Question

Выполните действия. 50 баллов. 8 класс.

Answer 1

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Answer 2

1) $\frac{ z^{2} -2z+4}{4 z^{2} -1} * \frac{2z^{2}+z }{ z^{3} +8} = \frac{ (z^{2} -2z+4)*z(2z+1)}{(2z-1)(2z+1)*(z+2)(z^{2} -2z+4)} = \frac{z}{(2z-1)(z+2)}$
2) $\frac{z}{(2z-1)(z+2)} - \frac{z+2}{2 z^{2} -z} = \frac{ z^{2}-( z+2)^{2} }{z(2z-1)(z+2)} = \frac{ z^{2} - z^{2} -4z-4}{z(2z-1)(z+2)}= -\frac{4(z+1)}{z(2z-1)(z+2)}$
3) $- \frac{4(z+1)}{z(2z-1)(z+2)}: \frac{7}{ z^{2} +2z} = -\frac{4(z+1)*z(z+2)}{z(2z-1)(z+2)*7} = -\frac{4(z+1)}{7(2z-1)}$
4) $- \frac{4(z+1)}{7(2z-1)} - \frac{10z+1}{7-14z} = \frac{4z+4}{7-14z} - \frac{10z+1}{7-14z} = \frac{4z+4-10z-1}{7-14z} = \frac{3-6z}{7-14z} = \frac{3(1-2z)}{7(1-2z)} =$$\frac{3}{7}$

Ответ: $\frac{3}{7}$