Question

Выполните вычитание дробей

с

Answer 1

$1)\; \; \frac{x-3}{3x^2}-\frac{7-x}{x^3}=\frac{(x-3)\cdot x-(7-x)\cdot 3}{3x^3}=\frac{x^2-3x-21+3x}{3x^3}=\frac{x^2-21}{3x^2}\\\\\\2)\; \; \frac{36}{a^2+3a}-\frac{12}{a}=\frac{36-12(a+3)}{a(a+3)}=\frac{36-12a-36}{a(a+3)}=-\frac{12a}{a(a+3)}$

$3)\; \; \frac{a^2}{a^2-25}-\frac{a}{a-5}=\frac{a^2}{(a-5)(a+5)}-\frac{a}{a-5}=\frac{a^2-a(a+5)}{(a-5)(a+5)}=-\frac{5a}{(a-5)(a+5)}\\\\\\4)\; \; \frac{16-x^2}{x^2+8x+16}=\frac{(4-x)(4+x)}{(x+4)^2}=\frac{4-x}{x+4}$

Answer 2

Вот:

Объяснение:

$\frac{x-3x-21+3x}{3x^{3} } = \frac{x^{2}-21}{3x^{3} }$

$\frac{36}{a(a+3)} -\frac{12}{a} = \frac{36-12a-36}{a^{2}+3a }=-\frac{12}{a+3}$

$\frac{a^{2} }{(a+5)(a-5)}-\frac{a}{a-5} = \frac{a^{2}-a^{2}-5a}{a^{2}-25} = -\frac{5a}{a^{2}-25}$

$\frac{(x+4)(x-4)}{(x+4)^{2}} =\frac{x-4}{x+4}$