Question

помогите решить пожалуйста​

Answer 1

2x²+8=0

2x²=-8

x²=-4

x∈∅

2x²-3x=0

x(2x-3)=0

x=0      2x-3=0

           x=3/2

5x²-10=0

x²=2

x=±√2

x²=0

x=0

3(x+1)²-27=0

3(x+1)²=27

(x+1)²=9

x+1=3   x+1=-3

x=2      x=-4

3x²+1=0

3x²=-1

x²=-1/3

x∈∅

2x²+8=0

2x²=-8

x²=-4

x∈∅

7x²-14=0

7x²=14

x²=2

x=±√2

6x²=0

x=0

x(4(x-1)+0)=0

x(4x-4)=0

x=0    4x-4=0

         x=1

Answer 2

Объяснение:

$2x^{2} + 8 = 0\\\x^{2} = \frac{8}{2} \\\x^{2} = 4\\x^{2} =+-\sqrt{4} \\x_{1} = -2\\ x_{2} = 2$

$2x^{2} -3x=0\\x(2x-3)=0\\x_{1} =0\\2x-3=0\\x=\frac{3}{2}\\x_{2} = 1,5$

$5x^{2} -10=0\\x^{2} =\frac{10}{5} \\x^{2} =2\\x_{1,2} = +-\sqrt{2}$

$x^{2} =0\\x=0$

$3(x+1)^{2} -27=0\\(3x+3)^{2} -27=0\\(9x^{2}+18x+9 )-27=0\\9x^{2}+18x+9-27=0\\9x^{2} +18x-18=0 \\: (9)\\x^{2} +2x-2=0\\D=2^{2} -4*(-2)=12=\sqrt{4*3} = 2\sqrt{3} \\\\x_{1,2} = \frac{-2+-\2\sqrt{3 } }{2}$

$3x^{2} +1=0\\x^{2} =\frac{1}{3} \\x = +-\sqrt{\frac{1}{3} }$

$2x^{2} +8=0\\x^{2} =\frac{8}{2} \\x^{2} =4\\x=+-\sqrt{4} \\x_{1,2} = -2; 2$