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AH
Akai Haruma
Giáo viên
12 tháng 7 2018

Lời giải:

\(P=\frac{\sqrt{x}-1}{\sqrt{x}+1}=\frac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\frac{2}{\sqrt{x}+1}\)

Ta thấy vì \(\sqrt{x}\geq 0, \forall x\geq 0\Rightarrow \sqrt{x}+1\geq 1\)

\(\Rightarrow \frac{2}{\sqrt{x}+1}\leq \frac{2}{1}=2\)

\(\Rightarrow P=1-\frac{2}{\sqrt{x}+1}\geq 1-2=-1\)

Vậy \(P_{\min}=-1\Leftrightarrow x=0\)

Ta có: \(P=A\cdot B\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

14 tháng 6 2018

\(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}=\dfrac{\sqrt{x}+2-3}{\sqrt{x}+2}=1-\dfrac{3}{\sqrt{x}+2}\ge1-\dfrac{3}{2}=-\dfrac{1}{2}\)

\("="\Leftrightarrow x=0\)

22 tháng 8 2021

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24 tháng 10 2023

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8 tháng 7 2023

\(a,P\) xác định \(\Leftrightarrow\left[{}\begin{matrix}x>0\\x\ne1\\x\ne4\end{matrix}\right.\)

\(b,P=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\\ =\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\\ =\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{x-1-x+4}\\ =\dfrac{1}{\sqrt{x}}.\dfrac{\sqrt{x}-2}{3}\\ =\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)

\(c,P=\dfrac{1}{4}\Leftrightarrow\dfrac{\sqrt{x}-2}{3\sqrt{x}}=\dfrac{1}{4}\\ \Leftrightarrow\dfrac{4\left(\sqrt{x}-2\right)-3\sqrt{x}}{12\sqrt{x}}=0\\ \Leftrightarrow4\sqrt{x}-8-3\sqrt{x}=0\\ \Leftrightarrow\sqrt{x}=8\\ \Leftrightarrow x=64\left(tmdk\right)\)

Vậy \(x=64\) thì \(P=\dfrac{1}{4}\)

7 tháng 12 2021

\(a,P=\dfrac{\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{2-\sqrt{x}}{\sqrt{x}}=\dfrac{-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{-2}{\sqrt{x}+2}\\ P=-\dfrac{3}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\\ \Leftrightarrow3\sqrt{x}+6=10\Leftrightarrow\sqrt{x}=\dfrac{4}{3}\Leftrightarrow x=\dfrac{16}{9}\left(tm\right)\)

8 tháng 12 2021

\(P=-\dfrac{3}{5}\) sao suy ra đc \(\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\) thế

 

7 tháng 7 2021

a) \(A=\sqrt{28}-\sqrt{63}+\dfrac{7+\sqrt{7}}{\sqrt{7}}-\sqrt{\left(\sqrt{7}+1\right)^2}\)

\(=2\sqrt{7}-3\sqrt{7}+\dfrac{\sqrt{7}\left(\sqrt{7}+1\right)}{\sqrt{7}}-\left|\sqrt{7}+1\right|\)

\(=-\sqrt{7}+\sqrt{7}+1-\sqrt{7}-1=-\sqrt{7}\)

\(B=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}-3}\right)\dfrac{4\sqrt{x}+12}{\sqrt{x}}\)

\(=\dfrac{\sqrt{x}-3+\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{4\left(\sqrt{x}+3\right)}{\sqrt{x}}=\dfrac{2\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{4\left(\sqrt{x}+3\right)}{\sqrt{x}}\)

\(=\dfrac{8}{\sqrt{x}-3}\)

b) \(A>B\Rightarrow-\sqrt{7}>\dfrac{8}{\sqrt{x}-3}\Rightarrow\dfrac{8}{\sqrt{x}-3}+\sqrt{7}< 0\)

\(\Rightarrow\dfrac{\sqrt{7x}+8-3\sqrt{7}}{\sqrt{x}-3}< 0\)

Ta có: \(\left\{{}\begin{matrix}8=\sqrt{64}\\3\sqrt{7}=\sqrt{63}\end{matrix}\right.\Rightarrow8-3\sqrt{7}>0\Rightarrow8-3\sqrt{7}+\sqrt{7x}>0\)

\(\Rightarrow\sqrt{x}-3< 0\Rightarrow\sqrt{x}< 3\Rightarrow x< 9\Rightarrow0< x< 9\)