Question

Помогите срочно!!!!! Пожалуйста!!! ​

Answer 1

Ответ:

$1)\ \ x^3+64=(x+4)(x^2-4x+16)\\\\(a-2c)^3=a^3-3a^2\cdot 2c+3a\cdot 4c^2-8c^3=a^3-6a^2c+12ac^2-8c^3\\\\\\2)\ \ 125x^3-225x^2y+135xy^2-27y^3=\\\\=(5x)^3-3\cdot (5x)^2\cdot (3y)+3\cdot (5x)\cdot (3y)^2-(3y)^3=(5x-3y)^3\\\\(a-2)(a^2+2a+4)=a^3-2^3=a^3-8\\\\\\3)\ \ (3x+4)^2-(3x-1)(3x+1)=41\\\\9x^2+24x+16-(9x^2-1)=41\\\\24x+17=41\\\\24x=24\\\\x=1$

$4)\ \ \left\{\begin{array}{l}x-y=3\\x^2-y^2=183\end{array}\right\ \ \left\{\begin{array}{l}x-y=3\\(x-y)(x+y)=183\end{array}\right\ \ \left\{\begin{array}{l}x-y=3\\3\cdot (x+y)=183\end{array}\right\\\\\\\left\{\begin{array}{l}x-y=3\\x+y=61\end{array}\right\ \oplus \ominus \left\{\begin{array}{l}2x=64\\-2y=-58\end{array}\right\ \ \left\{\begin{array}{l}x=32\\y=29\end{array}\right\\\\\\Otvet:\ \ x=32\ ,\ y=39$