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1. \(B=\left(\frac{\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}\right).\frac{\sqrt{x}-2}{2}=\frac{\sqrt{x}+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}-2}{2}=\frac{\sqrt{x}+1}{\sqrt{x}+2}\)
b,
\(B=\frac{\sqrt{x}+1}{\sqrt{x}+2}< \frac{2}{3}=>3\sqrt{x}+3< 2\sqrt{x}+4=>\sqrt{x}< 1=>0\le x< 1\)
Vậy ...
\(A=\left(\sqrt{x}-\frac{x+2}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}-4}{1-x}\right)\) \(ĐKXĐ:x\ge0;x\ne1;x\ne4\)
\(A=\left[\frac{\sqrt{x}\left(\sqrt{x}+1\right)-x-2}{\sqrt{x}+1}\right]:\left[\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}-4}{x-1}\right]\)
\(A=\frac{x+\sqrt{x}-x-2}{\sqrt{x}+1}:\left[\frac{x-\sqrt{x}+\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\)
\(A=\frac{\sqrt{x}-2}{\sqrt{x}+1}:\frac{x-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{\sqrt{x}-2}{\sqrt{x}+1}.\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(A=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
vậy \(A=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
b)theo bài ra: \(A=\frac{1}{\sqrt{x}}\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+2}=\frac{1}{\sqrt{x}}\)
\(\Leftrightarrow\left(\sqrt{x}-1\right).\sqrt{x}=\sqrt{x}+2\)
\(\Leftrightarrow x-\sqrt{x}-\sqrt{x}-2=0\)
\(\Leftrightarrow x-2\sqrt{x}-2=0\)
\(\Leftrightarrow x-2\sqrt{x}+1-3=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2-\left(\sqrt{3}\right)^2=0\)
\(\Leftrightarrow\left(\sqrt{x}-1-\sqrt{3}\right)\left(\sqrt{x}-1+\sqrt{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-1-\sqrt{3}=0\\\sqrt{x}-1+\sqrt{3}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=\sqrt{3}+1\\\sqrt{x}=1-\sqrt{3}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\left(\sqrt{3}+1\right)^2\\x=\left(1-\sqrt{3}\right)^2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3+2\sqrt{3}+1\\x=3-2\sqrt{3}+1\end{cases}}\)
vậy......
Ta có: \(B=\frac{\sqrt{x}}{x+1}-\frac{4\sqrt{x}+2}{x\sqrt{x}-2x+\sqrt{x}-2}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(x+1\right)\left(\sqrt{x}-2\right)}-\frac{4\sqrt{x}+2}{\left(x+1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-2\sqrt{x}-4\sqrt{x}-2}{\left(x+1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-6\sqrt{x}-2}{\left(x+1\right)\left(\sqrt{x}-2\right)}\)
Lời giải:
Có:
$\frac{\sqrt{x}+1}{\sqrt{x}+2}=\frac{\sqrt{x}+2-1}{\sqrt{x}+2}=1-\frac{1}{\sqrt{x}+2}$
Ta thấy:
$\sqrt{x}\geq 0, \forall x\geq 0; x\neq 4\Rightarrow \sqrt{x}+2\geq 2$
$\Rightarrow \frac{1}{\sqrt{x}+2}\leq \frac{1}{2}$
$\Rightarrow \frac{\sqrt{x}+1}{\sqrt{x}+2}=1-\frac{1}{\sqrt{x}+2}\geq 1-\frac{1}{2}=\frac{1}{2}$
Vậy GTNN của biểu thức là $\frac{1}{2}$ xảy ra khi $\sqrt{x}=0$ hay $x=0$