Question

Сократите дробь:
(a-b)^3/b^2-a^2
x^2-2x+1/x^2-1
5m^2+10mn+5n^2/15m^2-15n^2
m^3-m^3/m^2-n^2
3p^2-3q^3/6p^3-6q^3

Answer 1

$frac{ {(a - b)}^{3} }{ {b}^{2} - {a}^{2} } = - frac{ {(a - b)}^{3} }{(a - b)(a + b)} = frac{ {(a - b)}^{2} }{(a + b)}$

$frac{ {x}^{2} - 2x + 1}{ {x}^{2} - 1} = frac{ {(x - 1)}^{2} }{(x - 1)(x + 1)} = frac{x - 1}{x + 1}$

$frac{5 {m}^{2} + 10mn + 5 {n}^{2} }{15 {m}^{2} - 15 {n}^{2} } = frac{5( {m}^{2} + 2mn + {n}^{2}) }{15( {m}^{2} - {n}^{2} )} = frac{ {(m + n)}^{2} }{3(m - n)(m + n)} = frac{(m + n)}{3(m - n)}$

$frac{ {m}^{3} - {m}^{3} }{ {m}^{2} - {n}^{2} } = frac{0}{{m}^{2} - {n}^{2}} = 0 \ - - - - - - \ frac{ {m}^{3} - {n}^{3} }{ {m}^{2} - {n}^{2} } = frac{(m - n)( {m}^{2} + mn + {n}^{2} )}{(m - n)(m + n)} = frac{{m}^{2} + mn + {n}^{2} }{m + n}$

$frac{3 {p}^{2} - 3 {q}^{3} }{6 {p}^{3} - 6 {q}^{3} } = frac{3( {p}^{2} - {q}^{3}) }{6( {p}^{?} - {q}^{3} ) } = frac{{p}^{2} - {q}^{3}}{2( {p}^{?} - {q}^{3} ) }$