Question

решите плиз,мне надо срочно​

Answer 1

Объяснение:

a)

$x^2+x-c=(x-3)*(x-x_2)$

$\left \{ {{c=x_1*x_2} \atop {-(x_1+x_2)=1\ |*(-1)}} \right. \ \ \ \ \left \{ {{c=3*x_2} \atop {3+x_2=-1}} \right. \ \ \ \ \left \{ {{c=3*(-4)=-12} \atop {x_2=-4}} \right. .\ \ \ \ \Rightarrow\\x^2+x-12=(x-3)(x+4).$

b)

$2x^2-11x+c=2*(x-2)*(x-x_2)$

$\left \{ {{\frac{c}{2} =x_1*x_2\ |*2} \atop {-(x_1+x_2)=-\frac{11}{2} *|*(-1)}} \right. \ \ \ \ \left \{ {{c=2*3*x_2} \atop {3+x_2=5,5}} \right. \ \ \ \ \left \{ {{c=6*2,5=15} \atop {x_2=2,5}} \right. . \ \ \ \ \ \Rightarrow\\2x^2-11x+15=(x-3)(x-2,5).$

c)$7x^2+6x-c=7*(x+\frac{1}{7} )(x-x_2)\\\left \{ {{c=7*x_1*x_2 } \atop {-7*(x_1+x_2)=6\ |*(-1)}} \right. \ \ \ \ \left \{ {{c=7*(-\frac{1}{7})*x_2 } \atop {7*(-\frac{1}{7})+7*x_2=-6 }} \right. \ \ \ \ \left \{ {{c=-1*x_2} \atop {-1+7*x_2=-6}} \right. \ \ \ \ \left \{ {{c=-x_2} \atop {7x_2=-5\ |:7}} \right. \ \ \ \ \left \{ {{c=\frac{5}{7} } \atop {x_2=-\frac{5}{7} }} \right. \ \ \ \ \Rightarrow\\7x^2+6x+\frac{5}{7} =7*(x+\frac{1}{7})(x+\frac{5}{7} ).$d)

$42x^2+37x+c=42*(x+\frac{1}{6})(x-x_2) \\\left \{ {{c=42*x_1*x_2} \atop {-42*(x_1+x_2)=37\ |*(-1)}} \right.\ \ \ \ \left \{ {{c=42*(-\frac{1}{6})*x_2 } \atop {42*(-\frac{1}{6})+42*x_2=-37 }} \right. \ \ \ \ \left \{ {{c=-7*x_2} \atop {-7+42x_2=-37}} \right. \ \ \ \\\ \left \{ {{c=-7*x_2} \atop {42x_2=-30\ |:42}} \right. \ \ \ \ \left \{ {{c=-7*(-\frac{5}{7})=5 } \atop {x_2=-\frac{5}{7} }} \right. .\ \ \ \ \ \Rightarrow\\42x^2+37x+5=42*(x+\frac{1}{6})(x+\frac{5}{7}).$