Question

помогите пожалуйста (четвертое не надо)

Answer 1

Ответ:

Задание 1

А) $\frac{63a^6b^4}{c^7} *(-\frac{c^3}{36a^5b^7} )=-\frac{7a}{c^4} *\frac{1}{4b^3} =-\frac{7a}{4b^3c^4}$

Б) $\frac{54x^7}{y^{10}} : (18x^4y^6)=\frac{54x^7}{y^{10}} *\frac{1}{18x^4y^6} =\frac{3x^3}{y^{10}} *\frac{1}{y^6} =\frac{3x^3}{y^{16}}$

В) $\frac{4x+4t}{t} *\frac{3t^2}{x^2-t^2} =4(x+t)*\frac{3t}{(x-t)*(x+t)} =4*\frac{3t}{x-t} =\frac{12t}{x-t}$

Г) $\frac{6x-30}{x+8} :\frac{x^2-25}{2x+16} =\frac{6(x-5)}{x+8} *\frac{2x+16}{x^2-25} =\frac{6(x-5)}{x+8} *\frac{2(x+8)}{(x-5)*(x+5)} =6*\frac{2}{x+5} =\frac{12}{x+5}$

Задание 2

А) $\frac{2a}{a-2} +\frac{a+7}{8-4a} *\frac{32}{7a+a^2} =\frac{2a}{a-2} +\frac{a+7}{4(2-a)} *\frac{32}{a*(7+a)} =\frac{2a}{a-2} +\frac{1}{2-a}*\frac{8}{a} =\frac{2a}{a-2} +\frac{8}{a*(2-a)} =$

$=\frac{2a}{a-2} +\frac{8}{a*(-(a-2))} =\frac{2a}{a-2}-\frac{8}{a*(a-2)} =\frac{2a^2-8}{a*(a-2)} =\frac{2(a^2-4)}{a*(a-2)} =\frac{2(a-2)*(a+2)}{a*(a-2)} =$

$=\frac{2(a+2)}{a} =\frac{2a+4}{a}$

Б) $(\frac{x+1}{x-1} -\frac{x-1}{x+1} ):\frac{2x}{1-x^2} =\frac{(x+1)^2-(x-1)^2}{(x-1)*(x+1)} *\frac{1-x^2}{2x} =\frac{2*2x}{(x-1)*(x+1)} *\frac{(1-x)*(1+x)}{2x} =$

$=\frac{4x}{x-1} *\frac{-(x-1)*1}{2x} =2*(-1)=-2$