Задание 253

Решение:

• $${\largeа)}\ 1{,}1x-2{,}7y+0{,}8x-x+3y=1{,}1x+0{,}8x-x-2{,}7y+3y=(1{,}1+0{,}8-1)x+(-2{,}7+3)y=0{,}9x+0{,}3y$$

• $${\largeб)}\ 27a-3{,}1b+9a+3{,}1a+0{,}4b-a=27a+9a+3{,}1a-a-3{,}1b+0{,}4b=(27+9+3{,}1-1)a+(-3{,}1+0{,}4)b=38{,}1a-2{,}7b$$

• $${\largeв)}\ \frac{1}{3}x+\frac{2}{5}y-2x+1\frac{1}{4}y=\frac{1}{3}x-2x+\frac{2}{5}y+1\frac{1}{4}y=\left(\frac{1}{3}-1\frac{3}{3}\right)x+\left(\frac{8}{20}+1\frac{5}{20}\right)y=\left(-1\frac{2}{3}\right)x+1\frac{13}{20}y$$

• $${\largeг)}\ 15a-4x-5{,}6a+2{,}3x+a=15a-5{,}6a+a-4x+2{,}3x=(15-5{,}6+1)a+(-4+2{,}3)x=10{,}4a-1{,}7x$$

• $${\largeд)}\ 67{,}1a-\frac{1}{3}b+\frac{1}{5}a+2b+2{,}5a-7b=67{,}1a+\frac{1}{5}a+2{,}5a-\frac{1}{3}b+2b-7b=\left(67\frac{1}{10}+\frac{2}{10}+2\frac{5}{10}\right)a+\left(-\frac{1}{3}+2-7\right)b=69\frac{8}{10}a-5\frac{1}{3}b=69\frac{4}{5}a-5\frac{1}{3}b$$

• $${\largeе)}\ \frac{1}{4}b-7x-3{,}2b+2\frac{3}{4}x+b+0{,}6x=\frac{1}{4}b-3{,}2b+b-7x+2\frac{3}{4}x+0{,}6x=(0{,}25-3{,}2+1)b+(-7+2{,}75+0{,}6)x=(-1{,}95)b-3{,}65x$$

• $${\largeж)}\ xyx-2x^2y+2x-3x=x^2y-2x^2y+2x-3x=(1-2)x^2y+(2-3)x={-}x^2y-x$$

• $${\largeз)}\ ba^2-3a^3+7aba+3a^2-8a^2b=a^2b+7a^2b-8a^2b-3a^3+3a^2=(1+7-8)a^2b-3a^3+3a^2=0-3a^3+3a^2={-}3a^3+3a^2$$