Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1) Vì \(a,b>0\)\(\Rightarrow\)\(\sqrt{ab}>0\)
\(\Leftrightarrow\)\(2\sqrt{ab}>0\)
\(\Leftrightarrow\)\(a+b+2\sqrt{ab}>a+b\)
\(\Leftrightarrow\)\(\left(\sqrt{a}+\sqrt{b}\right)^2>a+b\)
\(\Leftrightarrow\)\(\sqrt{a}+\sqrt{b}>\sqrt{a+b}\)
Vậy \(\sqrt{a}+\sqrt{b}>\sqrt{a+b}\)
1. Ta có: \(\left(\sqrt{a+b}\right)^2=a+b\)
\(\left(\sqrt{a}+\sqrt{b}\right)^2=a+2\sqrt{ab}+b\)
Vì \(a>0\), \(b>0\)\(\Rightarrow\sqrt{ab}>0\)\(\Rightarrow2\sqrt{ab}>0\)
\(\Rightarrow a+b< a+2\sqrt{ab}+b\)
\(\Rightarrow\left(\sqrt{a+b}\right)^2< \left(\sqrt{a}+\sqrt{b}\right)^2\)
mà \(\hept{\begin{cases}\sqrt{a+b}>0\\\sqrt{a}+\sqrt{b}>0\end{cases}}\)\(\Rightarrow\sqrt{a+b}< \sqrt{a}+\sqrt{b}\)( đpcm )
Với mọi n nguyên dương ta có:
\(\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)=1\Rightarrow\frac{1}{\sqrt{n+1}+\sqrt{n}}=\sqrt{n+1}-\sqrt{n}\)
Với k nguyên dương thì
\(\frac{1}{\sqrt{k-1}+\sqrt{k}}>\frac{1}{\sqrt{k+1}+\sqrt{k}}\Rightarrow\frac{2}{\sqrt{k-1}+\sqrt{k}}>\frac{1}{\sqrt{k-1}+\sqrt{k}}+\frac{1}{\sqrt{k+1}+\sqrt{k}}=\sqrt{k}-\sqrt{k-1}+\sqrt{k+1}-\sqrt{k}\)
\(=\sqrt{k+1}-\sqrt{k-1}\)(*)
Đặt A = vế trái. Áp dụng (*) ta có:
\(\frac{2}{\sqrt{1}+\sqrt{2}}>\sqrt{3}-\sqrt{1}\)
\(\frac{2}{\sqrt{3}+\sqrt{4}}>\sqrt{5}-\sqrt{3}\)
...
\(\frac{2}{\sqrt{79}+\sqrt{80}}>\sqrt{81}-\sqrt{79}\)
Cộng tất cả lại
\(2A=\frac{2}{\sqrt{1}+\sqrt{2}}+\frac{2}{\sqrt{3}+\sqrt{4}}+....+\frac{2}{\sqrt{79}+\sqrt{80}}>\sqrt{81}-1=8\Rightarrow A>4\left(đpcm\right)\)
3.
Theo bất đẳng thức cô si ta có:
\(\sqrt{b-1}=\sqrt{1.\left(b-1\right)}\le\frac{1+b-1}{2}=\frac{b}{2}\Rightarrow a.\sqrt{b-1}\le\frac{a.b}{2}\)
Tương tự \(\Rightarrow b.\sqrt{a-1}\le\frac{a.b}{2}\Rightarrow a.\sqrt{b-1}+b.\sqrt{a-1}\le a.b\)
Dấu "=" xảy ra khi và chỉ khi \(a=b=2\)
Lời giải:
a) Ta thấy: \(a+b-2\sqrt{ab}=(\sqrt{a}-\sqrt{b})^2\geq 0, \forall a,b>0\)
\(\Rightarrow a+b\geq 2\sqrt{ab}>0\Rightarrow \frac{1}{a+b}\le \frac{1}{2\sqrt{ab}}\).
Vì $a> b$ nên dấu bằng không xảy ra . Tức \(\frac{1}{a+b}< \frac{1}{2\sqrt{ab}}\)
Ta có đpcm
b)
Áp dụng kết quả phần a:
\(\frac{1}{3}=\frac{1}{1+2}< \frac{1}{2\sqrt{2.1}}\)
\(\frac{1}{5}=\frac{1}{3+2}< \frac{1}{2\sqrt{2.3}}\)
\(\frac{1}{7}=\frac{1}{4+3}< \frac{1}{2\sqrt{4.3}}\)
.....
