308 Вправа 308Алгебра 8 клас Бевз 2025 · § 7 · стор. 65

308. Спрости вираз.

а) ${\displaystyle (\frac{x}{y^2}-\frac{1}{x}):(\frac{1}{x}+\frac{1}{y})}$;

1) ${\displaystyle \frac{x^{\mathrm{\setminus }x}}{y^2}-\frac{1^{\mathrm{\setminus }{y}^2}}{x}=}$${\displaystyle \frac{x^2-y^2}{x{y}^2}=}$${\displaystyle \frac{(x-y)(x+y)}{x{y}^2}}$;

2) ${\displaystyle \frac{1^{\mathrm{\setminus }y}}{x}+\frac{1^{\mathrm{\setminus }x}}{y}=}$${\displaystyle \frac{y+x}{xy}}$;

3) ${\displaystyle \frac{(x-y)(x+y)}{x{y}^2}:\frac{x+y}{xy}=}$${\displaystyle \frac{(x-y)(x+y)}{x{y}^2}\cdot \frac{xy}{x+y}}$=

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

б) ${\displaystyle (\frac{a}{b}+\frac{a^2}{b^2}):(\frac{b^2}{a^2}+\frac{b}{a})}$;

1) ${\displaystyle \frac{a^{\mathrm{\setminus }b}}{b}+\frac{a^2}{b^2}=}$${\displaystyle \frac{ab+a^2}{b^2}=}$${\displaystyle \frac{a(a+b)}{b^2}}$;

2) ${\displaystyle \frac{b^2}{a^2}+\frac{b^{\mathrm{\setminus }a}}{a}=}$${\displaystyle \frac{b^2+ba}{a^2}=}$${\displaystyle \frac{b(b+a)}{a^2}}$;

3) ${\displaystyle \frac{a(a+b)}{b^2}:\frac{b(a+b)}{a^2}=}$${\displaystyle \frac{a(a+b)}{b^2}\cdot \frac{a^2}{b(a+b)}=}$${\displaystyle \frac{a(a+b)\cdot a^2}{b^2\cdot b(a+b)}}$=

=${\displaystyle \frac{a^3\cancel{\left(a+b\right)}}{b^3\cancel{\left(a+b\right)}}=}$${\displaystyle \frac{a^3}{b^3}}$;

в) ${\displaystyle (\frac{m}{m+1}+1):(1-\frac{3m^2}{1-m^2})}$;

1) ${\displaystyle \frac{m}{m+1}+\frac{1^{\mathrm{\setminus }m+1}}{1}=}$${\displaystyle \frac{m+m+1}{m+1}=}$${\displaystyle \frac{2m+1}{m+1}}$;

2) ${\displaystyle 1-\frac{3m^2}{1-m^2}=}$${\displaystyle \frac{1^{\mathrm{\setminus }\left(1-m\right)\left(m+1\right)}}{1}-\frac{3m^2}{(1-m)(m+1)}}$=

=${\displaystyle \frac{(1-m)(m+1)-3m^2}{(1-m)(m+1)}=}$${\displaystyle \frac{1-m^2-3m^2}{(1-m)(m+1)}=}$${\displaystyle \frac{1-4m^2}{(1-m)(m+1)}}$=

=${\displaystyle \frac{(2m+1)(1-2m)}{(1-m)(m+1)}=}$${\displaystyle -\frac{(2m-1)(2m+1)}{1-m^2}}$;

3) ${\displaystyle \frac{2m+1}{m+1}:\frac{(2m-1)(2m+1)}{(m-1)(m+1)}=}$${\displaystyle \frac{2m+1}{m+1}\cdot \frac{(m-1)(m+1)}{(2m-1)(2m+1)}}$=

=${\displaystyle \frac{\cancel{\left(2m+1\right)}\cancel{\left(m+1\right)}(m-1)}{\cancel{\left(2m+1\right)}(2m-1)\cancel{\left(m+1\right)}}=}$${\displaystyle \frac{m-1}{2m-1}}$;

г) ${\displaystyle (\frac{3a+1}{3a-1}-\frac{3a-1}{3a+1}):\frac{12a}{15a-5}}$;

1) ${\displaystyle \frac{{3a+1}^{\mathrm{\setminus }3a+1}}{3a-1}-\frac{{3a-1}^{\mathrm{\setminus }3a-1}}{3a+1}}$=

=${\displaystyle \frac{(3a+1)(3a+1)-(3a-1)(3a-1)}{(3a-1)(3a+1)}}$=

=${\displaystyle \frac{9a^2+6a+1-9a^2+6a-1}{(3a-1)(3a+1)}=}$${\displaystyle \frac{12a}{(3a-1)(3a+1)}=}$${\displaystyle \frac{12a}{9a^2-1}}$;

2) ${\displaystyle \frac{12a}{(3a-1)(3a+1)}:\frac{12a}{15a-5}=}$${\displaystyle \frac{12a}{(3a-1)(3a+1)}\cdot \frac{15a-5}{12a}}$=

=${\displaystyle \frac{12a\cdot (15a-5)}{(3a-1)(3a+1)\cdot 12a}=}$${\displaystyle \frac{5\cdot \cancel{12}\cancel{a}\cancel{\left(3a-1\right)}}{\cancel{12}\cancel{a}(3a+1)\cancel{\left(3a-1\right)}}=}$${\displaystyle \frac{5}{3a+1}}$.