Пожалуйста помогите (даю 36 баллов). Created by gulbahor213. algebra-ru

Пожалуйста помогите (даю 36 баллов)...

gulbahor213 @gulbahor213

July 2022 1 50 Report

Пожалуйста помогите
(даю 36 баллов)

Answers & Comments

Ляляляля109

90.

$(\frac{2m+1}{2m-1}-\frac{2m-1}{2m+1} ):\frac{4m}{10m-5} =\frac{(2m+1)^{2}(2m-1)^{2}}{(2m-1)(2m+1)} *\frac{5(2m-1)}{4m} =\frac{8m}{(2m-1)(2m+1)} *\frac{5(2m-1)}{4m} =\\\\=\frac{10}{2m+1}$

$(\frac{z+6}{3z+9}-\frac{1}{z+3} ):\frac{z+2}{27z} =(\frac{z+6}{3(z+3)}-\frac{3}{3(z+3)} )*\frac{27z}{z+2} =\frac{z+3}{3(z+3)} *\frac{27}{z+2} =\frac{1}{3} *\frac{27}{z+2} =\frac{9}{z+2}$

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

$\frac{m-2}{m-5} :(\frac{m^{2}+24}{m^{2}-25} -\frac{4}{m-5} )=\frac{m-2}{m-5} :\frac{m^{2}+24-4m-20}{(m-5)(m+5)} =\frac{m-2}{m-5} *\frac{(m-5)(m+5)}{(m-2)^{2}} =\frac{m+5}{m-2}$

91.

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

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

$(\frac{c+d}{c} -\frac{2c}{c-d} )*\frac{d-c}{c^{2}+d^{2}} =\frac{c^{2}-d^{2}-2c^{2}}{c(c-d)} *\frac{d-c}{c^{2}+d^{2}} =\frac{-(c^{2}+d^{2})}{-c(d-c)} *\frac{d-c}{c^{2}+d^{2}} =\frac{1}{c}$

$(\frac{2c}{c+d}+\frac{d-c}{c} )*\frac{c+d}{c^{2}+d^{2}} =\frac{2c^{2}+d^{2}-c^{2}}{c(c+d)} *\frac{c+d}{c^{2}+d^{2}} =\frac{c^{2}+d^{2}}{c(c+d)} *\frac{c+d}{c^{2}+d^{2}} =\frac{1}{c}$

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