Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
ĐKXĐ: \(x>0;x\ne1\)
\(A=\left(\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right)\frac{\left(x-1\right)^2}{2}\)
\(=\left(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right)\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}\)
\(=\frac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}=-\sqrt{x}\left(\sqrt{x}-1\right)=\sqrt{x}\left(1-\sqrt{x}\right)\)
Khi \(0< x< 1\Rightarrow0< \sqrt{x}< 1\Rightarrow0< 1-\sqrt{x}< 1\)
\(\Rightarrow\sqrt{x}\left(1-\sqrt{x}\right)>0\)
\(A=\sqrt{x}-x=-\left(x-\sqrt{x}+\frac{1}{4}\right)+\frac{1}{4}=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)
\(A_{max}=\frac{1}{4}\) khi \(\sqrt{x}=\frac{1}{2}\Rightarrow x=\frac{1}{4}\)
\(\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\frac{x^2-2x+1}{2}\)
a)
Đkxđ:\(\left\{{}\begin{matrix}x-1\ne0\\x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ge0\end{matrix}\right.\)
\(=\)\(\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\frac{\left(x-1\right)^2}{2}\)
\(=\frac{x\sqrt{x}+2x+\sqrt{x}-2x-4\sqrt{x}-2-x\sqrt{x}+\sqrt{x}-2x+2}{\left(x-1\right)\left(x+2\sqrt{x}+1\right)}.\frac{\left(x-1\right)^2}{2}\)
\(=\frac{-2\sqrt{x}-2x}{\left(x-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(x-1\right)^2}{2}\)
\(=\frac{-2\sqrt{x}\left(1+\sqrt{x}\right)}{\left(x-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(x-1\right)^2}{2}\)
\(=\frac{-2\sqrt{x}\left(x-1\right)}{2\left(\sqrt{x}+1\right)}=\frac{-2\sqrt{x}\left(x-1\right)}{2\sqrt{x}+2}\)
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 \(a=b=2\)