Положительное число, квадрат которого равен a a a
Например, уравнение вида x 2 = 9 x^2 = 9 x 2 = 9 3 3 3 − 3 -3 − 3 a = x \sqrt a=x a = x − 3 -3 − 3
Свойства арифметического квадратного корня
a = x → x 2 = a , \sqrt a=x \rightarrow x^2=a, a = x → x 2 = a , a ≥ 0 a\geq0 a ≥ 0 a ⋅ a = a \sqrt a\cdot\sqrt a=a a ⋅ a = a a 2 = a \sqrt{a^2}=a a 2 = a a b = a ⋅ b , \sqrt{ab}=\sqrt a\cdot\sqrt b, a b = a ⋅ b , a ≥ 0 , a\geq0, a ≥ 0 , b ≥ 0 b\geq0 b ≥ 0 a b = a b , \sqrt{\frac{a}{b}}=\frac{\sqrt a}{\sqrt b}, b a = b a , a ≥ 0 , a\geq0, a ≥ 0 , b ≥ 0 b\geq0 b ≥ 0 ( a ) n = a n , a ≥ 0 \left(\sqrt a\right)^n=\sqrt{a^n},\ a\geq0 ( a ) n = a n , a ≥ 0 ( a ) − n = 1 a n \left(\sqrt a\right)^{-n}=\frac{1}{\sqrt{a^n}} ( a ) − n = a n 1
При сравнении арифметических корней нужно знать, что чем больше подкоренное выражение, тем больше корень этого числа.
Вычислить выражение
98 ⋅ 12 − 3 16 2 : 16 3 ⋅ 2 : 3 \frac{\sqrt{98}\cdot{\sqrt{12}}^{-3}}{\sqrt{{16}^2}:\sqrt{{16}^3}\cdot\sqrt2:\sqrt3} 1 6 2 : 1 6 3 ⋅ 2 : 3 9 8 ⋅ 1 2 − 3
Решение
Преобразуем подкоренное выражение 98 \sqrt{98} 9 8
98 = 7 ⋅ 7 ⋅ 2 = 7 ⋅ 7 ⋅ 2 = 7 2 ⋅ 2 = 7 2 \sqrt{98}=\sqrt{7\cdot7\cdot2}=\sqrt{7\cdot7}\cdot\sqrt2=\sqrt{7^2}\cdot\sqrt2=7\sqrt2 9 8 = 7 ⋅ 7 ⋅ 2 = 7 ⋅ 7 ⋅ 2 = 7 2 ⋅ 2 = 7 2
На основании свойств 7 и 4 упростим 12 − 3 {\sqrt{12}}^{-3} 1 2 − 3
12 − 3 = 1 12 3 = 1 ( 4 ⋅ 3 ) 3 = 1 4 3 ⋅ 3 3 = 1 4 2 ⋅ 4 1 ⋅ 3 2 ⋅ 3 1 = 1 4 ⋅ 4 ⋅ 3 ⋅ 3 = 1 24 ⋅ 3 {\sqrt{12}}^{-3}=\frac{1}{\sqrt{{12}^3}}=\frac{1}{\sqrt{{(4\cdot3)}^3}}=\frac{1}{\sqrt{4^3}\cdot\sqrt{3^3}}=\frac{1}{\sqrt{4^2}\cdot\sqrt{4^1}\cdot\sqrt{3^2}\cdot\sqrt{3^1}}=\frac{1}{4\cdot\sqrt4\cdot3\cdot\sqrt3}=\frac{1}{24\cdot\sqrt3} 1 2 − 3 = 1 2 3 1 = ( 4 ⋅ 3 ) 3 1 = 4 3 ⋅ 3 3 1 = 4 2 ⋅ 4 1 ⋅ 3 2 ⋅ 3 1 1 = 4 ⋅ 4 ⋅ 3 ⋅ 3 1 = 2 4 ⋅ 3 1
Получили
