Помогите пожалуйста!!!!!

Приложения:

Rakvium: Ого. Лучше бы такое отдельными вопросами кидать, - быстрее будет, но я возьмусь :)

Ответы

Ответ дал: Rakvium

Решение заданий с первого изображения:

$1. (\frac{z}{t}-\frac{t}{z})*\frac{5tz}{z-t}=(\frac{z*z}{z*t}-\frac{t*t}{z*t})*\frac{5tz}{z-t}=\\\\=\frac{z^2-t^2}{zt}*\frac{5tz}{z-t}=\frac{z^2-t^2}{z-t}=\frac{(z-t)(z+t)}{z-t}=z+t.$

$2.(\frac{5x^2}{x-3})^2*(\frac{x^2-6x+9}{5x^2})^2-x^2=(\frac{5x^2}{x-3})^2*(\frac{(x-3)^2}{5x^2})^2-x^2=\\\\\frac{(5x^2)^2}{(x-3)^2}*\frac{(x-3)^4}{(5x^2)^2}-x^2=(x-3)^2-x^2=x^2-6x+9-x^2=6x-9.$

$3.(\frac{x}{x-3}-\frac{2}{x+3}):\frac{4x^2+4x+24}{x^2-9}=\\\\=(\frac{x}{x-3}-\frac{2}{x+3})*\frac{x^2-9}{4x^2+4x+24}=\\\\=(\frac{x(x+3)}{(x-3)(x+3)}-\frac{2(x-3)}{(x+3)(x-3)})*\frac{x^2-9}{4x^2+4x+24}=\\\\=(\frac{x(x+3)-2(x-3)}{(x+3)(x-3)})*\frac{x^2-9}{4(x^2+x+6)}=\\\\\\=\frac{x^2+3x-2x+6}{(x+3)(x-3)}*\frac{x^2-9}{4(x^2+x+6)}=\\\\\\=\frac{x^2+x+6}{x^2-9}*\frac{x^2-9}{4(x^2+x+6)}=\frac{1}{4}.$

$4.\frac{5}{5x-2}-\frac{2}{x+3}:\frac{5x-2}{9x^2-1}=\\\\=\frac{5}{5x-2}-\frac{2}{x+3}*\frac{9x^2-1}{5x-2}=\\\\=\frac{5}{5x-2}-\frac{2*(3x-1)(3x+1)}{(x+3)(5x-2)}=\\\\=\frac{5(x+3)}{(x+3)(5x-2)}-\frac{2*(3x-1)(3x+1)}{(x+3)(5x-2)}=\\\\=\frac{5(x+3)-2*(9x^2-1)}{(x+3)(5x-2)}=\\\\=\frac{5x+15-18x^2+2}{(x+3)(5x-2)}=\\\\=\frac{-18x^2+5x+17}{5x^2+13x-6}.$

$5.\frac{472,472b^3a}{c^5}*\frac{c^2}{1,1b}*\frac{c^3}{0,7b}:\frac{1,3a}{c}+28bc=\\\\=\frac{472,472b^3a*c^2*c^3*c}{c^5*1,1b*0,7b*1,3a}+28bc=\\\\=\frac{472,472bc}{1,001}+28bc=472bc+28bc=500bc.$

$6.(\frac{a+b}{a^2-ab}-\frac{2b}{a^2-b^2})*\frac{b^2-a^2}{1+\frac{b^2}{a^2}}=\\\\=(\frac{a+b}{a(a-b)}-\frac{2b}{(a-b)(a+b)})*\frac{(b-a)(b+a)}{1+\frac{b^2}{a^2}}=\\\\=(\frac{(a+b)(a+b)}{a(a-b)(a+b)}-\frac{2ab}{a(a-b)(a+b)})*\frac{(b-a)(b+a)}{1+\frac{b^2}{a^2}}=\\\\=\frac{a^2+b^2}{a(a-b)(a+b)}*\frac{(b-a)(b+a)}{1+\frac{b^2}{a^2}}=\\\\=\frac{a^2+b^2}{a(a-b)(a+b)}*\frac{-(a-b)(b+a)}{1+\frac{b^2}{a^2}}=\\\\=\frac{a^2+b^2}{a}*\frac{-1}{1+\frac{b^2}{a^2}}=$

