Question

163. Представьте в виде многочлена или рациональной дроби:
а ) ( n+ 1/2)²
б) ( а/б - б/а )²
в) ( х/у + 1 )² + ( х/у - 1)²
г) ( p/q + q/p)² - ( p/q - q/p)²

ПРОШУ ПОМОГИТЕ
ДАЮ 90 БАЛЛОВ ​

Answer 1

$\displaystyle\bf\\1)\\\\\Big(n+\frac{1}{2} \Big)^{2} =n^{2} +2\cdot n\cdot \frac{1}{2} +\Big(\frac{1}{2} \Big)^{2}=n^{2}+n+\frac{1}{4} \\\\2)\\\\\Big(\frac{a}{b} -\frac{b}{a} \Big)^{2}=\Big(\frac{a}{b} \Big)^{2} -2\cdot \frac{a}{b} \cdot\frac{b}{a} +\Big(\frac{b}{a} \Big)^{2}=\frac{a^{2} }{b^{2} }-2+\frac{b^{2} }{a^{2} } \\\\3)\\\\\Big(\frac{x}{y}+1\Big)^{2} -\Big(\frac{x}{y} -1\Big)^{2} =$

$\displaystyle\bf\\=\Big(\frac{x}{y} \Big)^{2} +2\cdot \frac{x}{y} \cdot 1+1^{2} +\Big(\frac{x}{y} \Big)^{2} -2\cdot \frac{x}{y} \cdot 1+1^{2} =\\\\\\=\frac{x^{2} }{y^{2} } +\frac{x^{2} }{y^{2} } +2=\frac{2x^{2} }{y^{2} } +2\\\\4)\\\\\Big(\frac{p}{q} +\frac{q}{p} \Big)^{2} -\Big(\frac{p}{q} -\frac{q}{p} \Big)^{2} =$

$\displaystyle\bf\\=\Big(\frac{p}{q} \Big)^{2} +2\cdot\frac{p}{q} \cdot\frac{q}{p} +\Big(\frac{q}{p}\Big)^{2} -\Big[\Big(\frac{p}{q} \Big)^{2} -2\cdot\frac{p}{q} \cdot\frac{q}{p} +\Big(\frac{q}{p}\Big)^{2}\Big] =\\\\\\=\frac{p^{2} }{q^{2} } +2+\frac{q^{2} }{p^{2} } -\frac{p^{2} }{q^{2} } +2-\frac{q^{2} }{p^{2} }=4$