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a)
Q=a√a2−b2−(1+a√a2−b2):ba−√a2−b2=a√a2−b2−a2−(a2−b2)b√a2−b2=a√a2−b2−a2−a2+b2b√a2−b2=a−b√a2−
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 \(Q=\dfrac{3b-b}{\sqrt{9b^2-b^2}}=\dfrac{2b}{\sqrt{8b^2}}=\dfrac{2b}{2\sqrt{2}\cdot b}=\dfrac{1}{\sqrt{2}}\)
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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 : \(Q=\sqrt{\frac{3b-b}{3b+b}}=\sqrt{\frac{2b}{4b}}=\sqrt{\frac{1}{2}}\)
\(a,Q=\frac{a}{\sqrt{a^2-b^2}}-\left(1+\frac{a}{\sqrt{a^2-b^2}}\right):\left(\frac{b}{a-\sqrt{a^2-b^2}}\right)\)
\(=\frac{a}{\sqrt{a^2-b^2}}-\left(\frac{\sqrt{a^2-b^2}+a}{\sqrt{a^2-b^2}}\right):\frac{b}{a-\sqrt{a^2+b^2}}\)
\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{a+\sqrt{a^2-b^2}}{\sqrt{a^2-b^2}}.\frac{a-\sqrt{a^2-b^2}}{b}\)\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{a^2-\left(a^2-b^2\right)}{b\sqrt{a^2-b^2}}\)
\(=\frac{ab-a^2+a^2-b^2}{b\sqrt{a^2-b^2}}\)
\(=\frac{b\left(a-b\right)}{b\sqrt{a^2-b^2}}=\frac{\left(a-b\right)}{\sqrt{\left(a-b\right)\left(a+b\right)}}=\frac{\sqrt{a-b}}{\sqrt{a+b}}\)
\(b.\frac{\sqrt{3b-b}}{\sqrt{3b+b}}=\frac{\sqrt{2b}}{\sqrt{4b}}=\frac{\sqrt{2}.\sqrt{b}}{2\sqrt{b}}=\frac{\sqrt{2}}{2}\)
:") 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 :
\(Q=\frac{\sqrt{3b-b}}{\sqrt{3b+b}}=\frac{\sqrt{2b}}{\sqrt{4b}}=\sqrt{\frac{2b}{4b}}=\sqrt{\frac{1}{2}}=\frac{\sqrt{2}}{2}\)
1/ đkxđ: a > 0; a khác 1
a/ A= (\(\dfrac{\sqrt{a}}{2\sqrt{a}}-\dfrac{1}{2\sqrt{a}}\))\(\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-1}{2\sqrt{a}}\cdot\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)^2-\sqrt{a}\left(\sqrt{a}+1\right)^2}{a-1}\)
\(=\dfrac{1}{2\sqrt{a}}\cdot\dfrac{a\sqrt{a}-2a+\sqrt{a}-a\sqrt{a}-2a-\sqrt{a}}{a-1}\)
\(=\dfrac{1}{2\sqrt{a}}\cdot\dfrac{-4a}{a-1}=-\dfrac{2\sqrt{a}}{a-1}=\dfrac{2\sqrt{a}}{a+1}\)
b/+) A = 4
\(\Leftrightarrow\dfrac{2\sqrt{a}}{a+1}=4\)\(\Leftrightarrow2\sqrt{a}=4a+4\)
=> Không có gt a nào t/m
+) \(A>-6\)
\(\Leftrightarrow\dfrac{2\sqrt{a}}{a+1}>-6\)
\(\Leftrightarrow2\sqrt{a}>-6a-6\)
\(\Leftrightarrow6a+2\sqrt{a}+6>0\) (luôn đúng vì a > 0)
=> bpt có nghiệm với mọi a > 0
vậy........
c/ \(a^2-3=0\Leftrightarrow\left[{}\begin{matrix}a=\sqrt{3}\left(tm\right)\\a=-\sqrt{3}\left(ktmđkxđ\right)\end{matrix}\right.\)
Với a = \(\sqrt{3}\) ta có:
\(A=\dfrac{2\sqrt{3}}{\sqrt{3}+1}=\dfrac{2\sqrt{3}\left(\sqrt{3}-1\right)}{3-1}=\dfrac{2\sqrt{3}\left(\sqrt{3}-1\right)}{2}=\sqrt{3}\left(\sqrt{3}-1\right)=3-\sqrt{3}\)
a: \(G=\left(\sqrt{a^2+a\sqrt{a^2-b^2}}-\sqrt{a^2-a\sqrt{a^2-b^2}}\right)^2\)
\(=a^2+a\sqrt{a^2-b^2}+a^2-a\sqrt{a^2-b^2}-2\cdot\sqrt{a^4-a^2\left(a^2-b^2\right)}\)
\(=2a^2-2\cdot\sqrt{a^4-a^4+a^2b^2}=2a^2-2ab\)
\(A=\left(\sqrt{\dfrac{a}{b}}+\sqrt{\dfrac{b}{a}}-2\right)\)
\(=\dfrac{a}{\sqrt{ab}}+\dfrac{b}{\sqrt{ab}}-\dfrac{2\sqrt{ab}}{\sqrt{ab}}=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{ab}}\)
\(P=\dfrac{2a^2-2ab}{2a\sqrt{ab}}:\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{ab}}\)
\(=\dfrac{2a\left(a-b\right)}{2a\sqrt{ab}}\cdot\dfrac{\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)^2}=\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
b: Khi \(a=7+4\sqrt{3};b=7-4\sqrt{3}\) thì
\(P=\dfrac{2+\sqrt{3}+2-\sqrt{3}}{2+\sqrt{3}-2+\sqrt{3}}=\dfrac{4}{2\sqrt{3}}=\dfrac{2}{\sqrt{3}}\)
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
\(Q=\dfrac{\sqrt{3b-b}}{\sqrt{3b+b}}=\dfrac{\sqrt{2b}}{\sqrt{4b}}=\dfrac{1}{\sqrt{2}}\)