7 2 ⋅ 1 12 c d o t 4 ⋅ 3 16 2 : 16 3 ⋅ 2 : 3 = 7 2 24 ⋅ 3 16 2 16 3 ⋅ 2 3 \frac{7\sqrt2\cdot\frac{1}{12cdot\sqrt4\cdot\sqrt3}}{\sqrt{{16}^2}:\sqrt{{16}^3}\cdot\sqrt2:\sqrt3}=\frac{\frac{7\sqrt2}{24\cdot\sqrt3}}{\frac{\sqrt{{16}^2}}{\sqrt{{16}^3}}\cdot\frac{\sqrt2}{\sqrt3}} 1 6 2 : 1 6 3 ⋅ 2 : 3 7 2 ⋅ 1 2 c d o t 4 ⋅ 3 1 = 1 6 3 1 6 2 ⋅ 3 2 2 4 ⋅ 3 7 2
Преобразуем 16 2 16 3 \frac{\sqrt{{16}^2}}{\sqrt{{16}^3}} 1 6 3 1 6 2
16 2 16 3 = 16 2 16 3 = 1 16 = 1 4 \frac{\sqrt{{16}^2}}{\sqrt{{16}^3}}=\sqrt{\frac{{16}^2}{{16}^3}}=\sqrt{\frac{1}{16}}=\frac{1}{4} 1 6 3 1 6 2 = 1 6 3 1 6 2 = 1 6 1 = 4 1
Получили
7 2 24 ∙ 3 1 4 ⋅ 2 3 = 7 2 24 3 2 4 3 = 7 2 24 3 ⋅ 4 3 2 \frac{\frac{7\sqrt2}{24\bullet\sqrt3}}{\frac{1}{4}\cdot\frac{\sqrt2}{\sqrt3}}=\frac{\frac{7\sqrt2}{24\sqrt3}}{\frac{\sqrt2}{4\sqrt3}}=\frac{7\sqrt2}{24\sqrt3}\cdot\frac{4\sqrt3}{\sqrt2} 4 1 ⋅ 3 2 2 4 ∙ 3 7 2 = 4 3 2 2 4 3 7 2 = 2 4 3 7 2 ⋅ 2 4 3
Выполним сокращения
7 2 24 3 ⋅ 4 3 2 = 7 ⋅ 4 6 ⋅ 4 = 7 6 \frac{7\sqrt2}{24\sqrt3}\cdot\frac{4\sqrt3}{\sqrt2}=\frac{7\cdot4}{6\cdot4}=\frac{7}{6} 2 4 3 7 2 ⋅ 2 4 3 = 6 ⋅ 4 7 ⋅ 4 = 6 7
Ответ: 98 ⋅ 12 − 3 16 2 : 16 3 ⋅ 2 : 3 = 7 6 \frac{\sqrt{98}\cdot{\sqrt{12}}^{-3}}{\sqrt{{16}^2}:\sqrt{{16}^3}\cdot\sqrt2:\sqrt3}=\frac{7}{6} 1 6 2 : 1 6 3 ⋅ 2 : 3 9 8 ⋅ 1 2 − 3 = 6 7
Вычислить выражение при x = 2 x = 2 x = 2 x 2 − 1 ⋅ ( x + 5 ) 2 ( x + 5 ) ⋅ ( x 2 − 1 ) 4 : ( x − 1 ) ( x + 1 ) \frac{\sqrt{x^2-1}\cdot\sqrt{{(x+5)}^2}}{(x+5)\cdot\sqrt{\left(x^2-1\right)^4}:\sqrt{(x-1)(x+1)}} ( x + 5 ) ⋅ ( x 2 − 1 ) 4 : ( x − 1 ) ( x + 1 ) x 2 − 1 ⋅ ( x + 5 ) 2
Решение
Упростим знаменатель, для чего свернем выражение
( x − 1 ) ( x + 1 ) = x 2 − 1 \left(x-1\right)\left(x+1\right)=x^2-1 ( x − 1 ) ( x + 1 ) = x 2 − 1
Таким образом,
