154° Вправа 154Алгебра 8 клас Тарасенкова 2025 · § 4 · стор. 48

154. Спростіть вираз.

1) ${\displaystyle \frac{1^{\mathrm{\setminus }2x^2}}{5x}+\frac{3x^2-1}{10x^3}=}$${\displaystyle \frac{1\cdot 2x^2}{10x^3}+\frac{3x^2-1}{10x^3}=}$${\displaystyle \frac{1\cdot 2x^2+3x^2-1}{10x^3}=}$${\displaystyle \frac{5x^2-1}{10x^3}}$;

2) ${\displaystyle \frac{1^{\mathrm{\setminus }3x}}{4x^2{y}^2}-\frac{2^{\mathrm{\setminus }2y}}{6x^{3}y}=}$${\displaystyle \frac{1\cdot 3x}{12x^3{y}^2}-\frac{2\cdot 2y}{12x^3{y}^2}=}$${\displaystyle \frac{1\cdot 3x-2\cdot 2y}{12x^3{y}^2}=}$${\displaystyle \frac{3x-4y}{12x^3{y}^2}}$;

3) ${\displaystyle \frac{5^{\mathrm{\setminus }x}}{5+x}+\frac{3x-7}{(5+x)x}=}$${\displaystyle \frac{5\cdot x}{x(x+5)}+\frac{3x-7}{x(x+5)}=}$${\displaystyle \frac{5\cdot x+3x-7}{x(x+5)}=}$${\displaystyle \frac{8x-7}{x\left(x+5\right)}}$;

4) ${\displaystyle \frac{10x-1}{x(5-x)}-\frac{2^{\mathrm{\setminus }x}}{5-x}=}$${\displaystyle \frac{10x-1}{x(5-x)}-\frac{2\cdot x}{x(5-x)}=}$${\displaystyle \frac{10x-1-2\cdot x}{x(5-x)}=}$${\displaystyle \frac{8x-1}{x\left(5-x\right)}}$;

5) ${\displaystyle 3y+3=}$${\displaystyle 3(y+1)}$;

${\displaystyle \frac{4y}{3(y+1)}+\frac{{1-y}^{\mathrm{\setminus }3}}{y+1}=}$${\displaystyle \frac{4y}{3(y+1)}+\frac{(1-y)\cdot 3}{3(y+1)}=}$${\displaystyle \frac{4y+(1-y)\cdot 3}{3(y+1)}=}$${\displaystyle \frac{4y+3-3y}{3(y+1)}=}$${\displaystyle \frac{y+3}{3\left(y+1\right)}}$;

6) ${\displaystyle \frac{{4c}^{\mathrm{\setminus }{c}^2}}{c-1}+\frac{2-5c^3}{c^2(c-1)}=}$${\displaystyle \frac{4c\cdot c^2}{c^2(c-1)}+\frac{2-5c^3}{c^2(c-1)}=}$${\displaystyle \frac{4c\cdot c^2+2-5c^3}{c^2(c-1)}=}$${\displaystyle \frac{4c^3+2-5c^3}{c^2(c-1)}=}$${\displaystyle \frac{2-c^3}{c^2\left(c-1\right)}}$.