$=\frac{a(a^2+b^2)}{a*a}*\frac{-1}{1+\frac{b^2}{a^2}}=\\=-\frac{a(a^2+b^2)}{a^2(1+\frac{b^2}{a^2})}=-\frac{a(a^2+b^2)}{a^2+b^2}=-a.$

$7.\frac{\frac{a+b}{a}*(\frac{a}{b}-\frac{a}{a+b})}{\frac{m+b}{m}*(\frac{m}{b}-\frac{m}{m-b})}=\frac{\frac{(a+b)*a}{a*b}-1}{\frac{(m+b)*m}{mb}-1}=\frac{\frac{a+b}{b}-1}{\frac{m+b}{b}-1}=\frac{b(\frac{a+b}{b}-1)}{b(\frac{m+b}{b}-1)}=\frac{a+b-b}{m+b-b}=\frac{a}{m}.$

Решение заданий со второго изображения:

$1)(\frac{a+3}{a-3}+\frac{a-3}{a+3}):\frac{3a^2+27}{9-a^2}=\\=\frac{(a+3)(a+3)+(a-3)(a-3)}{(a-3)(a+3)}*\frac{(3-a)(3+a)}{3(a^2+9)}=\\=-\frac{a^2+6a+9+a^2-6a+9}{(a-3)(a+3)}*\frac{(a-3)(a+3)}{3(a^2+9)}=\\=-\frac{2a^2+18}{3(a^2+9)}=-\frac{2(a^2+9)}{3(a^2+9)}=-\frac{2}{3}.$

$2)\frac{3a}{a-4}-\frac{a+2}{5a-20}*\frac{240}{a^2+2a}=\\=\frac{3a}{a-4}-\frac{a+2}{5(a-4)}*\frac{240}{a(a+2)}=\\=\frac{3a}{a-4}-\frac{48}{a(a-4)}=\\=\frac{3a*a}{a(a-4)}-\frac{48}{a(a-4)}=\\=\frac{3a*a-48}{a(a-4)}=\frac{3(a^2-16)}{a(a-4)}=\frac{3(a-4)(a+4)}{a(a-4)}=\frac{3a+12}{a}=3+\frac{12}{a}.$

Доказательство тождества с третьего изображения:

$\frac{8m^3}{(m^2-64)^2}:(\frac{1}{(m+8)^2}+\frac{2}{m^2-64}+\frac{1}{(m-8)^2})=2m\\\frac{8m^3}{(m-8)^2(m+8)^2}:(\frac{1}{(m+8)^2}+\frac{2}{(m-8)(m+8)}+\frac{1}{(m-8)^2})=2m\\\frac{8m^3}{(m-8)^2(m+8)^2}:(\frac{(m-8)^2}{(m+8)^2(m-8)^2}+\frac{2(m-8)(m+8)}{(m-8)^2(m+8)^2}+\frac{(m+8)^2}{(m-8)^2(m+8)^2})=2m\\\frac{8m^3}{(m-8)^2(m+8)^2}:(\frac{(m-8)^2+2(m-8)(m+8)+(m+8)^2}{(m-8)^2(m+8)^2})=2m\\\frac{8m^3}{(m-8)^2(m+8)^2}:(\frac{((m-8)+(m+8))^2}{(m-8)^2(m+8)^2})=2m\\$

$\frac{8m^3}{(m-8)^2(m+8)^2}:(\frac{(2m)^2}{(m-8)^2(m+8)^2})=2m\\\frac{8m^3}{(m-8)^2(m+8)^2}:\frac{4m^2}{(m-8)^2(m+8)^2}=2m\\\frac{8m^3}{(m-8)^2(m+8)^2}*\frac{(m-8)^2(m+8)^2}{4m^2}=2m\\\frac{8m^3}{4m^2}=2m\\2m=2m.$