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\(a,Q=\frac{a}{\sqrt{a^2-b^2}}-\left(1+\frac{a}{\sqrt{a^2-b^2}}\right):\left(\frac{b}{a-\sqrt{a^2-b^2}}\right)\)
\(=\frac{a}{\sqrt{a^2-b^2}}-\left(\frac{\sqrt{a^2-b^2}+a}{\sqrt{a^2-b^2}}\right):\frac{b}{a-\sqrt{a^2+b^2}}\)
\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{a+\sqrt{a^2-b^2}}{\sqrt{a^2-b^2}}.\frac{a-\sqrt{a^2-b^2}}{b}\)\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{a^2-\left(a^2-b^2\right)}{b\sqrt{a^2-b^2}}\)
\(=\frac{ab-a^2+a^2-b^2}{b\sqrt{a^2-b^2}}\)
\(=\frac{b\left(a-b\right)}{b\sqrt{a^2-b^2}}=\frac{\left(a-b\right)}{\sqrt{\left(a-b\right)\left(a+b\right)}}=\frac{\sqrt{a-b}}{\sqrt{a+b}}\)
\(b.\frac{\sqrt{3b-b}}{\sqrt{3b+b}}=\frac{\sqrt{2b}}{\sqrt{4b}}=\frac{\sqrt{2}.\sqrt{b}}{2\sqrt{b}}=\frac{\sqrt{2}}{2}\)
:") Làm bừa nhezzz
a) \(Q=\frac{a}{\sqrt{a^2-b^2}}-\left(1+\frac{a}{\sqrt{a^2-b^2}}\right):\frac{b}{a-\sqrt{a^2}-b^2}\)
\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{a+\sqrt{a^2-b^2}}{\sqrt{a^2-b^2}}.\frac{a-\sqrt{a^2-b^2}}{b}\)
\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{a^2-\left(\sqrt{a^2-b^2}\right)^2}{b.\left(\sqrt{a^2-b^2}\right)}\)
\(=\frac{a}{\sqrt{a^2-b^2}}-\left(\frac{a^2-\left(a^2-b^2\right)}{b.\left(\sqrt{a^2-b^2}\right)}\right)\)
\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{b^2}{b\sqrt{a^2-b^2}}\)
\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{b}{\sqrt{a^2-b^2}}\)
\(=\frac{a-b}{\sqrt{a^2-b^2}}=\frac{a-b}{\sqrt{\left(a-b\right)\left(a+b\right)}}=\frac{\sqrt{a-b}}{\sqrt{a+b}}\)
b) Thay a = 3b vào , ta được :
\(Q=\frac{\sqrt{3b-b}}{\sqrt{3b+b}}=\frac{\sqrt{2b}}{\sqrt{4b}}=\sqrt{\frac{2b}{4b}}=\sqrt{\frac{1}{2}}=\frac{\sqrt{2}}{2}\)
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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: \(=\dfrac{\sqrt{a}-1}{\sqrt{a}\left(a-\sqrt{a}+1\right)}\cdot\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{1}\)
\(=a-1\)
b: \(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\left(\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}+\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}\right)\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\dfrac{\sqrt{ab}+b+\sqrt{ab}-b}{\sqrt{a}\left(a-b\right)}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{1}{\sqrt{a}}\)
c: \(=\dfrac{a\sqrt{b}+b}{a-b}\cdot\sqrt{\dfrac{ab+b^2-2b\sqrt{ab}}{a^2+2a\sqrt{b}+b}}\cdot\left(\sqrt{a}+\sqrt{b}\right)\)
\(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\cdot\sqrt{\dfrac{b\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(a+\sqrt{b}\right)^2}}\)
\(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{a+\sqrt{b}}=b\)
a)\(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right)\):\(\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a-1}\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a-1}\right)}\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{\left(a-1\right)\left(a-4\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{a-1-a+4}\)
\(=\dfrac{1}{\sqrt{a}}.\dfrac{\sqrt{a}-2}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\) ĐKXĐ: \(x>0\) \(a\ne4\) \(a\ne1\)
b) \(Q>0\)
\(\Leftrightarrow\dfrac{\sqrt{a}-2}{3\sqrt{a}}>0\)
mà \(3\sqrt{a}>0\) (Kết hợp ĐKXĐ \(a>0\))
\(\Leftrightarrow\sqrt{a}-2>0\)
\(\Leftrightarrow\sqrt{a}>2\)
\(\Leftrightarrow a>4\) (Thỏa mãn ĐKXĐ)
Vậy \(a>4\) thì \(Q>0\)
____♫ Chúc bạn học tốt ♫____
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 \(Q=\dfrac{3b-b}{\sqrt{9b^2-b^2}}=\dfrac{2b}{\sqrt{8b^2}}=\dfrac{2b}{2\sqrt{2}\cdot b}=\dfrac{1}{\sqrt{2}}\)