Question

Помогите пожалуйста после завтра экзамен.
Формула разности квадратов.
8. Сократите дробь:
а) a-b/ b^2-a^2
б) x^2 - 16/ 16 - 4x
в) 4x^2 - 1/ y - 2xy
г) 2z^2 - 8/ z^2 - 2z
д) ab^2 - ac^2 / c^2 - bc
е) x^4 - x^2 y^2 / ay^2 - axy
ж) a^4 x^4 - 4 / 2x + a^2 x^3
з) 81a^2 - b^2 c^4 / b^2 c^2 - 9ab

Answer 1

Формула разности квадратов:
${a}^{2} - {b}^{2} = (a - b)(a + b)$
а)
$\frac{a - b}{ {b}^{2} - {a}^{2} } = \frac{a -b }{(b - a)(b + a)} = \frac{ - (b - a)}{(b - a)(b + a)} = - \frac{1}{a + b}$
б)
$\frac{ {x}^{2} - 16 }{16 - 4x} = \frac{(x - 4)(x + 4)}{4(4 - x)} = \frac{ (x - 4)(x + 4)}{ - 4(x - 4)} = - \frac{x + 4}{4}$
в)
$\frac{4 {x}^{2} - 1}{y- 2xy} = \frac{(2x - 1)(2x + 1)}{y(1 - 2x)} = \frac{(2x - 1)(2x + 1)}{ - y(2x - 1)} = - \frac{2x + 1}{y}$
г)
$\frac{2 {z}^{2} - 8}{ {z}^{2} - 2z } = \frac{2( {z}^{2} - 4)}{z(z - 2) } = \frac{2(z - 2)(z + 2)}{z(z - 2)} = \frac{2(z + 2)}{z}$
д)
$\frac{a {b}^{2} - a {c}^{2} }{ {c}^{2} - bc } = \frac{ a({b}^{2} - {c}^{2} )}{c(c - b)} = \frac{a(b - c)(b + c)}{ - c(b - c)} = - \frac{a(b + c)}{c}$
е)
$\frac{ {x}^{4} - {x}^{2} {y}^{2} }{a {y}^{2} - axy } = \frac{ {x}^{2}( {x}^{2} - {y}^{2}) }{ay(y - x)} = \frac{ {x}^{2} (x - y)(x + y)}{ay(y - x)} = \frac{ {x}^{2}(x - y)(x + y) }{ - ay(x - y)} = - \frac{ {x}^{2}(x + y) }{ay}$
ж)
$\frac{ {a}^{4} {x}^{4} - 4 }{2x + {a}^{2} {x}^{3} } = \frac{( {a}^{2} {x}^{2} - 2)( {a}^{2} {x}^{2} + 2) }{x(2 + {a}^{2} {x}^{2}) } = \frac{ {a}^{2} {x}^{2} - 2 }{x}$
з)
$\frac{81 {a}^{2} - {b}^{2} {c}^{4} }{ {b}^{2} {c}^{2} - 9ab} = \frac{(9a - b {c}^{2})(9a + b {c}^{2} ) }{b(b {c}^{2} - 9a) } = \frac{(9a - b {c}^{2})(9a + b {c}^{2}) }{ - b(9a - b {c}^{2} )} = - \frac{9a + b {c}^{2} }{ b}$