Question

номер 173!!!!!!!!пжпжжпжпж

Answer 1

Объяснение:

надеюсь понятно

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Answer 2

Ответ:

$\displaystyle 1)\ \ \frac{a^2-1}{a^2-1}=\frac{(a-1)(a+1)}{(a-2)(a+2)}\\\\\\a=1:\ \frac{(1-1)(1+1)}{(1-2)(1+2)}=0\\\\a=5:\ \frac{4\cdot 6}{3\cdot 7}=\frac{24}{21}\\\\a=-5:\ \frac{-6\cdot (-4)}{-7\cdot (-3)}=\frac{24}{21}\\\\a=6:\ \frac{5\cdot 7}{4\cdot 8}=\frac{35}{32}$

$\displaystyle 2)\ \ \frac{4-t^2}{16-t^2}=\frac{4-t^2}{(4-t^2)(4+t^2)}=\frac{1}{4+t^2}\\\\\\t=-1:\ \frac{1}{4+1}=\frac{1}{5}\\\\t=2:\ \frac{1}{4+4}=\frac{1}{8}\\\\t=-2:\ \frac{1}{4+4}=\frac{1}{8}\\\\t=3:\ \frac{1}{4+9}=\frac{1}{13}$

$\displaystyle 3)\ \ \frac{x^2-2x+1}{25+10x+x^2}=\frac{(x-1)^2}{(x+5)^2}\\\\\\x=0:\ \frac{(0-1(^2}{(0+5)^2}=\frac{1}{25}\\\\t=1:\ \frac{0}{36}=0\\\\t=2:\ \frac{1}{7^2}=\frac{1}{49}\\\\t=5:\ \frac{4^2}{10^2}=\frac{16}{100}=\frac{4}{25}$

$\displaystyle 4)\ \ \frac{u^2-12u+36}{4+4u+u^2}=\frac{(u-6)^2}{(u+2)^2}\\\\\\u=0:\ \frac{6^2}{2^2}=9\\\\u=1:\ \frac{5^2}{3^2}=\frac{25}{9}\\\\u=2:\ \frac{4^2}{4^2}=1\\\\u=6:\ \frac{0}{8^2}=0$