Question

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Answer 1

Ответ:

Решение представлено на фотографиях

Answer 2

Ответ:

$1)\ \ x=-5\ ,\ \ y=-0,5\\\\\\x^2-\dfrac{x^3-4xy^2}{x^3-2x^2y+xy^2}\cdot \dfrac{x^2-2xy+y^2}{x-2y}=x^2-\dfrac{x\, (x^2-4y^2)}{x(x^2-2xy+y^2)}\cdot \dfrac{(x-y)^2}{x-2y}=\\\\\\=x^2-\dfrac{x\, (x-2y)(x+2y)}{x\, (x-y)^2}\cdot \dfrac{(x-y)^2}{x-2y}=x^2-(x+2y)=x^2-x-2y=\\\\\\=25+5+1=31$

$2)\ \ n=\dfrac{1}{3}\\\\\dfrac{3}{2}-\dfrac{3n^2-6n+3}{2n^2+2n+2}:\dfrac{n-1}{n^5n^2+n}=\dfrac{3}{2}-\dfrac{3\, (n^2-2n+1)}{2\, (n^2+n+1)}\cdot \dfrac{n\, (n^2+n+1)}{n-1}=\\\\\\=\dfrac{3}{2}-\dfrac{3\, (n-1)^2\cdot n}{2\, (n-1)}=\dfrac{3}{2}-\dfrac{3\, (n-1)\cdot n}{2}=\dfrac{3-3\, (n^2-n)}{2}=\\\\\\=\dfrac{3\, (1-n^2+n)}{2}=-\dfrac{3}{2}\, (n^2-n+1)=-\dfrac{3}{2}\, \Big(\dfrac{1}{9}-\dfrac{1}{3}+1\Big)=-\dfrac{3}{2}\cdot \dfrac{1-3+9}{9}=\\\\\\=-\dfrac{3}{2}\cdot \dfrac{7}{9}=-\dfrac{7}{6}=-1\dfrac{1}{6}$

$3)\ \ a=3\dfrac{1}{4}=\dfrac{13}{4}\ ,\ \ \ \ b=-0,75=-\dfrac{3}{4}\\\\\\\Big(\dfrac{3}{a-b}-\dfrac{3a}{b^2-a^2}\Big):\dfrac{6a+3b}{a^2+2ab+b^2}=\Big(\dfrac{3}{a-b}+\dfrac{3a}{(a-b)(a+b)}\Big):\dfrac{3\, (2a+b)}{(a+b)^2}=\\\\\\=\dfrac{3(a+b)+3a}{(a-b)(a+b)}:\dfrac{3\, (2a+b)}{(a+b)^2}=\dfrac{3\, (2a+b)}{(a-b)(a+b)}\cdot \dfrac{(a+b)^2}{3\, (2a+b)}=\dfrac{a+b}{a-b}=\\\\\\=\dfrac{\frac{13}{4}-\frac{3}{4}}{\frac{13}{4}+\frac{3}{4}}=\dfrac{10}{16}=\dfrac{5}{8}=0,625$