x 2 − 1 ⋅ ( x + 5 ) 2 ( x + 5 ) ⋅ ( x 2 − 1 ) 4 : x 2 − 1 = x 2 − 1 ⋅ ( x + 5 ) 2 ⋅ x 2 − 1 ( x + 5 ) ⋅ ( x 2 − 1 ) 4 \frac{\sqrt{x^2-1}\cdot\sqrt{{(x+5)}^2}}{(x+5)\cdot\sqrt{\left(x^2-1\right)^4}:\sqrt{x^2-1}}=\frac{\sqrt{x^2-1}\cdot\sqrt{{(x+5)}^2}\cdot\sqrt{x^2-1}}{(x+5)\cdot\sqrt{\left(x^2-1\right)^4}} ( x + 5 ) ⋅ ( x 2 − 1 ) 4 : x 2 − 1 x 2 − 1 ⋅ ( x + 5 ) 2 = ( x + 5 ) ⋅ ( x 2 − 1 ) 4 x 2 − 1 ⋅ ( x + 5 ) 2 ⋅ x 2 − 1
По свойству 2
x 2 − 1 ⋅ x 2 − 1 = x 2 − 1 \sqrt{x^2-1}\cdot\sqrt{x^2-1}=x^2-1 x 2 − 1 ⋅ x 2 − 1 = x 2 − 1
Значит,
x 2 − 1 ⋅ ( x + 5 ) 2 ( x + 5 ) ⋅ ( x 2 − 1 ) 4 \frac{x^2-1\cdot\sqrt{{(x+5)}^2}}{(x+5)\cdot\sqrt{\left(x^2-1\right)^4}} ( x + 5 ) ⋅ ( x 2 − 1 ) 4 x 2 − 1 ⋅ ( x + 5 ) 2
По свойству 3
( x + 5 ) 2 = ( x + 5 ) \sqrt{{(x+5)}^2}=(x+5) ( x + 5 ) 2 = ( x + 5 )
Получим
x 2 − 1 ⋅ ( x + 5 ) ( x + 5 ) ⋅ ( x 2 − 1 ) 4 \frac{x^2-1\cdot(x+5)}{(x+5)\cdot\sqrt{\left(x^2-1\right)^4}} ( x + 5 ) ⋅ ( x 2 − 1 ) 4 x 2 − 1 ⋅ ( x + 5 )
Разобьем корень на слагаемые
( x 2 − 1 ) 4 = ( x 2 − 1 ) 2 ⋅ ( x 2 − 1 ) 2 = ( x 2 − 1 ) ⋅ ( x 2 − 1 ) \sqrt{\left(x^2-1\right)^4}=\sqrt{\left(x^2-1\right)^2}\cdot\sqrt{\left(x^2-1\right)^2}=\left(x^2-1\right)\cdot\left(x^2-1\right) ( x 2 − 1 ) 4 = ( x 2 − 1 ) 2 ⋅ ( x 2 − 1 ) 2 = ( x 2 − 1 ) ⋅ ( x 2 − 1 )
Упростим выражение, выполнив сокращения
x 2 − 1 ⋅ ( x + 5 ) ( x + 5 ) ⋅ ( x 2 − 1 ) ⋅ ( x 2 − 1 ) = x 2 − 1 ( x 2 − 1 ) ⋅ ( x 2 − 1 ) = 1 ( x 2 − 1 ) \frac{x^2-1\cdot(x+5)}{(x+5)\cdot\left(x^2-1\right)\cdot\left(x^2-1\right)}=\frac{x^2-1}{\left(x^2-1\right)\cdot\left(x^2-1\right)}=\frac{1}{\left(x^2-1\right)} ( x + 5 ) ⋅ ( x 2 − 1 ) ⋅ ( x 2 − 1 ) x 2 − 1 ⋅ ( x + 5 ) = ( x 2 − 1 ) ⋅ ( x 2 − 1 ) x 2 − 1 = ( x 2 − 1 ) 1
Заменим х известным значением
1 ( x 2 − 1 ) = 1 ( 2 2 − 1 ) = 1 3 \frac{1}{\left(x^2-1\right)}=\frac{1}{\left(2^2-1\right)}=\frac{1}{3} ( x 2 − 1 ) 1 = ( 2 2 − 1 ) 1 = 3 1
Ответ: При x = 2 x = 2 x = 2 x 2 − 1 ⋅ ( x + 5 ) 2 ( x + 5 ) ⋅ ( x 2 − 1 ) 4 : ( x − 1 ) ( x + 1 ) = 1 3 \frac{\sqrt{x^2-1}\cdot\sqrt{{(x+5)}^2}}{(x+5)\cdot\sqrt{\left(x^2-1\right)^4}:\sqrt{(x-1)(x+1)}}=\frac{1}{3} ( x + 5 ) ⋅ ( x 2 − 1 ) 4 : ( x − 1 ) ( x + 1 ) x 2 − 1 ⋅ ( x + 5 ) 2 = 3 1
