Verified answer

Объяснение:

№1

213,25х - 49 5/7у - 215 5/6х + 50у =

(213,25х - 215 5/6х) + (-49 5/7у + 50у) =

(213 25/100х - 215 5/6х) + (-49 5/7у + 49 7/7у) =

(213 1/4х - 215 5/6х) + (2/7у) =

(213 3/12х - 215 10/12х) + 2/7у =

-2 7/12х + 2/7у

№3

-59,5с + 44 5/6d - 46 2/9d + 57 2/7c =

(-59,5c + 57 2/7c) + (44 5/6d - 46 2/9d) =

(-59 1/2c + 57 2/7c) + (44 15/18d - 46 4/18d) =

(-59 7/14c + 57 4/14c) + (44 15/18d - 45 22/18d) =

-2 3/14c + (-1 7/18d) = -2 3/14c - 1 7/18d

Ответ:

Преобразовать (упростить) выражения.

[tex]\displaystyle 1)\ \ 213,25x-49\frac{5}{7}\, y-215\frac{5}{6}\, x+50y=\Big(213\frac{1}{4}\, x-215 \frac{5}{6}\, x\Big) +\Big(50y-49\frac{5}{7}\, y\Big)=\\\\\\=\Big(213+\frac{1}{4}-215-\frac{5}{6}\Big)\, x+\Big(49+\frac{7}{7}- 49-\frac{5}{7}\Big)\, y=\Big(-2+\frac{3-10}{12}\Big)\, x+\frac{2}{7}\, y=\\\\\\=\Big (-2-\frac{7}{12}\Big)\, x+\frac{2}{7}\, y=-\frac{31}{12}\, x+\frac{2}{7}\, y=\bf -2\frac{7}{12}\, x+\frac{2}{7}\, y[/tex]  

[tex]\displaystyle 3)\ \ -59,5c+44\frac{5}{6}\, d-46\frac{2}{9}\, d+57\frac{2}{7}\, c=\Big(-59\frac{1}{2}+57\frac{2}{7}\Big)\, c+\Big(44\frac{5}{6}-46\frac{2}{9}\Big)\, d=\\\\\\=\Big(-59-\frac{1}{2}+57+\frac{2}{7}\Big)\, c+\Big(44+\frac{5}{6}-46-\frac{2}{9}\Big)\, d=\\\\\\=\Big(-2-\frac{1}{2}+\frac{2}{7}\Big)\, c+\Big(-2+\frac{5}{6}-\frac{2}{9}\Big)\, d=\\\\\\=\Big(-2+\frac{2\cdot 2-1\cdot 7}{14}\Big)\, c+\Big(-2+\frac{5\cdot 3-2\cdot 2}{18}\Big)\, d=[/tex]  

[tex]\displaystyle=\Big(-2-\frac{3}{14}\Big)\, c+\Big(-2+\frac{11}{18}\Big)\, d=-\frac{31}{14}\, c-\frac{25}{18}\, d=\bf -2\frac{3}{14}\, c-1\frac{7}{18}\, d[/tex]