Question

РЕБЯТА ПОМОГИТЕ ПОЖАЛУЙСТА!!!!!

Answer 1

а) $\frac{3x}{x-4}- \frac{x+8}{x-4}=\frac{3x-x-8}{x-4}= \frac{2x-8}{x-4}= \frac{2(x-4)}{x-4}=2$

б) $\frac{b-a}{ab} +\frac{b-a}{b^{2}} =\frac{b(b-a)}{ab^{2}} +\frac{a(b-a)}{ab^{2}}= \frac{b^{2}-ab+ab-a^{2}}{ab^{2}} =\frac{b^{2}-a^{2}}{ab^{2}}= \frac{(b-a)(b+a)}{ab^{2}}$

в) $\frac{12x^{3}}{25}* \frac{5}{4x^{4}} =\frac{3}{5x}$

г) $-\frac{4a}{5c^{2}}:(- \frac{a}{10c^{4}})=\frac{4a*10c^{4}}{5c^{2}*a} =\frac{4*2c^{2}}{1} =8c^{2}$

а) $\frac{4x}{x+1}- \frac{x-3}{x+1}= \frac{4x-x+3}{x+1}= \frac{3x+3}{x+1}= \frac{3(x+1)}{x+1}=3$

б) $\frac{a-b}{a^{2}}+ \frac{a-b}{ab}= \frac{b(a-b)}{a^{2}b}+ \frac{a(a-b)}{a^{2}b}= \frac{(b-a)(b+a)}{a^{2}b}$

в) $\frac{2}{3x^{3}} *\frac{9x^{4}}{4}= \frac{2*9x^{4}}{3x^{3}*4}= \frac{3x}{2}$

г) $-\frac{3a^{3}}{2b^{2}}:(- \frac{a}{4b^{2}})= \frac{3a^{3}*4b^{2}}{2b^{2}*a}=3a^{2}*2=6a^{2}$