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11 tháng 11 2018
\(R=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}+\dfrac{1}{\sqrt{x}}+1\ge2+1=3\)
NT
0
19 tháng 3 2021
a) Với \(x>0;x\ne1\), ta có:
\(P=\left(\frac{x-2}{x+2\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right).\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(P=\left[\frac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\frac{1}{\sqrt{x}+2}\right].\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(P=\left[\frac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\right].\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(P=\frac{x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{\sqrt{x}-1}{\sqrt{x}}.\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(P=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}}\)
Vậy với \(x>0,x\ne1\)thì \(P=\frac{\sqrt{x}+1}{\sqrt{x}}\)
19 tháng 3 2021
\(P=\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(\Rightarrow2P=\frac{2\sqrt{x}+2}{\sqrt{x}}\)
\(2P=2\sqrt{x}+5\Leftrightarrow\frac{2\sqrt{x}+2}{\sqrt{x}}=2\sqrt{x}+5\left(ĐKXĐ:x\ne0\right)\left(1\right)\)
Mà theo đề bài : \(x>0\)nên phương trình luôn được xác định.
\(\left(1\right)\Leftrightarrow\frac{2\sqrt{x}+2}{\sqrt{x}}=\frac{\sqrt{x}\left(2\sqrt{x}+5\right)}{\sqrt{x}}\)
\(\Rightarrow2\sqrt{x}+2=\sqrt{x}\left(2\sqrt{x}+5\right)\)
\(\Leftrightarrow2\sqrt{x}+2=2x+5\sqrt{x}\)
\(\Leftrightarrow2\sqrt{x}+2-2x-5\sqrt{x}\)
\(\Leftrightarrow-2x-3\sqrt{x}+2=0\Leftrightarrow2x+3\sqrt{x}-2=0\)
\(\Leftrightarrow\left(2\sqrt{x}-1\right)\left(\sqrt{x}+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2\sqrt{x}-1=0\\\sqrt{x}+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2\sqrt{x}=1\\\sqrt{x}=-2\left(vn\right)\end{cases}}\Leftrightarrow2\sqrt{x}=1\)
\(\Leftrightarrow\sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}\left(TMĐK:x>0;x\ne1\right)\)
Vậy \(2P=2\sqrt{x}+5\Leftrightarrow x=\frac{1}{4}\)
3 tháng 9 2022
1: \(=3\left(x+\dfrac{2}{3}\sqrt{x}+\dfrac{1}{3}\right)\)
\(=3\left(x+2\cdot\sqrt{x}\cdot\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{2}{9}\right)\)
\(=3\left(\sqrt{x}+\dfrac{1}{3}\right)^2+\dfrac{2}{3}>=3\cdot\dfrac{1}{9}+\dfrac{2}{3}=1\)
Dấu '=' xảy ra khi x=0
2: \(=x+3\sqrt{x}+\dfrac{9}{4}-\dfrac{21}{4}=\left(\sqrt{x}+\dfrac{3}{2}\right)^2-\dfrac{21}{4}>=-3\)
Dấu '=' xảy ra khi x=0
3: \(A=-2x-3\sqrt{x}+2< =2\)
Dấu '=' xảy ra khi x=0
5: \(=x-2\sqrt{x}+1+1=\left(\sqrt{x}-1\right)^2+1>=1\)
Dấu '=' xảy ra khi x=1
22 tháng 11 2022
c: \(=\sqrt{\dfrac{4}{16-6\sqrt{7}}}+\sqrt{7}\)
\(=\dfrac{2}{3-\sqrt{7}}+\sqrt{7}\)
\(=3+2\sqrt{7}\)
d: \(=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-5\sqrt{x}-2}{x-4}\)
\(=\dfrac{3x-6\sqrt{x}}{x-4}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
30 tháng 7 2018
4 , Ta có :
\(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x-9}{x-9}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}+\dfrac{2\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}-\dfrac{3\left(x-3\right)}{x-9}\)
\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x+9}{x-9}\)
\(=\dfrac{3\sqrt{x}+9}{x-9}\)
\(=\dfrac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{3}{\sqrt{x}-3}\)