Корень произвольной степени $n$ Число, n n n a a a
Корень n n n
если n n n
если n n n
Кубический корень в отличие от квадратного существует при a ≥ 0 a\geq0 a ≥ 0 a < 0 a<0 a < 0
Свойства корня n n n
a n = x → x n = a \sqrt[n]{a}=x\ \rightarrow x^n=a n a = x → x n = a a n ∙ a m = a n + m и л и a n : a m = a n − m \sqrt[n]{a}\bullet\sqrt[m]{a}=\sqrt[n+m]{a} или \sqrt[n]{a}:\sqrt[m]{a}=\sqrt[n-m]{a} n a ∙ m a = n + m a и л и n a : m a = n − m a a n n = a \sqrt[n]{a^n}=a n a n = a a b n = a n ⋅ b n \sqrt[n]{ab}=\sqrt[n]{a}\cdot\sqrt[n]{b} n a b = n a ⋅ n b a b n = a n b n \sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}} n b a = n b n a ( a n ) m = a m n \left(\sqrt[n]{a}\right)^m=\sqrt[n]{a^m} ( n a ) m = n a m ( a n ) − m = 1 a m n \left(\sqrt[n]{a}\right)^{-m}=\frac{1}{\sqrt[n]{a^m}} ( n a ) − m = n a m 1 a m n = a m n \sqrt[n]{a^m}=a^\frac{m}{n} n a m = a n m
Для удобства корни степени n n n
Упростить выражение
25 3 ⋅ 7 5 x y 5 3 ⋅ 14 3 ⋅ 5 7 3 ⋅ 25 3 x y 14 2 3 ⋅ 5 \frac{\sqrt[3]{25}\cdot\sqrt[5]{7}xy}{\sqrt[3]{5}\cdot\sqrt[3]{14}}\cdot\frac{5\sqrt{7^3}\cdot\sqrt[3]{25}xy}{\sqrt[3]{{14}^2}\cdot5} 3 5 ⋅ 3 1 4 3 2 5 ⋅ 5 7 x y ⋅ 3 1 4 2 ⋅ 5 5 7 3 ⋅ 3 2 5 x y
Решение
Представим подкоренные выражения в виде произведения простых множителей
5 ⋅ 5 3 ⋅ 7 5 x y 5 3 ⋅ 7 ⋅ 2 3 ⋅ 5 7 3 ⋅ 5 ⋅ 5 3 ( 7 ⋅ 2 ) 2 3 ⋅ 5 x y \frac{\sqrt[3]{5\cdot5}\cdot\sqrt[5]{7}xy}{\sqrt[3]{5}\cdot\sqrt[3]{7\cdot2}}\cdot\frac{5\sqrt{7^3}\cdot\sqrt[3]{5\cdot5}}{\sqrt[3]{\left(7\cdot2\right)^2}\cdot5}xy 3 5 ⋅ 3 7 ⋅ 2 3 5 ⋅ 5 ⋅ 5 7 x y ⋅ 3 ( 7 ⋅ 2 ) 2 ⋅ 5 5 7 3 ⋅ 3 5 ⋅ 5 x y
По свойству 4 выполним следующие действия
