a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b: Thay x=0 vào B, ta được:
\(B=\dfrac{5\cdot0}{0-1}-\dfrac{26}{0+3}-\dfrac{20}{0+2\cdot0-3}\)
\(=0-\dfrac{26}{3}-\dfrac{20}{-3}\)
\(=-\dfrac{26}{3}+\dfrac{20}{3}\)
=-2
c: Ta có: \(B=\dfrac{5\sqrt{x}}{\sqrt{x}-1}-\dfrac{26}{\sqrt{x}+3}-\dfrac{20}{x+2\sqrt{x}-3}\)
\(=\dfrac{5x+15\sqrt{x}-26\sqrt{x}+26-20}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{5x-11\sqrt{x}+6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{5\sqrt{x}-6}{\sqrt{x}+3}\)
a, B xác định khi:
\(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-1\ne0\\x+2\sqrt{x}-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\\x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b, Khi x=0:
\(B=\dfrac{5\sqrt{x}}{\sqrt{x}-1}-\dfrac{26}{\sqrt{x}+3}-\dfrac{20}{x+2\sqrt{x}-3}\)
\(=\dfrac{5\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{26\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{20}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{5x+15\sqrt{x}-26\sqrt{x}+26-20}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{5x-11\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(5\sqrt{x}-6\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{5\sqrt{x}-6}{\sqrt{x}+3}=-2\)
d: Để \(B=-\dfrac{1}{4}\) thì \(\dfrac{5\sqrt{x}-6}{\sqrt{x}+3}=\dfrac{-1}{4}\)
\(\Leftrightarrow20\sqrt{x}-24=-\sqrt{x}-3\)
\(\Leftrightarrow21\sqrt{x}=21\)
hay x=1(loại)
e: Để \(B=\dfrac{\sqrt{x}+2}{5}\) thì \(\dfrac{5\sqrt{x}-6}{\sqrt{x}+3}=\dfrac{\sqrt{x}+2}{5}\)
\(\Leftrightarrow x+5\sqrt{x}+6=25\sqrt{x}-30\)
\(\Leftrightarrow x-20\sqrt{x}+36=0\)
\(\Leftrightarrow x-18\sqrt{x}-2\sqrt{x}+36=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-18\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=324\left(nhận\right)\end{matrix}\right.\)
giúp em với ạ. Em cảm ơn ạ