30 tháng 7 2018
2 , Ta có :
\(\dfrac{x\sqrt{x}+1}{x-1}-\dfrac{x-1}{\sqrt{x}+1}=\dfrac{x\sqrt{x}+1}{x-1}-\dfrac{\left(x-1\right)\left(\sqrt{x}-1\right)}{x-1}\)
\(=\dfrac{x\sqrt{x}+1}{x-1}-\dfrac{x\sqrt{x}-x-\sqrt{x}+1}{x-1}\)
\(=\dfrac{x\sqrt{x}+1-x\sqrt{x}+x+\sqrt{x}-1}{x-1}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
26 tháng 5 2022
a: \(A=\dfrac{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}-\dfrac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}-3}-6\)
\(=-\sqrt{x}+3-\sqrt{x}+3-6=-2\sqrt{x}\)
b: \(\left(\dfrac{2\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right):\left(\dfrac{2\sqrt{x}}{\sqrt{x}+1}-1\right)\)
\(=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x+1\right)}-\dfrac{1}{\sqrt{x}+1}\right):\dfrac{2\sqrt{x}-\sqrt{x}-1}{\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}-x-1}{\left(\sqrt{x}+1\right)\left(x+1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{1}{x+1}\)
g: \(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)\left(\dfrac{x-1}{\sqrt{x}+1}-2\right)\)
\(=\dfrac{\sqrt{x}+1+\sqrt{x}-1}{x-1}\cdot\left(\sqrt{x}-1-2\right)\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{x-1}\)
9 tháng 4 2018
\(A=B:C\)
\(C=\dfrac{x+\sqrt{x}}{x-2\sqrt{x}+1}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\)
\(B=\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{1}{1-\sqrt{x}}+\dfrac{2-x}{x-\sqrt{x}}=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(B=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}.\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
\(\left\{{}\begin{matrix}x>0;\ne1\\A=\dfrac{x}{\sqrt{x}-1}\end{matrix}\right.\)
4 tháng 8 2017
b) \(B=\dfrac{x-\sqrt{x}}{1-\sqrt{x}}-\dfrac{x\sqrt{x}}{\sqrt{x}}=\dfrac{\sqrt{x}\left(x-\sqrt{x}\right)-x\sqrt{x}\left(1-\sqrt{x}\right)}{\sqrt{x}\left(1-\sqrt{x}\right)}\) = \(\dfrac{x\sqrt{x}-x-x\sqrt{x}+x^2}{\sqrt{x}-x}=\dfrac{x^2-x}{\sqrt{x}-x}\)
c) \(C=\dfrac{x+2\sqrt{x}}{\sqrt{x}-x}-\dfrac{x\sqrt{x}}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}+1\right)\left(x+2\sqrt{x}\right)-x\sqrt{x}\left(\sqrt{x}-x\right)}{\left(\sqrt{x}-x\right)\left(\sqrt{x}+1\right)}=x+2\sqrt{x}-x\sqrt{x}\)
\(d,D=\dfrac{x+2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{x-4}=\dfrac{x+2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\) \(\dfrac{\left(x+2\sqrt{x}\right)\left(\sqrt{x}-2\right)+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{x+7\sqrt{x}-2}{\sqrt{x}+2}\)
e) \(E=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{\sqrt{x}-24}{x-9}=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)+\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\) = \(\dfrac{2\sqrt{x}-24}{\sqrt{x}+3}\)
F) F = \(\dfrac{3}{\sqrt{x}+5}+\dfrac{20-2\sqrt{x}}{x-25}=\dfrac{3\left(\sqrt{x}-5\right)+20-2\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\dfrac{23-2\sqrt{x}}{\sqrt{x}+5}\)
Lời giải:
ĐKXĐ: $x>0$
Áp dụng BĐT Cô-si:
$x+2\geq 2\sqrt{2x}$
$\Rightarrow x+2\sqrt{x}+2\geq \sqrt{x}(2+2\sqrt{2})$
$\Rightarrow \frac{x+2\sqrt{x}+2}{\sqrt{x}}\geq 2+2\sqrt{2}$
$\Rightarrow P=-\frac{x+2\sqrt{x}+2}{\sqrt{x}}\leq -(2+2\sqrt{2})$
Vậy $P_{\max}=-(2+2\sqrt{2})$ khi $x=2$