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\(x^2-1=\frac{1}{4}\left(a^2+\frac{1}{a^2}+2\right)-1=\frac{1}{4}\left(a-\frac{1}{a}\right)^2\)
\(\Rightarrow\sqrt{x^2-1}=\frac{1}{2}\left(a-\frac{1}{a}\right)\)
Tương tự \(\sqrt{y^2-1}=\frac{1}{2}\left(b-\frac{1}{b}\right)\)
\(A=\frac{\frac{1}{4}\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)-\frac{1}{4}\left(a-\frac{1}{a}\right)\left(b-\frac{1}{b}\right)}{\frac{1}{4}\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)+\frac{1}{4}\left(a-\frac{1}{a}\right)\left(b-\frac{1}{b}\right)}=\frac{ab+\frac{a}{b}+\frac{b}{a}+\frac{1}{ab}-ab-\frac{1}{ab}+\frac{a}{b}+\frac{b}{a}}{ab+\frac{a}{b}+\frac{b}{a}+\frac{1}{ab}+ab+\frac{1}{ab}-\frac{a}{b}-\frac{b}{a}}\)
\(=\frac{\frac{a}{b}+\frac{b}{a}}{ab+\frac{1}{ab}}=\frac{a^2+b^2}{a^2b^2+1}\)
b/ \(B=\frac{\left(\sqrt{a+bx}+\sqrt{a-bx}\right)^2}{a+bx-\left(a-bx\right)}=\frac{a+\sqrt{a^2-b^2x^2}}{bx}\)
\(a^2-b^2x^2=a^2-\frac{4a^2m^2}{\left(1+m^2\right)^2}=\frac{a^2\left(m^4+2m^2+1\right)-4a^2m^2}{\left(1+m^2\right)^2}=\frac{a^2\left(1-m^2\right)^2}{\left(1+m^2\right)^2}\)
\(\Rightarrow B=\left(a+\frac{a\left(1-m^2\right)}{1+m^2}\right).\left(\frac{1+m^2}{2am}\right)=\frac{a+am^2+a-am^2}{2am}=\frac{1}{m}\)
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Tham khảo:
\(x=\dfrac{1}{a}.\sqrt{\dfrac{2a}{b}-1}\Rightarrow ax=\sqrt{\dfrac{2a}{b}-1}\)
\(\Rightarrow\left\{{}\begin{matrix}1+ax=\dfrac{\sqrt{2a-b}+\sqrt{b}}{\sqrt{b}}\\1-ax=\dfrac{\sqrt{b}-\sqrt{2a-b}}{\sqrt{b}}\end{matrix}\right.\)
\(\Rightarrow\dfrac{1-ax}{1+ax}=\dfrac{\sqrt{b}-\sqrt{2a-b}}{\sqrt{b}+\sqrt{2a-b}}=\dfrac{\left(\sqrt{b}-\sqrt{2a-b}\right)^2}{2\left(b-a\right)}\)
Lại có:
\(\dfrac{1+bx}{1-bx}=\dfrac{a+\sqrt{2ab-b^2}}{a-\sqrt{2ab-b^2}}=\dfrac{a^2-\left(2ab-b^2\right)}{\left(a-\sqrt{2ab-b^2}\right)^2}=\dfrac{\left(a-b\right)^2}{\left(a-\sqrt{2ab-b^2}\right)^2}\)
\(\Rightarrow\sqrt{\dfrac{1+bx}{1-bx}}=\dfrac{b-a}{a-\sqrt{2ab-b^2}}\)
\(\Rightarrow A=\dfrac{1-ax}{1+ax}.\sqrt{\dfrac{1+bx}{1-bx}}=\dfrac{\left(\sqrt{b}-\sqrt{2a-b}\right)^2}{2a-2\sqrt{2ab-b^2}}=\dfrac{2a-2\sqrt{2ab-b^2}}{2a-2\sqrt{2ab-b^2}}=1\)
d: \(=-\left(x+\sqrt{x}-12\right)=-\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)\)