5 3 ⋅ 5 3 ⋅ 7 5 x y 5 3 ⋅ 7 3 ⋅ 2 3 ⋅ 5 7 3 ⋅ 5 3 ⋅ 5 3 7 ⋅ 2 ⋅ 7 ⋅ 2 3 ⋅ 5 x y = 5 3 ⋅ 5 3 ⋅ 7 5 x y 5 3 ⋅ 7 3 ⋅ 2 3 ⋅ 5 7 3 ⋅ 5 3 ⋅ 5 3 7 3 ⋅ 2 3 ⋅ 7 3 ⋅ 2 3 ⋅ 5 x \frac{\sqrt[3]{5}\cdot\sqrt[3]{5}\cdot\sqrt[5]{7}xy}{\sqrt[3]{5}\cdot\sqrt[3]{7}\cdot\sqrt[3]{2}}\cdot\frac{5\sqrt{7^3}\cdot\sqrt[3]{5}\cdot\sqrt[3]{5}}{\sqrt[3]{7\cdot2\cdot7\cdot2}\cdot5xy}=\frac{\sqrt[3]{5}\cdot\sqrt[3]{5}\cdot\sqrt[5]{7}xy}{\sqrt[3]{5}\cdot\sqrt[3]{7}\cdot\sqrt[3]{2}}\cdot\frac{5\sqrt{7^3}\cdot\sqrt[3]{5}\cdot\sqrt[3]{5}}{\sqrt[3]{7}\cdot\sqrt[3]{2}\cdot\sqrt[3]{7}\cdot\sqrt[3]{2}\cdot5}x 3 5 ⋅ 3 7 ⋅ 3 2 3 5 ⋅ 3 5 ⋅ 5 7 x y ⋅ 3 7 ⋅ 2 ⋅ 7 ⋅ 2 ⋅ 5 x y 5 7 3 ⋅ 3 5 ⋅ 3 5 = 3 5 ⋅ 3 7 ⋅ 3 2 3 5 ⋅ 3 5 ⋅ 5 7 x y ⋅ 3 7 ⋅ 3 2 ⋅ 3 7 ⋅ 3 2 ⋅ 5 5 7 3 ⋅ 3 5 ⋅ 3 5 x
Сократим числитель и знаменатель
5 3 ⋅ 5 3 ⋅ 7 5 x y 5 3 ⋅ 7 3 ⋅ 2 3 ⋅ 5 7 3 ⋅ 5 3 ⋅ 5 3 7 3 ⋅ 2 3 ⋅ 7 3 ⋅ 2 3 ⋅ 5 x y = 5 3 ⋅ 7 5 x y 7 3 ⋅ 2 3 ⋅ 7 3 ⋅ 5 3 ⋅ 5 3 7 3 ⋅ 2 3 ⋅ 7 3 ⋅ 2 3 x y \frac{\sqrt[3]{5}\cdot\sqrt[3]{5}\cdot\sqrt[5]{7}xy}{\sqrt[3]{5}\cdot\sqrt[3]{7}\cdot\sqrt[3]{2}}\cdot\frac{5\sqrt{7^3}\cdot\sqrt[3]{5}\cdot\sqrt[3]{5}}{\sqrt[3]{7}\cdot\sqrt[3]{2}\cdot\sqrt[3]{7}\cdot\sqrt[3]{2}\cdot5xy}=\frac{\sqrt[3]{5}\cdot\sqrt[5]{7}xy}{\sqrt[3]{7}\cdot\sqrt[3]{2}}\cdot\frac{\sqrt{7^3}\cdot\sqrt[3]{5}\cdot\sqrt[3]{5}}{\sqrt[3]{7}\cdot\sqrt[3]{2}\cdot\sqrt[3]{7}\cdot\sqrt[3]{2}}xy 3 5 ⋅ 3 7 ⋅ 3 2 3 5 ⋅ 3 5 ⋅ 5 7 x y ⋅ 3 7 ⋅ 3 2 ⋅ 3 7 ⋅ 3 2 ⋅ 5 x y 5 7 3 ⋅ 3 5 ⋅ 3 5 = 3 7 ⋅ 3 2 3 5 ⋅ 5 7 x y ⋅ 3 7 ⋅ 3 2 ⋅ 3 7 ⋅ 3 2 7 3 ⋅ 3 5 ⋅ 3 5 x y
По свойству 2
7 5 7 3 = 7 5 − 3 = 7 2 5 3 ⋅ 7 2 x y 2 3 ⋅ 7 3 ⋅ 5 3 ⋅ 5 3 7 3 ⋅ 2 3 ⋅ 7 3 ⋅ 2 3 x \frac{\sqrt[5]{7}}{\sqrt[3]{7}}=\sqrt[5-3]{7}=\sqrt[2]{7}
\frac{\sqrt[3]{5}\cdot\sqrt[2]{7}xy}{\sqrt[3]{2}}\cdot\frac{\sqrt{7^3}\cdot\sqrt[3]{5}\cdot\sqrt[3]{5}}{\sqrt[3]{7}\cdot\sqrt[3]{2}\cdot\sqrt[3]{7}\cdot\sqrt[3]{2}}x 3 7 5 7 = 5 − 3 7 = 2 7 3 2 3 5 ⋅ 2 7 x y ⋅ 3 7 ⋅ 3 2 ⋅ 3 7 ⋅ 3 2 7 3 ⋅ 3 5 ⋅ 3 5 x
По свойству 8 перейдем от корней к степеням
