Cho A=\(\frac{\sqrt{x}}{x-\sqrt{x}+1}\)
Chứng minh: 0<=A<=1
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21 tháng 9 2020
a)\(G=\left(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\right):\frac{\sqrt{x}-1}{2}\)
\(=\frac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{2}{\sqrt{x}-1}\)
\(=\frac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{2}{\sqrt{x}-1}\)
\(=\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{2}{\sqrt{x}-1}\)
\(=\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{2}{\sqrt{x}-1}\)
\(=\frac{2}{x+\sqrt{x}+1}\)
b) \(x+\sqrt{x}+1>0\Rightarrow G>0\)
\(x+\sqrt{x}+1>0+0+1=1\)
\(\Rightarrow\frac{2}{x+\sqrt{x}+1}< \frac{2}{1}=2\Rightarrow G< 2\)
\(\Rightarrow O< G< 2\)
NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 9 2019
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 \(\sqrt{x}=\frac{1}{2}\Rightarrow x=\frac{1}{4}\)
12 tháng 8 2019
a) A= \(\left(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\right):\left(\frac{\sqrt{x}-1}{2}\right)\) (x ≥ 0; x ≠ 4)
= \(\left(\frac{x+2}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}-1\right)\cdot\sqrt{x}}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}\right):\frac{\sqrt{x}-1}{2}\)
=\(\left(\frac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}\right)\cdot\frac{2}{\sqrt{x}-1}\)
=\(\left(\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}\right)\cdot\frac{2}{\sqrt{x}-1}\)
= \(\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}\cdot\frac{2}{\sqrt{x}-1}\)
=\(\frac{2}{x+\sqrt{x}+1}\)
b) Ta có: x ≥ 0 ⇒ \(\sqrt{x}\) ≥ 0
⇒x+\(\sqrt{x}\)+1 ≥ 1 > 0
mà 2 > 0
⇒ A > 0 (1)
Ta có:
\(x+\sqrt{x}+1\) ≥ 1
⇒ \(\frac{1}{x+\sqrt{x}+1}\) ≤ 1
⇒\(\frac{2}{x+\sqrt{x}+1}\) ≤ 2
⇒A ≤ 2 (2)
Từ (1) và (2) => 0 < A ≤ 2
1 tháng 8 2019
a) ĐKXĐ: x\(\ge0,x\ne1\)
A = \(\frac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\frac{\sqrt{x}-1}{2}\)
= \(\frac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x +\sqrt{x}+1\right)}.\frac{2}{\sqrt{x}-1}\)
= \(\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{2}{\sqrt{x}-1}\)
= \(\frac{2}{x+\sqrt{x}+1}\)
b) Ta có x\(\ge0,x\ne1\) =>\(x+\sqrt{x}+1>0\Rightarrow\frac{2}{x+\sqrt{x}+1}>0\)
=> A>0 (1)
Mặt khác \(x\ge0,x\ne1\Rightarrow x+\sqrt{x}+1\ge1\)
\(\Rightarrow\frac{2}{x+\sqrt{x}+1}\le2\) \(\Rightarrow A\ge2\) (2)
Từ (1) và (2) => \(0< A\le2\)
11 tháng 2 2020
ĐKXĐ:\(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
a) \(D=\left[\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right]:\frac{\sqrt{x}-1}{2}\)
\(=\frac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\frac{\sqrt{x}-1}{2}=\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{2}{\sqrt{x}-1}=\frac{2}{x+\sqrt{x}+1}\)
b) Ta có: \(x+\sqrt{x}+1>1\)
Suy ra \(D=\frac{2}{x+\sqrt{x}+1}< 2\)
\(\Rightarrow0< D< 2\)