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NT
1
27 tháng 6 2017
\(A=\frac{x}{2}+\sqrt{1-x-2x^2}\)
\(2A-x=2\sqrt{1-x-2x^2}\)\(\left(A\ge\frac{x}{2}\right)\)
\(\Leftrightarrow4A^2-4Ax+x^2=4-4x-8x^2\)
\(\Leftrightarrow9x^2+\left(4-4A\right)x+4A^2-4=0\)
Để phương trình có nghiệm thì
\(\Delta'=\left(2-2A\right)^2-9.\left(4A^2-4\right)\ge0\)
\(\Rightarrow A\le1\)
KN
11 tháng 3 2020
a) \(A=\frac{4}{\sqrt{x}+3}+\frac{2x-\sqrt{x}-13}{x-9}-\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(=\frac{4\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{2x-\sqrt{x}-13}{x-9}-\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{4\sqrt{x}-12}{x-9}+\frac{2x-\sqrt{x}-13}{x-9}-\frac{x+3\sqrt{x}}{x-9}\)
\(=\frac{4\sqrt{x}-12+2x-\sqrt{x}-13-x-3\sqrt{x}}{x-9}\)
\(=\frac{x-25}{x-9}\)
b) \(P=\frac{A}{B}=\frac{\frac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}}{\frac{\sqrt{x}+5}{\sqrt{x}-3}}\)
\(=\frac{\sqrt{x}-5}{\sqrt{x}+3}\)
\(\sqrt{P}< \frac{1}{3}\Rightarrow\sqrt{\frac{\sqrt{x}-5}{\sqrt{x}+3}}< \frac{1}{3}\)
\(\Rightarrow\frac{\sqrt{x}-5}{\sqrt{x}+3}< \frac{1}{9}\Leftrightarrow9\sqrt{x}-45< \sqrt{x}+3\)
\(\Leftrightarrow8\sqrt{x}< 48\Leftrightarrow\sqrt{x}< 6\Rightarrow0\le x< 36\)
11 tháng 3 2020
\(a,\)\(A=\frac{4}{\sqrt{x}+3}+\frac{2x-\sqrt{x}-13}{x-9}=\frac{4\left(\sqrt{x}-3\right)+2x-\sqrt{x}-13}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{4\sqrt{x}-12+2x-\sqrt{x}-13}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)\(=\frac{2x+3\sqrt{x}-1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(b,P=\frac{A}{B}=\frac{2x+3\sqrt{x}-1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\frac{\sqrt{x}+5}{\sqrt{x}-3}\)
\(=\frac{2x+3\sqrt{x}-1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\left(\sqrt{x}-3\right)}{\sqrt{x}+5}=\frac{2x+3\sqrt{x}-1}{\sqrt{x}+5}\)
Để \(\sqrt{P}< \frac{1}{3}\Rightarrow\frac{2x+3\sqrt{x}-1}{\sqrt{x}+5}< \frac{1}{3}\)
\(\Rightarrow\frac{2x+3\sqrt{x}-1}{\sqrt{x}+5}-\frac{1}{3}< 0\)
\(\Rightarrow\frac{3\left(2x+3\sqrt{x}-1\right)-\sqrt{x}-5}{3\left(\sqrt{x}+5\right)}< 0\)
\(\Rightarrow6x+9\sqrt{x}-3-\sqrt{x}-5< 0\)( do \(3\left(\sqrt{x}+5\right)>0\))
\(\Rightarrow6x-8\sqrt{x}-8< 0\Rightarrow3x-4\sqrt{x}-4< 0\)
\(\Rightarrow3x-6\sqrt{x}+2\sqrt{x}-4< 0\)
\(\Rightarrow3\sqrt{x}\left(\sqrt{x}-2\right)+2\left(\sqrt{x}-2\right)< 0\)
\(\Rightarrow\left(\sqrt{x}-2\right)\left(3\sqrt{x}+2\right)< 0\)
Vì \(3\sqrt{x}+2>0\Rightarrow\sqrt{x}-2< 0\)
\(\Rightarrow\sqrt{x}< 2\Rightarrow x< 4\)
Vậy để \(\sqrt{P}< \frac{1}{3}\)thì \(0\le x< 4\)
MN
8 tháng 7 2020
a) \(ĐKXĐ:\hept{\begin{cases}x>0\\x\ne9\end{cases}}\)
