Ôn tập chương 1: Căn bậc hai. Căn bậc ba

a: Để A nguyên thì \(\sqrt{x}-3⋮\sqrt{x}+1\)

\(\Leftrightarrow\sqrt{x}+1\in\left\{1;2;4\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{0;1;3\right\}\)

hay \(x\in\left\{0;1;9\right\}\)

a: Ta có: \(A=\left(1-\dfrac{2\sqrt{x}-2}{x-1}\right):\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{x\sqrt{x}+1}\right)\)

\(=\dfrac{x-1-2\sqrt{x}+2}{x-1}:\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)

\(=\dfrac{x-\sqrt{x}+1}{\sqrt{x}-1}\)

Bài 7:

Ta có: \(P=\left(\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right):\left(1+\dfrac{\sqrt{x}}{x+1}\right)\)

\(=\dfrac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{x+1}{x+\sqrt{x}+1}\)

\(=\dfrac{2\sqrt{x}}{x\sqrt{x}-1}\)

đề sai kìa bn

\(a,E=\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}:\dfrac{\sqrt{x}+1+x}{\sqrt{x}\left(\sqrt{x}+1\right)}\\ E=\dfrac{1}{\sqrt{x}+1}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\\ b,E=\dfrac{2}{7}\Leftrightarrow7\sqrt{x}=2x+2\sqrt{x}+2\\ \Leftrightarrow2x-5\sqrt{x}+2=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2\\\sqrt{x}=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=\dfrac{1}{4}\left(tm\right)\end{matrix}\right.\\ c,\dfrac{1}{E}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}+\dfrac{1}{\sqrt{x}}+1\ge2\sqrt{\dfrac{\sqrt{x}}{\sqrt{x}}}+1=3\\ \Leftrightarrow E\le\dfrac{1}{3}\\ d,E\le\dfrac{1}{3}\left(cm.câu.C\right)\\ E_{max}=\dfrac{1}{3}\Leftrightarrow x=1\)