163 Вправа 163Алгебра 8 клас Тарасенкова 2025 · § 4 · стор. 50

163. Спростіть вираз.

1) ${\displaystyle \frac{{3x-1}^{\mathrm{\setminus }3}}{2}+\frac{x-2}{6}-\frac{{2x-3}^{\mathrm{\setminus }2}}{3}=}$${\displaystyle \frac{(3x-1)\cdot 3}{6}+\frac{x-2}{6}-\frac{(2x-3)\cdot 2}{6}=}$${\displaystyle \frac{(3x-1)\cdot 3+x-2-(2x-3)\cdot 2}{6}}$;${\displaystyle \frac{9x-3+x-2-4x+6}{6}=}$${\displaystyle \frac{6x+1}{6}}$;

2) ${\displaystyle \frac{{2a-3b}^{\mathrm{\setminus }a}}{5a{b}^2}-\frac{{2b-3a}^{\mathrm{\setminus }b}}{5a^{2}b}=}$${\displaystyle \frac{(2a-3b)\cdot a}{5a^2{b}^2}-\frac{(2b-3a)\cdot b}{5a^2{b}^2}=}$${\displaystyle \frac{(2a-3b)\cdot a-(2b-3a)\cdot b}{5a^2{b}^2}}$;${\displaystyle \frac{2a^2-3ab-2b^2+3ab}{5a^2{b}^2}=}$${\displaystyle \frac{2a^2-2b^2}{5a^2{b}^2}}$;

3) ${\displaystyle 2xy-3y^2=}$${\displaystyle y(2x-3y)}$; ${\displaystyle 3xy-2x^2=}$${\displaystyle -x(2x-3y)}$;

${\displaystyle \frac{{x-y}^{\mathrm{\setminus }x}}{y(2x-3y)}+\frac{{y-x}^{\mathrm{\setminus }y}}{x(2x-3y)}=}$${\displaystyle \frac{(x-y)\cdot x}{xy(2x-3y)}+\frac{(y-x)\cdot y}{xy(2x-3y)}=}$${\displaystyle \frac{(x-y)\cdot x+(y-x)\cdot y}{xy(2x-3y)}}$;${\displaystyle \frac{x^2-xy-xy+y^2}{xy(2x-3y)}=}$${\displaystyle \frac{{\left(x-y\right)}^2}{xy\left(2x-3y\right)}}$;

4) ${\displaystyle \frac{{3y-x}^{\mathrm{\setminus }5x}}{21x^2{y}^2}-\frac{{5x-y}^{\mathrm{\setminus }3y}}{35x^{3}y}=}$${\displaystyle \frac{(3y-x)\cdot 5x}{105x^3{y}^2}-\frac{(5x-y)\cdot 3y}{105x^3{y}^2}=}$${\displaystyle \frac{(3y-x)\cdot 5x-(5x-y)\cdot 3y}{105x^3{y}^2}}$;${\displaystyle \frac{15xy-5x^2-15xy+3y^2}{105x^3{y}^2}=}$${\displaystyle \frac{3y^2-5x^2}{105x^3{y}^2}}$.