Предмет: Алгебра, автор: ПомогитеPLZ

Помогите пожалуйста номер 176

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Ответы

Автор ответа: JuliaKovalchook
0
1)  frac{ x^{2} +14x+24}{x-2} =0 \  left { {{x^{2} +14x+24=0} atop {x-2 neq 0}}  \ right.  \ left { {{x^{2} +14x+24=0} atop {x neq 2}}  \ right. \   x^{2} +14x+24=0 \ x_{1}+x_{1}=-14 \  x_{1} x_{2}=24    \ \  x_{1}=-12 \  x_{2}=-2 \  x neq 2

2)  frac{ x^{2} +12x-13}{x-1}=0 \  left { {{x^{2} +12x-13=0} atop {x-1 neq 0}} right.   \ left { {{x^{2} +12x-13=0} atop {x neq 1}} right.   \ x^{2} +12x-13=0 \  \ x_{1}+x_{1}=-12 \ x_{1} x_{2}=-13 \  \ x_{1}=-13 \ x_{2}=1 \ x neq 1 \  \ x=-13

3)  frac{ 3x^{2}-4x-20 }{x+4} =0 \  left { {{3x^{2}-4x-20=0} atop {x+4}  neq 0}} right.    \  left { {{3x^{2}-4x-20=0} atop {x}  neq -4}} right.  \  \ D=(-4)^{2}-4*(-20)*3=16+240=256 \ x_{1}= frac{-(-4)- sqrt{256} }{2*3} = frac{4-16}{6} = frac{-12}{6} =-2 \  x_{2}= frac{-(-4)+ sqrt{256} }{2*3} = frac{4+16}{6} = frac{20}{6} =3 frac{2}{6}=3 frac{1}{3}   \  \ x_{1}=-2 \ x_{2}=3 frac{1}{3} \ x neq  -4

4)  frac{6 x^{2} -5x-4}{5-x} =0 \   left { {{6 x^{2} -5x-4=0} atop {5-x neq 0}} right.  \  left { {{6 x^{2} -5x-4=0} atop {x neq 5}} right.  \ 6 x^{2} -5x-4=0 \ \  D=(-5)^{2}-4*(-4)*6=25+96=121 \ x_{1}= frac{-(-5)+ sqrt{121} }{2*6} = frac{5+11}{12} = frac{16}{12}= frac{4}{3}=1 frac{1}{3}   \ x_{2}= frac{-(-5)- sqrt{121} }{2*6} = frac{-6}{12} = - 0.5 \  \ x_{1}=1 frac{1}{3} \ x_{2}=-0,5 \ x neq 5
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