5 1 3 ⋅ 7 1 2 x y 2 1 3 ⋅ 7 3 2 ⋅ 5 1 3 ⋅ 5 1 3 7 1 3 ⋅ 2 1 3 ⋅ 7 1 3 ⋅ 2 1 3 ⋅ x y = 5 1 3 ⋅ 7 1 2 x y 2 1 3 ⋅ 7 3 2 ⋅ 5 2 3 7 2 3 ⋅ 2 2 3 x y \frac{5^\frac{1}{3}\cdot7^\frac{1}{2}xy}{2^\frac{1}{3}}\cdot\frac{7^\frac{3}{2}\cdot5^\frac{1}{3}\cdot5^\frac{1}{3}}{7^\frac{1}{3}\cdot2^\frac{1}{3}\cdot7^\frac{1}{3}\cdot2^\frac{1}{3}\cdot x y}=\frac{5^\frac{1}{3}\cdot7^\frac{1}{2}xy}{2^\frac{1}{3}}\cdot\frac{7^\frac{3}{2}\cdot5^\frac{2}{3}}{7^\frac{2}{3}\cdot2^\frac{2}{3}}xy 2 3 1 5 3 1 ⋅ 7 2 1 x y ⋅ 7 3 1 ⋅ 2 3 1 ⋅ 7 3 1 ⋅ 2 3 1 ⋅ x y 7 2 3 ⋅ 5 3 1 ⋅ 5 3 1 = 2 3 1 5 3 1 ⋅ 7 2 1 x y ⋅ 7 3 2 ⋅ 2 3 2 7 2 3 ⋅ 5 3 2 x y
По свойству степеней 7 3 2 : 7 2 3 = 7 3 2 + 2 3 = 7 5 6 7^\frac{3}{2}:7^\frac{2}{3}=7^{\frac{3}{2}+\frac{2}{3}}=7^\frac{5}{6} 7 2 3 : 7 3 2 = 7 2 3 + 3 2 = 7 6 5
5 1 3 ⋅ 7 1 2 x y 2 1 3 ⋅ 7 5 6 ⋅ 5 2 3 x y 2 2 3 = 5 1 3 ⋅ 7 1 2 x y ⋅ 7 5 6 ⋅ 5 2 3 x y 2 1 3 + 2 3 = 5 3 3 ⋅ 7 4 3 ( x y ) 2 2 3 3 = 5 2 7 4 3 ( x y ) 2 \frac{5^\frac{1}{3}\cdot7^\frac{1}{2}xy}{2^\frac{1}{3}}\cdot\frac{7^\frac{5}{6}\cdot5^\frac{2}{3}xy}{2^\frac{2}{3}}=\frac{5^\frac{1}{3}\cdot7^\frac{1}{2}xy\cdot7^\frac{5}{6}\cdot5^\frac{2}{3}xy}{2^{\frac{1}{3}+\frac{2}{3}}}=\frac{5^\frac{3}{3}\cdot7^\frac{4}{3}\left(xy\right)^2}{2^\frac{3}{3}}=\frac{5}{2}7^\frac{4}{3}\left(xy\right)^2 2 3 1 5 3 1 ⋅ 7 2 1 x y ⋅ 2 3 2 7 6 5 ⋅ 5 3 2 x y = 2 3 1 + 3 2 5 3 1 ⋅ 7 2 1 x y ⋅ 7 6 5 ⋅ 5 3 2 x y = 2 3 3 5 3 3 ⋅ 7 3 4 ( x y ) 2 = 2 5 7 3 4 ( x y ) 2
Ответ: 25 3 ⋅ 7 5 x y 5 3 ⋅ 14 3 + 5 7 3 ⋅ 25 3 x y 14 2 3 ⋅ 5 = 5 2 7 4 3 ( x y ) 2 \frac{\sqrt[3]{25}\cdot\sqrt[5]{7}xy}{\sqrt[3]{5}\cdot\sqrt[3]{14}}+\frac{5\sqrt{7^3}\cdot\sqrt[3]{25}xy}{\sqrt[3]{{14}^2}\cdot5}=\frac{5}{2}7^\frac{4}{3}\left(xy\right)^2 3 5 ⋅ 3 1 4 3 2 5 ⋅ 5 7 x y + 3 1 4 2 ⋅ 5 5 7 3 ⋅ 3 2 5 x y = 2 5 7 3 4 ( x y ) 2
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