ESPRESSIONE CON I RADICALI!!

[yolo.]

[ √(a+4+4/a) - √(a+2+1/a) : √1/a^5]

[carlogiannini]

[math]\sqrt{\frac{a^2+4a+4}{a}}=\sqrt{\frac{(a+2)^2}{a}}=\frac{\sqrt{(a+2)^2}}{\sqrt{a}}=\frac{(a+2)}{\sqrt{a}}[/math]

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[math]\sqrt{\frac{a^2+2a+1}{a}}=\sqrt{\frac{(a+1)^2}{a}}=\frac{\sqrt{(a+1)^2}}{\sqrt{a}}=\frac{(a+1)}{\sqrt{a}}[/math]

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.Risolvendo:
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[math][\sqrt{\frac{a^2+4a+4}{a}}-\sqrt{\frac{a^2+2a+1}{a}}]\sqrt{a^5}[/math]

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[math][\sqrt{\frac{(a+2)^2}{a}}-\sqrt{\frac{(a+1)^2}{a}}]a^{2}\sqrt{a}[/math]

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[math][\frac{(a+2)}{\sqrt{a}}-\frac{(a+1)}{\sqrt{a}}]a^2\sqrt{a}[/math]

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[math]\frac{(a+2)-(a+1)}{\sqrt{a}}a^2\sqrt{a}[/math]

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[math]\frac{a+2-a-1}{1}a^2(1)=\frac{1}{1}a^2=a^2[/math]

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Aggiunto 7 minuti più tardi:

N.B.
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[math]\sqrt{a^5}=\sqrt{aaaaa}=\sqrt{(aa)(aa)a}=\sqrt{a^2a^2a}[/math]

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[math]\sqrt{a^2}\sqrt{a^2}\sqrt{a}=aa\sqrt{a}=a^2\sqrt{a}[/math]