Cho \(M=\frac{3\sqrt{x}+1}{\sqrt{x}-3}\) . Tìm x để M>0 , M<0
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NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 10 2019
ĐKXĐ: \(x\ge0;x\ne1\)
\(A=\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right).\frac{\left(x-1\right)^2}{2}\)
\(=\frac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}=-\sqrt{x}\left(\sqrt{x}-1\right)\)
\(=\sqrt{x}\left(1-\sqrt{x}\right)\)
\(0< x< 1\Rightarrow\left\{{}\begin{matrix}\sqrt{x}>0\\1-\sqrt{x}>0\end{matrix}\right.\) \(\Rightarrow\sqrt{x}\left(1-\sqrt{x}\right)>0\Rightarrow A>0\)
\(A< 0\Leftrightarrow\sqrt{x}\left(1-\sqrt{x}\right)< 0\Leftrightarrow1-\sqrt{x}< 0\Rightarrow x>1\)
\(A>-2\Leftrightarrow\sqrt{x}\left(1-\sqrt{x}\right)+2>0\Leftrightarrow-x+\sqrt{x}+2>0\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(2-\sqrt{x}\right)>0\Leftrightarrow2-\sqrt{x}>0\Rightarrow x< 4\)
Kết hợp ĐKXĐ \(\Rightarrow\left\{{}\begin{matrix}0\le x< 4\\x\ne1\end{matrix}\right.\)
\(A< -2x\Leftrightarrow\sqrt{x}-x< -2x\Leftrightarrow x+\sqrt{x}< 0\) (vô nghiệm \(\forall x\ge0\))
\(A>2\sqrt{x}\Leftrightarrow\sqrt{x}-x>2\sqrt{x}\Leftrightarrow x+\sqrt{x}< 0\) giống như trên
\(A=-x+\sqrt{x}=-x+\sqrt{x}-\frac{1}{4}+\frac{1}{4}=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)
\(A_{max}=\frac{1}{4}\) khi \(\sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 9 2019
ĐKXĐ: \(x>0;x\ne\left\{4;9\right\}\)
\(P=\left(\frac{-\left(\sqrt{x}+2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{4x}{x-4}\right).\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}\)
\(=\left(\frac{-x-4\sqrt{x}-4+x-4\sqrt{x}+4+4x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right).\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}\)
\(=\left(\frac{4\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right).\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)
\(=\frac{4x\left(\sqrt{x}-2\right)^2}{3-\sqrt{x}}\)
Hình như bạn ghi nhầm đề
NH
2 tháng 10 2019
a, C = \(\left(\frac{\sqrt{x}}{3+\sqrt{x}}-\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)
\(=\left[\frac{\sqrt{x}}{3+\sqrt{x}}-\frac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right]:\left[\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{1}{\sqrt{x}}\right]\)
\(=\frac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}:\frac{3\sqrt{x}+1-\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{3\sqrt{x}-x+x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}:\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{3\sqrt{x}+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}:\frac{2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(3+\sqrt{x}\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\)
đk: x>=0
với x>= 0 => \(3\sqrt{x}+1>0\)với mọi x
=> M>0 <=>\(\sqrt{x}-3>0\Leftrightarrow\sqrt{x}>3\Leftrightarrow x>9\)
M<0 <=> \(\sqrt{x}-3<0\Leftrightarrow\sqrt{x}<3\Leftrightarrow x<9\)