Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(A=a-\sqrt{a}=\sqrt{a}\left(\sqrt{a}-1\right)\)
Vì a>1 nên \(\sqrt{a}-1>0\)
=>A>0
hay \(a>\sqrt{a}\)
b: \(A=a-\sqrt{a}=\sqrt{a}\left(\sqrt{a}-1\right)\)
Vì a<1 nên \(\sqrt{a}-1< 0\)
=>A<0
hay \(a< \sqrt{a}\)
\(a^2=\dfrac{\sqrt{2}}{4}\left(1-a\right)\)
\(\Rightarrow a^4=\dfrac{1}{8}\left(1-a\right)^2\)
\(\Rightarrow a^4+a+1=\dfrac{1}{8}\left(1-a\right)^2+a+1=\dfrac{1}{8}\left(a^2+6a+9\right)=\dfrac{1}{8}\left(a+3\right)^2\)
\(\Rightarrow\sqrt{a^4+a+1}-a^2=\sqrt{\dfrac{1}{8}\left(3+a\right)^2}-a^2=\dfrac{\sqrt{2}}{4}\left(a+3\right)-\dfrac{\sqrt{2}}{4}\left(1-a\right)=\dfrac{\sqrt{2}}{2}\left(a+1\right)\)
\(\Rightarrow\dfrac{a+1}{\sqrt{a^4+a+1}-a^2}=\dfrac{a+1}{\dfrac{\sqrt{2}}{2}\left(a+1\right)}=\sqrt{2}\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Leftrightarrow ab+bc+ca=0\)
Cần cm:
\(\sqrt{a+b}=\sqrt{a+c}+\sqrt{b+c}\\ \Leftrightarrow a+b=a+b+2c+2\sqrt{\left(a+c\right)\left(b+c\right)}\\ \Leftrightarrow2c+2\sqrt{ab+ac+bc+c^2}=0\\ \Leftrightarrow2c+2\sqrt{c^2}=0\\ \Leftrightarrow2c+2\left|c\right|=0\\ \Leftrightarrow2c-2c=0\left(c< 0\right)\\ \Leftrightarrow0=0\left(luôn.đúng\right)\)
Vậy đẳng thức đc cm
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 )
\(\left\{{}\begin{matrix}\sqrt{a}=x\\\sqrt{b}=y\end{matrix}\right.\)
\(bdt\Leftrightarrow x\left(\frac{x}{y}-1\right)\ge y\left(1-\frac{y}{x}\right)\Leftrightarrow\frac{x^2}{y}-x\ge y-\frac{y^2}{x}\)
\(\Leftrightarrow\frac{x^2}{y}+\frac{y^2}{x}-x-y\ge0\)
bđt này hiển nhiên đúng theo Cauchy-Schwarz:
\(\frac{x^2}{y}+\frac{y^2}{x}\ge\frac{\left(x+y\right)^2}{x+y}=x+y\Rightarrow\frac{x^2}{y}+\frac{y^2}{x}-x-y\ge0\)
\("="\Leftrightarrow x=y\Rightarrow a=b\)