Найдите корни уравнения
а) ( х-1 )(х+4) = 0
б) ( х+4 )(2х-1) = 0
в) ( 6-х)(5-2х) = 0
г) ( 2+5у)(2у+7) = 0
д) -2х(х-3) = 0
е) у(3у+9) = 0

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Ответ:

luba20144444444444

15.10.2020 01:00

Объяснение:

а) ( х-1 )(х+4) = 0

$x^{2} +4x-x-4=0\\x^{2}+ 3x-4=0\\D=b^{2} -4ac=3^{2} -4*1*(-4)=9+16=25\\x_{1,2} =\frac{-b+(-)\sqrt{D} }{2a} \\\\x_{1} =\frac{-b+\sqrt{D} }{2a} =\frac{-3+\sqrt{25} }{2*1} =\frac{-3+5}{2} =\frac{2}{2} =1\\\\x_{2} =\frac{-b-\sqrt{D} }{2a} =\frac{-3-\sqrt{25} }{2*1} =\frac{-3-5}{2} =\frac{-8}{2} =-4$

ответ.(1;-4)

б) ( х+4 )(2х-1) = 0

$2x^{2} -x+8x-4=0\\2x^{2}+7x-4=0\\D=b^{2} -4ac=7^{2}-4*2*(-4)=49+32=81\\\\x_{1} =\frac{-b+\sqrt{D} }{2a} =\frac{-7+\sqrt{81} }{2*2} =\frac{-7+9}{4} =\frac{2}{4} =\frac{1}{2} \\\\x_{2} =\frac{-b-\sqrt{D} }{2a} =\frac{-7-\sqrt{81} }{2*2} =\frac{-7-9}{4} =\frac{-16}{4} =-4$

ответ.(1/2;-4)

в) ( 6-х)(5-2х) = 0

$30-12x-5x+2x^{2} \\2x^{2} -17x+30\\D=b^{2} -4ac=(-17)^{2} -4*2*30=289-250=49\\x_{1} =\frac{17+\sqrt{D} }{2*2} =\frac{17+\sqrt{49} }{4} =\frac{17+7}{4} =\frac{24}{4} =6\\\\x_{2} =\frac{17-\sqrt{D} }{2*2} =\frac{17-\sqrt{49} }{4} =\frac{17-7}{4} =\frac{10}{4} =\frac{5}{2} =2\frac{1}{2} , and 2,5$

ответ.(6;2,5)

г) ( 2+5у)(2у+7) = 0

$4y+14+10y^{2} +35y=0\\10y^{2} +39y+14=0\\D=b^{2} -4ac=39^{2} -4*10*14=1521-560=961\\y_{1} =\frac{-39+\sqrt{961} }{2*10} =\frac{-39+31}{20} =\frac{-8}{20} =- \frac{2}{5} \\y_{2} =\frac{-39-\sqrt{961} }{2*10} =\frac{-39-31}{20} =\frac{-70}{20} =-\frac{7}{2} =-3,5$

ответ. (-2/5;-3,5)

д) -2х(х-3) = 0

$-2x^{2} +6x=0\\D=b^{2} -4ac=36^{2} -4*(-2)*0=36\\x_{1} =\frac{-6+\sqrt{36} }{2*(-2)} =\frac{-6+6}{-4} =0\\\\x_{2} =\frac{-6-\sqrt{36} }{2*(-2)} =\frac{-6-6}{-4} =\frac{-12}{-4} =3$

ответ.(0;3)

у(3у+9) = 0

$3y^{2} +9y=0\\D=b^{2} -4ac=9^{2} -4*3*0=81\\\\y_{1} =\frac{-b+\sqrt{D} }{2a} =\frac{-9+\sqrt{81} }{2*3} =\frac{-9+9}{6} =0\\\\y_{2} =\frac{-b-\sqrt{D} }{2a} =\frac{-9-\sqrt{81} }{2*3} =\frac{-9-9}{6}=\frac{-18}{6} =-3$