\(C=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)
\(\Leftrightarrow C=\frac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{9-x}:\frac{3\sqrt{x}+1-\sqrt{x}+3}{x-3\sqrt{x}}\)
\(\Leftrightarrow C=\frac{3\sqrt{x}+9}{9-x}:\frac{2\sqrt{x}+4}{x-3\sqrt{x}}\)
\(\Leftrightarrow C=\frac{3}{3-\sqrt{x}}\cdot\frac{x-3\sqrt{x}}{2\sqrt{x}+4}\)
\(\Leftrightarrow C=\frac{-3}{2\sqrt{x}+4}\)
b) Để \(-\frac{3}{2\sqrt{x}+4}< -1\)
\(\Leftrightarrow\frac{1+2\sqrt{x}}{2\sqrt{x}+4}< 0\)
Vì \(\hept{\begin{cases}1+2\sqrt{x}>0\\2\sqrt{x}+4>0\end{cases}\Leftrightarrow C>0}\)
Vậy để C <-1 <=> \(x\in\varnothing\)
c) \(A=\frac{1}{\sqrt{3}-\sqrt{2}}=\sqrt{3}+\sqrt{2}\)
\(\Leftrightarrow A^2=3+2+2\sqrt{5}=5+2\sqrt{5}\)
\(B=\sqrt{5}+1\)
\(\Leftrightarrow B^2=5+1+2\sqrt{5}=6+2\sqrt{5}\)
Vì \(5+2\sqrt{5}< 6+2\sqrt{5}\)
\(\Leftrightarrow A^2< B^2\)
\(\Leftrightarrow A< B\)
Vậy \(\frac{1}{\sqrt{3}-\sqrt{2}}< \sqrt{5}+1\)
13 tháng 7 2018
\(x=9\Rightarrow\sqrt{x}=3\Rightarrow A=\frac{3+2}{3-5}=\frac{5}{-2}=-\frac{5}{2}\\ \)
\(B=\frac{3}{\sqrt{x}+5}+\frac{20-2\sqrt{x}}{x-25}=\frac{3.\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right).\left(\sqrt{x}-5\right)}+\frac{20-2\sqrt{x}}{\left(x+\sqrt{5}\right).\left(x-\sqrt{5}\right)}\)
\(=\frac{3\sqrt{x}-15+20-2\sqrt{x}}{\left(\sqrt{x}+5\right).\left(\sqrt{x}-5\right)}=\frac{\sqrt{x}+5}{\left(\sqrt{x}+5\right).\left(\sqrt{x}-5\right)}=\frac{1}{\sqrt{x}-5}\)
\(A=B.\left|x-4\right|\Leftrightarrow\left|x-4\right|=A:B=\frac{\sqrt{x}+2}{\sqrt{x}-5}:\frac{1}{\sqrt{x}-5}=\sqrt{x}+2\)
\(\Rightarrow\left(x-4\right)^2=\left(\sqrt{x}+2\right)^2\Leftrightarrow x^2-8x+16=x+4\sqrt{x}+4\)
\(\Leftrightarrow x^2-9x-4\sqrt{x}+12=0\Leftrightarrow x.\left(x-9\right)-4.\left(\sqrt{x}-3\right)=0\)
\(\Leftrightarrow x.\left(\sqrt{x}-3\right).\left(\sqrt{x}+3\right)-4.\left(\sqrt{x}-3\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-3\right).\left(x\sqrt{x}+3x-4\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-3\right).\left(\left(x\sqrt{x}-x\right)+\left(4x-4\right)\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-3\right).\left(x.\left(\sqrt{x}-1\right)+4.\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-3\right).\left(\sqrt{x}-1\right).\left(x+4\sqrt{x}+4\right)=0\Leftrightarrow\left(\sqrt{x}-3\right).\left(\sqrt{x}-1\right).\left(\sqrt{x}+2\right)^2=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}-3=0\\\sqrt{x}-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=9\\x=1\end{cases}}}\)(Vì \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+2\ge2\Rightarrow\left(\sqrt{x}+2\right)^2\ge4>0\))
5 tháng 2 2022
a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)