Cho P= \(\frac{\sqrt{x}-3}{\sqrt{x}+3}\)với \(x\ge0,x\ne9\)
Tìm x thỏa mãn \(P\left(\sqrt{x}+3\right)\ge x+5\sqrt{x-1}-\sqrt{x}-2\)
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31 tháng 8 2023
a: \(P=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\cdot\dfrac{\sqrt{x}-7+\sqrt{x}+1}{\sqrt{x}+1}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\cdot\dfrac{2\sqrt{x}-6}{\sqrt{x}+1}\)
\(=\dfrac{-3\sqrt{x}-3}{\sqrt{x}+1}\cdot\dfrac{2}{\sqrt{x}+3}=-\dfrac{6}{\sqrt{x}+3}\)
b: P>=-1/2
=>P+1/2>=0
=>\(\dfrac{-6}{\sqrt{x}+3}+\dfrac{1}{2}>=0\)
=>\(\dfrac{-12+\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}>=0\)
=>căn x-9>=0
=>x>=81
c: căn x+3>=3
=>6/căn x+3<=6/3=2
=>-6/căn x+3>=-2
Dấu = xảy ra khi x=0
2 tháng 7 2015
câu 2:\(\Leftrightarrow\frac{\sqrt{x}-2}{\sqrt{x}+1}.\left(\sqrt{x}+1\right)=m\left(x+1\right)-2\Leftrightarrow\sqrt{x}-2-mx-m+2=0\Leftrightarrow\sqrt{x}=m\left(x+1\right)\Leftrightarrow m=\frac{\sqrt{x}}{x+1}\)
vì x>=0 =>x+1>0 \(\sqrt{x}\ge0\)=> m phải >=0
\(x\ne4\Rightarrow x+1\ne5;\sqrt{x}\ne2\Rightarrow m\ne\frac{2}{5}\)
\(x\ne9\Rightarrow x+1\ne10;\sqrt{x}\ne3\Rightarrow m\ne\frac{3}{10}\)
17 tháng 10 2021
\(a,A=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\\ A=\dfrac{x-6\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{x-6\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)
B
1 tháng 8 2018
\(\left(\dfrac{3x-3\sqrt{x}-3}{x+\sqrt{x}-2}+\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+2}\right)\) : \(\dfrac{1}{\sqrt{x}+2}\)
=\(\dfrac{3x-3\sqrt{x}-3+\sqrt{x}+2-\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+2)}\) .\(\sqrt{x}+2\)
=\(\dfrac{(3x-3\sqrt{x})(\sqrt{x}+2)}{(\sqrt{x}-1)(\sqrt{x}+2)}\)
=\(\dfrac{3\sqrt{x}(\sqrt{x}-1)(\sqrt{x}+2)}{(\sqrt{x}-1)(\sqrt{x}+2)}\) =\(3\sqrt{x}\)
18 tháng 9 2021
a) \(D=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}-1}\)
\(=\dfrac{-3\sqrt{x}+3}{\sqrt{x}+3}.\dfrac{1}{\sqrt{x}-1}=\dfrac{-3}{\sqrt{x}+3}\)
b) \(D=-\dfrac{3}{\sqrt{x}+3}< -\dfrac{1}{4}\)
\(\Leftrightarrow12>\sqrt{x}+3\Leftrightarrow\sqrt{x}< 9\)
\(\Leftrightarrow0\le x< 81\) và \(x\ne9\)
18 tháng 9 2021
a) D=\(\left(\dfrac{2\sqrt{x}.\left(\sqrt{x}-3\right)+\sqrt{x}.\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-3\right)}\right)\) \(:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(\Leftrightarrow D=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right).\left(\sqrt{x}+3\right)}\) \(.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(\Leftrightarrow D=\dfrac{-3-3\sqrt{x}}{\sqrt{x}+3}.\dfrac{1}{\sqrt{x}+1}\)
\(\Leftrightarrow D=\dfrac{-3.\left(\sqrt{x}+1\right)}{\sqrt{x}+3}.\dfrac{1}{\sqrt{x}+1}\)
\(\Leftrightarrow D=\dfrac{-3}{\sqrt{x}+3}\)
b) Để D\(< \dfrac{-1}{4}\) \(\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}< \dfrac{-1}{4}\)
\(\Leftrightarrow12>\sqrt{x}+3\Leftrightarrow9>\sqrt{x}\Leftrightarrow81>x\ge0\)
NV
30 tháng 7 2016
a/ \(P=\left[1-\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]:\left[\frac{3-\sqrt{x}}{\sqrt{x}-2}+\frac{\sqrt{x}-2}{\sqrt{x}+3}-\frac{9x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]\)
\(=\left(1-\frac{\sqrt{x}}{\sqrt{x}+3}\right):\left[\frac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)+\left(\sqrt{x}-2\right)^2-9x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]\)
\(=\left(\frac{\sqrt{x}+3-\sqrt{x}}{\sqrt{x}+3}\right):\left[\frac{9-x+x-4\sqrt{x}+4-9x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]\)
\(=\frac{3}{\sqrt{x}+3}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{13-4\sqrt{x}-9x}\)
\(=\frac{3\sqrt{x}-6}{13-4\sqrt{x}-9x}\)
b/ \(P=1\Rightarrow\frac{3\sqrt{x}-6}{13-4\sqrt{x}-9x}=1\Rightarrow3\sqrt{x}-6=13-4\sqrt{x}-9x\)
\(\Rightarrow9x+7\sqrt{x}-19=0\)
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