Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1/ ĐKXĐ: \(\hept{\begin{cases}x>0\\x\ne4\end{cases}}\)
\(A=\left[\frac{x}{\sqrt{x}\left(x-4\right)}-\frac{6}{3\left(\sqrt{x}-2\right)}+\frac{1}{\sqrt{x}-2}\right]:\left(\frac{x-4+10-x}{\sqrt{x}+2}\right)\)
\(=\left[\frac{\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{2}{\sqrt{x}-2}+\frac{1}{\sqrt{x}-2}\right]:\left(\frac{6}{\sqrt{x}+2}\right)\)
\(=\frac{\sqrt{x}-2\left(\sqrt{x}+2\right)+\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\frac{\left(\sqrt{x}+2\right)}{6}\)
\(=\frac{-2}{\sqrt{x}-2}.\frac{1}{6}=-\frac{1}{3\left(\sqrt{x}-2\right)}\)
2/ Để \(A>2\Rightarrow\frac{-1}{3\left(\sqrt{x}-2\right)}>2\)\(\Rightarrow6\sqrt{x}-12+1>0\Rightarrow6\sqrt{x}-11>0\Rightarrow\sqrt{x}>\frac{11}{6}\)
\(\Rightarrow x>\frac{121}{36}\)
\(B=\frac{-2a\sqrt{a}+2a^2}{\left(\sqrt{a}-\right)\left(a-1\right)}\)
\(C=-x\sqrt{x}+x+\sqrt{x}-1\)
\(D=x-\sqrt{x}+1\)
\(P=\left(\frac{x+\sqrt{x}}{\sqrt{x}+1}+1\right):\left(\frac{x-\sqrt{x}}{\sqrt{x}-1}-1\right)\) \(ĐKXĐ:x\ne1\)
\(P=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}+1\right):\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-1\right)\)
\(P=\left(\sqrt{x}+1\right):\left(\sqrt{x}-1\right)\)
\(P=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
b) theo bài ra \(P=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
theo bài ra \(P< 1\)\(\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}-1}< 1\)
\(\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}-1}-1< 0\)
\(\Leftrightarrow\frac{\sqrt{x}+1-\sqrt{x}+1}{\sqrt{x}-1}< 0\)
\(\Leftrightarrow\frac{2}{\sqrt{x}-1}< 0\)
\(\Rightarrow\sqrt{x}-1< 0\)
\(\Rightarrow\sqrt{x}< 1\)
\(\Rightarrow x< 1\) ( tm ĐK \(x\ne1\))
vậy \(x< 1\)thì \(P< 1\)