Question

сократите дробь,помогите пожалуйста если можно с фоткой пожалуйста

Answer 1

$1) \frac{a ^{2}-4 }{a+2} = \frac{(a-2)(a+2)}{a+2} =a-2$

2) a² - 7a - 18 = 0
D = (-7)² - 4· 1 · (-18) = 49 + 72 = 121
$3) \sqrt{D} =11$

$a_{1} = \frac{7+11}{2} = \frac{18}{2} =9$

$a _{2} = \frac{7-11}{2} = \frac{-4}{2}=-2$

$a^{2} -7a-18=(a+2)(a-9)$

$\frac{a+2}{ a^{2}-7a-18 } = \frac{a+2}{(a-9)(a+2)} = \frac{1}{a-9}$

3) a² + 7a + 12 = 0
D = 7² - 4· 1 · 12 = 49 - 48 = 1
$\sqrt{D} =1$

$a _{1} = \frac{-7+1}{2} = \frac{-6}{2} =-3$

$a_{2} = \frac{-7-1}{2} = \frac{-8}{2} =-4$

a² + 7a + 12 = (a + 3)(a + 4)

a² + 6a + 8 = 0
D = 6² - 4· 1· 8 = 36 - 32 = 4
$\sqrt{D} =2$

$a _{1} = \frac{-6+2}{2}= \frac{-4}{2} =-2$

$a _{2} = \frac{-6-2}{2} = \frac{-8}{2} =-4$
a² + 6a + 8 = (a + 2)(a + 4)

$\frac{a ^{2}+7a+12 }{a ^{2} +6a+8} = \frac{(a+3)(a+4)}{(a+2)(a+4) } = \frac{a+3}{a+2}$

4) 2a² - 5a - 3 = 0
D = (-5)² - 4 · 2 · (-3) = 25 + 24 = 49

$\sqrt{D} =7$

$a_{1} = \frac{5+7}{4} = \frac{12}{4} =3$

$a _{2} = \frac{5-7}{4} = \frac{-2}{4} =- \frac{1}{2}$

2a² - 5a - 3 = $2(a-3)(a+ \frac{1}{2} )$

4a² - 6a - 4 = 0
D = (-6)² - 4· 4 · (-4) = 36+ 64 = 100

$\sqrt{D} =10$

$a _{1} = \frac{6+10}{8} = \frac{16}{8} =2$

$a _{2} = \frac{6-10}{8} = \frac{-4}{8} =- \frac{1}{2}$

4a² - 6a - 4 = $4(a-2)(a+ \frac{1}{2} )$

$\frac{2a ^{2} -5a-3}{4a ^{2}-6a-4 } = \frac{2(a-3)(a+ \frac{1}{2}) }{4(a-2)(a+ \frac{1}{2}) } = \frac{a-3}{2(a-2)} = \frac{a-3}{2a-4}$

5) -2a² + 3a + 2 = 0
D = 3² - 4 ·(-2) · 2 = 9 + 16 = 25
$\sqrt{D} =5$

$a _{1} = \frac{-3+5}{-4} = \frac{2}{-4} =- \frac{1}{2}$

$a _{2} = \frac{-3-5}{-4} = \frac{-8}{-4} =2$

-2a² + 3a + 2 = $-2(a+ \frac{1}{2} )(a-2)$

2a² + 5a + 2 = 0
D = 5² - 4· 2 · 2 = 25 - 16 = 9

$\sqrt{D } = 3$

$a _{1} = \frac{-5+3}{4} = \frac{-2}{4} =- \frac{1}{2}$

$a _{2} = \frac{-5-3}{4} = \frac{-8}{4} =-2$

2a² + 5a + 2 = $2(a+ \frac{1}{2})(a+2)$

$\frac{-2a ^{2}+3a+2 }{2 a^{2}+5a+2 } = \frac{-2(a+ \frac{1}{2})(a-2) }{2(a+ \frac{1}{2})(a+2) } =- \frac{(a-2)}{(a+2)}$

6) -5a² + 13a + 6 = 0

D = 13² - 4· (-5) · 6 = 169 + 120 = 289

$\sqrt{D} =17$

$a _{1} = \frac{-13+17}{-10}= \frac{4}{-10} =- \frac{2}{5}$

$a _{2} = \frac{-13-17}{-10} = \frac{-30}{-10} =3$

-5a² + 13a + 6 = $-5(a+ \frac{2}{5} )(a-3)$

5a² - 8a - 4 = 0
D = (-8)² - 4· 5 · (-4) = 64 + 80 = 144

$\sqrt{D} =12$

$a _{1} = \frac{8+12}{10} = \frac{20}{10} =2$

$a _{2} = \frac{8-12}{10} = \frac{-4}{10} =- \frac{2}{5}$

5a² - 8a - 4 = $5(a-2)(a+ \frac{2}{5})$

$\frac{-5a ^{2}+13a+6 }{5a ^{2}-8a-4 } = \frac{-5(a+ \frac{2}{5})(a-3) }{5(a-2)(a+ \frac{2}{5}) } =- \frac{(a-3)}{(a-2)}$