Question

СРОЧНО!!!
Знайдіть корені квадратного тричлена:

Answer 1

Ответ:

$1)\ -5;\ 2\\\\2)\ - \frac{ 1 }{ 2 };\ \frac{ 1 }{ 3 }$

Объяснение:

1)

$x^{2} + 3x - 10 =0\\\\ a=1 ,\ \ b=3 ,\ \ c=-10\\\\ D = b^2 - 4ac = 3^2 - 4\cdot1\cdot( - 10) = 9 + 40 = 49\\\\\sqrt{D} =\sqrt{49} = 7\\\\ x_1=\frac{-b-\sqrt{D}}{2a}=\frac{-3-7}{2\cdot1}=\frac{-10 }{2 }=-5\\\\ x_2=\frac{-b+\sqrt{D}}{2a}=\frac{-3+7}{2\cdot1}=\frac{4}{2}=2$

2)

$6x^{2} + x - 1 =0\\\\ a=6 ,\ \ b=1 ,\ \ c=-1\\\\ D = b^2 - 4ac = 1^2 - 4\cdot6\cdot( - 1) = 1 + 24 = 25\\\\\sqrt{D} =\sqrt{25} = 5\\\\ x_1=\frac{-b-\sqrt{D}}{2a}=\frac{-1-5}{2\cdot6}=\frac{-6 }{12 }=- \frac{ 1 }{ 2 } \\\\ x_2=\frac{-b+\sqrt{D}}{2a}=\frac{-1+5}{2\cdot6}=\frac{4}{12}= \frac{ 1 }{ 3 }$

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$(5x-\frac{10x}{x+1}):\frac{15x-15}{4x+4}=$

$\frac{5x(x+1)-10x}{x+1}:\frac{15(x-1)}{4(x+1)}=$

$\frac{5x^2+5x-10x}{x+1}\cdot\frac{4(x+1)}{15(x-1)}=$

$\frac{5x^2-5x}{1}\cdot\frac{4}{15(x-1)}=$

$\frac{5x(x-1)}{1}\cdot\frac{4}{15(x-1)}=$

$\frac{4x}{3}$