\dfrac{2a-2}{4\sqrt{a}}.\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)^2-\sqrt{a}\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
= \(\dfrac{2a-2}{4\sqrt{a}}.\dfrac{\sqrt{a}\left(a-2\sqrt{a}+1\right)-\sqrt{a}\left(a+2\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
= \(\dfrac{2a-2}{4\sqrt{a}}.\dfrac{a\sqrt{a}-2a+\sqrt{a}-a\sqrt{a}-2a-\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
= \(\dfrac{2\left(a-1\right)}{4\sqrt{a}}.\dfrac{-4a}{a-1}\) = \(-2\)
Bài 3:
a: \(=\dfrac{3+2\sqrt{2}}{1}-\dfrac{\sqrt{2}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}\)
\(=3+2\sqrt{2}-\sqrt{2}=3+\sqrt{2}\)
b: \(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{a-b}\cdot\sqrt{\dfrac{ab+b^2-2b\sqrt{ab}}{a^2+2a\sqrt{b}+b}}\)
\(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{\left(\sqrt{ab}-b\right)}{\left(a+\sqrt{b}\right)^2}\)
\(=\dfrac{\sqrt{b}}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{a+\sqrt{b}}=\dfrac{b}{a+\sqrt{b}}\)
c: \(=x+\sqrt{x}-2\sqrt{x}-1+1=x-\sqrt{x}\)
\(A=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{120}+\sqrt{121}}\)
\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{121}-\sqrt{120}\)
\(=\sqrt{121}-\sqrt{1}=11-1=10\)
Lại có: \(\dfrac{1}{\sqrt{k}}=\dfrac{2}{2\sqrt{k}}>\dfrac{2}{\sqrt{k+1}+\sqrt{k}}\left(k>1\right)\)
\(\Leftrightarrow\dfrac{1}{\sqrt{k}}>\dfrac{2\left(\sqrt{k+1}-\sqrt{k}\right)}{k+1-k}=2\left(\sqrt{k+1}-\sqrt{k}\right)\)
Áp dụng đánh giá trên vào B ta có:
\(B>1+2\left(\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{36}-\sqrt{35}\right)\)
\(=1+2\left(\sqrt{36}-\sqrt{2}\right)>1+2\left(6-1\right)=10\)
Suy ra \(A=10< B\Rightarrow A< B\)
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 \(A>B\)