Mathematics

Question

23 number please
thank youu​
23 number please thank youu​

1 Answer


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  • Step-by-step explanation:

    [tex]\dfrac{\left(x^2-\frac{1}{y^2}\right)^x\left(x-\frac{1}{y}\right)^{y-x}}{\left(y^2-\frac{1}{x^2}\right)^y\left(y+\frac{1}{x}\right)^{x-y}}\\\\=\dfrac{\left(\frac{x^2y^2-1}{y^2}\right)^x\left(\frac{xy}{y}-\frac{1}{y}\right)^{y-x}}{\left(\frac{x^2y^2}{x^2}\right)^y\left(\frac{xy}{x}+\frac{1}{x}\right)^{x-y}}[/tex]

    [tex]=\dfrac{\left(\frac{x^2y^2-1}{y^2}\right)^x\left(\frac{xy-1}{y}\right)^y\left(\frac{xy-1}{y}\right)^{-x}}{\left(\frac{x^2y^2}{x^2}\right)^y\left(\frac{xy+1}{x}\right)^x\left(\frac{xy+1}{x}\right)^{-y}}\\\\=\dfrac{\left(\frac{x^2y^2-1}{y^2}\right)^x\left(\frac{xy-1}{y}\right)^y\left(\frac{y}{xy-1}\right)^x}{\left(\frac{x^2y^2-1}{x^2}\right)^y\left(\frac{xy+1}{x}\right)^x\left(\frac{x}{xy+1}\right)^y}[/tex]

    [tex]=\left(\dfrac{x^2y^2-1}{y^2}\cdot\dfrac{y}{xy-1}\cdot\dfrac{x}{xy+1}\right)^x\left(\dfrac{xy-1}{y}\cdot\dfrac{x^2}{x^2y^2-1}\cdot\dfrac{xy+1}{x}\right)^y[/tex]

    [tex]=\left(\dfrac{(xy-1)(xy+1)xy}{y^2(xy-1)(xy+1)}\right)^x\left(\dfrac{x^2(xy-1)(xy+1)}{xy(xy-1)(xy+1)}\right)^y\\\\=\left(\dfrac{x}{y}\right)^x\left(\dfrac{x}{y}\right)^y=\left(\dfrac{x}{y}\right)^{x+y}[/tex]

    Used:

    [tex]a^{-n}=\left(\dfrac{1}{a}\right)^n\\\\(a\cdot b)^n=a^n\cdot b^n\\\\\left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\\\\a^2-b^2=(a-b)(a+b)\\\\a^n\cdot a^m=a^{n+m}[/tex]