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a: \(Q=\dfrac{a}{\sqrt{a^2-b^2}}-\dfrac{\sqrt{a^2-b^2}+a}{\sqrt{a^2-b^2}}\cdot\dfrac{a-\sqrt{a^2-b^2}}{b}\)
\(=\dfrac{ab}{b\left(\sqrt{a^2-b^2}\right)}-\dfrac{a^2-\left(a^2-b^2\right)}{b\sqrt{a^2-b^2}}\)
\(=\dfrac{ab-a^2+a^2-b^2}{b\sqrt{a^2-b^2}}=\dfrac{ab-b^2}{b\sqrt{a^2-b^2}}=\dfrac{a-b}{\sqrt{a^2-b^2}}\)
b: Khi a=3b thì \(Q=\dfrac{3b-b}{\sqrt{9b^2-b^2}}=\dfrac{2b}{\sqrt{8b^2}}=\dfrac{2b}{2\sqrt{2}\cdot b}=\dfrac{1}{\sqrt{2}}\)
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-1}{\sqrt{x}}\)
b: Thay \(x=3+2\sqrt{2}\) vào P, ta được:
\(P=\dfrac{2\sqrt{2}+2}{\sqrt{2}+1}=2\)
a) Vì khi a>0 và \(a\notin\left\{4;1\right\}\) thì \(\left\{{}\begin{matrix}\sqrt{a}-1\ne0\\\sqrt{a}\ne0\\\sqrt{a}-2\ne0\end{matrix}\right.\)
nên Q xác định
b) Ta có: \(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
Để Q dương thì \(\sqrt{a}-2>0\)
\(\Leftrightarrow a>4\)
Kết hợp ĐKXĐ, ta được: a>4
a. Đề là \(Q=\frac{a}{\sqrt{a^2-b^2}}-\left(1+\frac{a}{\sqrt{a^2-b^2}}\right):\frac{b}{a-\sqrt{a^2-b^2}}\) ?
\(\Leftrightarrow Q=\frac{a}{\sqrt{a^2-b^2}}-\frac{a+\sqrt{a^2-b^2}}{\sqrt{a^2-b^2}}.\frac{a-\sqrt{a^2-b^2}}{b}\)
\(\Leftrightarrow Q=\frac{a}{\sqrt{a^2-b^2}}-\frac{\left(a+\sqrt{a^2-b^2}\right)\left(a-\sqrt{a^2-b^2}\right)}{b\sqrt{a^2-b^2}}\)
\(\Leftrightarrow Q=\frac{a}{\sqrt{a^2-b^2}}-\frac{a^2-\left(a^2-b^2\right)}{b\sqrt{a^2-b^2}}\)
\(\Leftrightarrow Q=\frac{a}{\sqrt{a^2-b^2}}-\frac{b^2}{b\sqrt{a^2-b^2}}\)
\(\Leftrightarrow Q=\frac{a}{\sqrt{a^2-b^2}}-\frac{b}{\sqrt{a^2-b^2}}\)
\(\Leftrightarrow Q=\frac{a-b}{\sqrt{\left(a-b\right)\left(a+b\right)}}=\sqrt{\frac{a-b}{a+b}}\)
b. Thay a = 3b vào Q, ta được : \(Q=\sqrt{\frac{3b-b}{3b+b}}=\sqrt{\frac{2b}{4b}}=\sqrt{\frac{1}{2}}\)
\(a,Q=\frac{a}{\sqrt{a^2-b^2}}-\left(1+\frac{a}{\sqrt{a^2-b^2}}\right):\left(\frac{b}{a-\sqrt{a^2-b^2}}\right)\)
\(=\frac{a}{\sqrt{a^2-b^2}}-\left(\frac{\sqrt{a^2-b^2}+a}{\sqrt{a^2-b^2}}\right):\frac{b}{a-\sqrt{a^2+b^2}}\)
\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{a+\sqrt{a^2-b^2}}{\sqrt{a^2-b^2}}.\frac{a-\sqrt{a^2-b^2}}{b}\)\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{a^2-\left(a^2-b^2\right)}{b\sqrt{a^2-b^2}}\)
\(=\frac{ab-a^2+a^2-b^2}{b\sqrt{a^2-b^2}}\)
\(=\frac{b\left(a-b\right)}{b\sqrt{a^2-b^2}}=\frac{\left(a-b\right)}{\sqrt{\left(a-b\right)\left(a+b\right)}}=\frac{\sqrt{a-b}}{\sqrt{a+b}}\)
\(b.\frac{\sqrt{3b-b}}{\sqrt{3b+b}}=\frac{\sqrt{2b}}{\sqrt{4b}}=\frac{\sqrt{2}.\sqrt{b}}{2\sqrt{b}}=\frac{\sqrt{2}}{2}\)
a. \(Q=\dfrac{a}{\sqrt{a^2-b^2}}-\left(1+\dfrac{a}{\sqrt{a^2-b^2}}\right):\dfrac{b}{a-\sqrt{a^2-b^2}}\)
\(=\dfrac{a}{\sqrt{a^2-b^2}}-\dfrac{a+\sqrt{a^2-b^2}}{\sqrt{a^2-b^2}}.\dfrac{a-\sqrt{a^2-b^2}}{b}\)
\(=\dfrac{a}{\sqrt{a^2-b^2}}-\dfrac{b}{\sqrt{a^2-b^2}}=\dfrac{a-b}{\sqrt{a^2-b^2}}=\dfrac{\sqrt{a-b}}{\sqrt{a+b}}\)
b. Thay \(a=3b\) vào \(Q\), ta được
\(Q=\dfrac{\sqrt{3b-b}}{\sqrt{3b+b}}=\dfrac{\sqrt{2b}}{\sqrt{4b}}=\dfrac{1}{\sqrt{2}}\)