\(\frac{1}{4021}=\frac{1}{2011+2010}< \frac{1}{2\sqrt{2011.2010}}\)
Do đó:
\(\frac{\sqrt{2}-\sqrt{1}}{3}+\frac{\sqrt{3}-\sqrt{2}}{5}+...+\frac{\sqrt{2011}-\sqrt{2010}}{4021}\)
\(< \frac{\sqrt{2}-\sqrt{1}}{2\sqrt{2.1}}+\frac{\sqrt{3}-\sqrt{2}}{2\sqrt{3.2}}+\frac{\sqrt{4}-\sqrt{3}}{2\sqrt{4.3}}+....+\frac{\sqrt{2011}-\sqrt{2010}}{2\sqrt{2011.2010}}\)
\(=\frac{1}{2}-\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}-\frac{1}{2\sqrt{3}}+...+\frac{1}{2\sqrt{2010}}-\frac{1}{2\sqrt{2011}}\)
\(=\frac{1}{2}-\frac{1}{2\sqrt{2011}}< \frac{1}{2}\) (đpcm)
a)
+) Ta có: \(\dfrac{1}{\sqrt{n}}=\dfrac{2}{2\sqrt{n}}>\dfrac{2}{\sqrt{n}+\sqrt{n+1}}=\dfrac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(\sqrt{n}+\sqrt{n+1}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\) \(=\dfrac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{n+1-n}\)
\(=2\left(\sqrt{n+1}-\sqrt{n}\right)\) (1)
+) Ta có:
\(\dfrac{1}{\sqrt{n}}=\dfrac{2}{2\sqrt{n}}< \dfrac{2}{\sqrt{n}+\sqrt{n-1}}=\dfrac{2\left(\sqrt{n}-\sqrt{n-1}\right)}{\left(\sqrt{n}+\sqrt{n-1}\right)\left(\sqrt{n}-\sqrt{n-1}\right)}\) \(=\dfrac{2\left(\sqrt{n}-\sqrt{n-1}\right)}{n-\left(n-1\right)}\)
\(=2\left(\sqrt{n}-\sqrt{n-1}\right)\) (2)
Từ (1) và (2) ⇒ đpcm
Học toán vui vẻ! 
Ta co:
\(a,a-\sqrt{a}=\sqrt{a}\left(\sqrt{a}-1\right);a>1\Leftrightarrow\sqrt{a}>1\Leftrightarrow\sqrt{a}-1>0\Rightarrow\left\{{}\begin{matrix}\sqrt{a}>1\\\sqrt{a}-1>0\end{matrix}\right.\Rightarrow\sqrt{a}\left(\sqrt{a}-1\right)>0\Leftrightarrow a-\sqrt{a}>0\Leftrightarrow a>\sqrt{a}\left(\text{đpcm}\right)\)
\(b,a-\sqrt{a}=\sqrt{a}\left(\sqrt{a}-1\right);a< 1\Rightarrow0< \sqrt{a}< 1\Rightarrow\sqrt{a}-1< 0\Rightarrow\sqrt{a}\left(\sqrt{a}-1\right)< 0\left(vì:0< \sqrt{a}< 1\right)\Leftrightarrow a-\sqrt{a}< 0\Leftrightarrow a< \sqrt{a}\left(\text{đpcm}\right)\)
\(\dfrac{1}{a}+\dfrac{1}{b}=1\)
\(\Leftrightarrow a+b=ab\)(*)
Xét
\(\sqrt{a+b}=\sqrt{a-1}+\sqrt{b-1}\)
\(\Leftrightarrow a+b=a+b-2+2\sqrt{ab-a-b+1}\)
\(\Leftrightarrow2=2\sqrt{ab-a-b+1}\)
\(\Leftrightarrow4=4\left(ab-a-b+1\right)\)
\(\Leftrightarrow4\left(ab-a-b\right)=0\Leftrightarrow ab-a-b=0\)
\(\Leftrightarrow ab=a+b\)(đúng với *)
\(\Rightarrow\)đpcm
Ta có\(\dfrac{1}{a}+\dfrac{1}{b}=1\Leftrightarrow\dfrac{a+b}{ab}=1\Leftrightarrow a+b=ab\Leftrightarrow ab-a-b=0\Leftrightarrow1=ab-a-b+1\Leftrightarrow1=a\left(b-1\right)-\left(b+1\right)\Leftrightarrow1=\left(a-1\right)\left(b-1\right)\Leftrightarrow1=\sqrt{\left(a-1\right)\left(b-1\right)}\Leftrightarrow2=2\sqrt{\left(a-1\right)\left(b-1\right)}\Leftrightarrow0=-2+2\sqrt{\left(a-1\right)\left(b-1\right)}\Leftrightarrow a+b=a-1+2\sqrt{\left(a-1\right)}\sqrt{\left(b-1\right)}+b-1\Leftrightarrow a+b=\left(\sqrt{a+1}+\sqrt{b+1}\right)^2\Leftrightarrow\sqrt{a+b}=\sqrt{\left(\sqrt{a+1}+\sqrt{b+1}\right)^2}\Leftrightarrow\sqrt{a+b}=\sqrt{a+1}+\sqrt{b+1}\left(đpcm\right)\)
a) \(\sqrt{a}+1>\sqrt{a+1}\)\(\Leftrightarrow\)\(a+2\sqrt{a}+1>a+1\)\(\Leftrightarrow\)\(2\sqrt{a}>0\)( luôn đúng \(\forall x>0\) )
b) \(a-1< a\)\(\Leftrightarrow\)\(\sqrt{a-1}< \sqrt{a}\)
c) \(\left(\sqrt{6}-1\right)^2=6-2\sqrt{6}+1>3-2\sqrt{3.2}+2=\left(\sqrt{3}-\sqrt{2}\right)^2\)
do \(\sqrt{6}-1>0;\sqrt{3}-\sqrt{2}>0\) nên \(\sqrt{6}-1>\sqrt{3}-\sqrt{2}\) ( đpcm )