реши номер 26
а) \(\dfrac{4a^2}{6ac}=\dfrac{2a}{3c}\)
б) \(\dfrac{7x^2y}{21xy^2}=\dfrac{x}{3y}\)
в) \(\dfrac{56m^2n^5}{35mn^5}=\dfrac{8m}{5}\)
г) \(\dfrac{25p^4q}{100p^5q}=\dfrac{1}{4p}\)
Обсуждение
4 сообщенияРеши №32
Реши №32
№ 32
а)
\[
\frac{15a^2-10ab}{3ab-2b^2}
=\frac{5a(3a-2b)}{b(3a-2b)}
=\frac{5a}{b}
=\frac{5\cdot(-2)}{-0{,}1}=100.
\]
б)
\[
\frac{9c^2-4d^2}{18c^2d-12cd^2}
=\frac{(3c-2d)(3c+2d)}{6cd(3c-2d)}
=\frac{3c+2d}{6cd}
=\frac{3}{2}.
\]
в)
\[
\frac{6x^2+12xy}{5xy+10y^2}
=\frac{6x(x+2y)}{5y(x+2y)}
=\frac{6x}{5y}
=\frac{6\cdot\frac23}{5\cdot(-0{,}4)}
=-2.
\]
г)
\[
\frac{x^2+6xy+9y^2}{4x^2+12xy}
=\frac{(x+3y)^2}{4x(x+3y)}
=\frac{x+3y}{4x}
=\frac{-0{,}2+3(-0{,}6)}{4(-0{,}2)}
=\frac52.
\]
Ответ: \(100;\ \frac32;\ -2;\ \frac52.\)
№ 32
а)
\[
\frac{15a^2-10ab}{3ab-2b^2}
=\frac{5a(3a-2b)}{b(3a-2b)}
=\frac{5a}{b}
\]
\[
\frac{5\cdot(-2)}{-0{,}1}=100
\]
б)
\[
\frac{9c^2-4d^2}{18c^2d-12cd^2}
=\frac{(3c-2d)(3c+2d)}{6cd(3c-2d)}
=\frac{3c+2d}{6cd}
\]
\[
\frac{3\cdot\frac23+2\cdot\frac12}
{6\cdot\frac23\cdot\frac12}
=\frac{3}{2}
\]
в)
\[
\frac{6x^2+12xy}{5xy+10y^2}
=\frac{6x(x+2y)}{5y(x+2y)}
=\frac{6x}{5y}
\]
\[
\frac{6\cdot\frac23}{5\cdot(-0{,}4)}=-2
\]
г)
\[
\frac{x^2+6xy+9y^2}{4x^2+12xy}
=\frac{(x+3y)^2}{4x(x+3y)}
=\frac{x+3y}{4x}
\]
\[
\frac{-0{,}2+3\cdot(-0{,}6)}{4\cdot(-0{,}2)}
=\frac{-2}{-0{,}8}=2{,}5
\]
Ответ: а) \(100\); б) \(\frac32\); в) \(-2\); г) \(2{,}5\).
