Question

100 баллов! Решите 3 задания:

Answer 1

Ответ:

1. 1) $x^{2}$+3x-40 = $x^{2}$+8x-5x-40 = x(x+8)-5(x+8) =(x+8)(x-5)

2) 6$x^{2}$+x-12 = 6$x^{2}$+9x-8x-12 = 3x(2x+3)-4(2x+3) = (2x+3)(3x-4)

2. 1) $x^{4}$-15$x^{2}$-16=0

Пусть $x^{2}$=t

$t^{2}$-15t-16=0

t=-1

t=16

Обратно

$x^{2}$=-1                                                   $x^{2}$=16

x∉R                                                    x=-4

                                                          x=4

2)$\frac{x^{2}+12}{x-3}$=$\frac{7x}{x-3}$                                       ОДЗ: x$\neq 3$

$x^{2}$+12=7x

$x^{2}$-7x+12=0

$x^{2}$-3x-4x+12=0

x(x-3)(x-4)=0

(x-3)(x-4)=0

x-3=0

x-4=0

x=3 x$\neq$3 по ОДЗ

x=4

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

Answer 2

$\displaystyle\bf\\1)\\\\x^{2}+3x-40=0\\\\D=3^{2} -4\cdot(-40)=9+160=169=13^{2} \\\\\\x_{1} =\frac{-3-13}{2} =\frac{-16}{2} =-8\\\\\\x_{2} =\frac{-3+13}{2} =\frac{10}{2} =5\\\\\\\boxed{x^{2} +3x-40=(x+8)(x-5)}\\\\\\2)\\\\6x^{2} +x-12=0\\\\D=1^{2} -4\cdot 6\cdot(-12)=1+288=289=17^{2} \\\\\\x_{1} =\frac{-1-17}{12} =\frac{-18}{12} =-1,5\\\\\\x_{2} =\frac{-1+17}{12} =\frac{16}{12} =1\frac{1}{3} \\\\\\6x^{2} +x-12=6\Big(x+1,5\Big)\Big(x-1\frac{1}{3} \Big)$

$\displaystyle\bf\\1)\\\\x^{4} -15x^{2} -16=0\\\\x^{2} =m \ \ , \ \ m\geq 0\\\\m^{2} -15m-16=0\\\\Teorema \ Vieta:\\\\m_{1}+ m_{2} =15\\\\m_{1} \cdot m_{2} =-16\\\\m_{1} =16 \ \ \ ; \ \ \ m_{2} =-1 < 0-neyd\\\\x^{2} =16\\\\x_{1,2} =\pm \ \sqrt{16} =\pm \ 4\\\\Otvet: \ -4 \ \ ; \ \ 4\\\\\\2) \\\\\frac{x^{2} +12}{x-3} =\frac{7x}{x-3} \\\\\\\left \{ {{x^{2} +12=7x} \atop {x-3\neq 0}} \right. \\\\\\\left \{ {{x^{2} -7x+12=0} \atop {x\neq 3}} \right.$

$\displaystyle\bf\\x^{2} -7x+12=0\\\\Teorema \ Vieta:\\\\x_{1} =3 -neyd\ \ ; \ \ x_{2} =4\\\\Otvet: 4\\\\\\3)\\\\5a^{2} +3a-2=0\\\\D=3^{2} -4\cdot 5\cdot(-2)=9+40=49=7^{2}\\\\\\a_{1} =\frac{-3-7}{10} =\frac{-10}{10} =-1\\\\\\a_{2} =\frac{-3+7}{10} =\frac{4}{10} =0,4\\\\\\5a^{2} +3a-2=5(a+1)(a-0,4)=(a+1)(5a-2)\\\\\\\frac{5a^{2} +3a-2}{a^{2} -1} =\frac{(a+1)(5a-2)}{(a+1)(a-1)} =\frac{5a-2}{a-1}$