Ответы и объяснения

2013-09-08T04:54:03+04:00
A) 2,5x^2 + 4x = 0
D = 16 - 4*2,5*0 = 16 - 0 = 16
 \sqrt{D} = 4
 x_{1}= \frac{-4+4}{5}=0
 x_{2}= \frac{-4-4}{5}=-1,6
б) 6y^2 - 0,24 = 0
D = 0 + 4*6*0,24 = 5,76
 \sqrt{D} =2,4
 y_{1}= \frac{0+2,4}{12}=0,2
 y_{2}= \frac{0-2,4}{12}=-0,2
в) 0,2t^2 - t - 4,8 = 0
D = 1 + 4*0.2*4.8 = 1 + 3,84 = 4,84
 \sqrt{D} = 2,2
 t_{1}= \frac{1+2,2}{0,4}=8
 t_{2}= \frac{1-2,2}{0,4}=-3
г) (10/3)*u^2 + 3u - 3 = 0
D = 9 + 4*(10/3)*3 = 9 + 120/3 = 9 + 40 = 49
 \sqrt{D} = 7
 u_{1}=  \frac{-3+7}{ \frac{20}{3} } =  \frac{4*3}{20}= \frac{3}{5} = 0,6
 u_{2}= \frac{-3-7}{ \frac{20}{3} }= \frac{-10*3}{20}= -\frac{3}{2}=-1,5