Question

сократите дробь
1)5a²-20/a²-4a+4
2)10ab²/13a²$b^{3}$
3)7a²b/14a²bd
4)m²-6m+9/4m²+36
6)2b²+25/2b+10
8)10x²y+20xy²/15xy
9)7(a+1)²z/14k²z²
10)3/15bc
11)12d²/24a²(a+1)³d²
12)a²-4/3a+6
13)12c²+9cd/9cd

Answer 1

Ответ:

$\displaystyle \frac{5a^2-20}{a^2-4a+4}=\frac{5(a-2)(a+2)}{(a-2)^2}=\frac{5(a+2)}{a-2}\\\\\\\frac{10ab^2}{13a^2b^3}=\frac{10}{13ab}\\\\\\\frac{7a^2b}{14a^2bd}=\frac{1}{2d}\\\\\\\frac{m^2-6m+9}{4m^2+36}=\frac{(m-3)^2}{4(m^2+9)}\ \ ne\ \ sokratitsya\\\\\\\star \ \ \frac{m^2-6m+9}{4m^2-36}=\frac{(m-3)^2}{4(m-3)(m+3)}=\frac{m-3}{4(m+3)}$

$\displaystyle \frac{2b^2+25}{2b+10}=\frac{2b^2+25}{2(b+5)}\\\\\\\frac{10x^2y+20xy^2}{15xy}=\frac{10xy(x+2y)}{15xy}=\frac{2(x+2y)}{3}\\\\\\\frac{7(a+1)^2z}{14\, k^2z^2}=\frac{(a+1)^2}{2\, k^2z}\\\\\\\frac{3}{15bc}=\frac{1}{5bc}\\\\\\\frac{12d^2}{24a^2(a+1)^3d^2}=\frac{1}{2a^2(a+1)^3}\\\\\\\frac{a^2-4}{3a+6}=\frac{(a-2)(a+2)}{3(a+2)}=\frac{a-2}{3}$

$\dfrac{12c^2+9cd}{9\, cd}=\dfrac{3c(4c+3d)}{9\, cd}=\dfrac{4c+3d}{3\, d}$

Answer 2

Ответ:

$\frac{5 {a}^{2} - 20}{ {a}^{2} - 4a + 4} = \frac{5( {a}^{2} - 4) }{ {(a - 2)}^{2} } = \frac{5(a - 2)(a + 2)}{({a - 2)}^{2} } = \frac{5(a + 2)}{(a - 2)}$

$\frac{10a {b}^{2} }{13 {a}^{2} {b}^{3} } = \frac{10}{13ab}$

$\frac{7 {a}^{2} b}{14 {a}^{2}bd } = \frac{1}{2d}$

$\frac{m²-6m+9}{4m²-36} = \frac{ {(m - 3)}^{2} }{4(m - 3)(m + 3)} = \frac{m - 3}{4(m + 3)} \\ </p><p>6) \frac{2b²+25}{2b+10} = \frac {2b²+25)}{2(b + 5)} \\ </p><p>8) \frac{10x²y+20xy²}{15xy} = \frac{ 10xy(x + 2y)}{15xy} = \frac{2(x + 2y)}{3}$

$9) \frac{ 7(a+1)²z}{14k²z²} = \frac{ {(a + 1)}^{2} }{2 {k}^{2} z} \\ </p><p>10) \frac{3}{15bc} = \frac{1}{5bc} </p><p>11) \frac {12d²}{24a²(a+1)³d²} = \frac{1}{2 {a}^{2} {(a + 1)}^{3} } \\ </p><p>12) \frac{ a²-4}{3a+6} = \frac{(a - 2)(a + 2)}{3(a + 2)} = \frac{a - 2}{3} \\ </p><p>13) \frac{12c²+9cd}{9cd} = \frac{3c(4c + 3d)}{9cd} = \frac{4c + 3d}{3d}$