1) x^2+x/4-3-7x/20=0,3

2) 1/2x^2-3x-9+3/x^2-x-6=x/2x^2+7x+6

Question

Алгебра

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24 апреля, 14:33

1) x^2+x/4-3-7x/20=0,3

2) 1/2x^2-3x-9+3/x^2-x-6=x/2x^2+7x+6

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Маруся · Answer 1

1) 0,3x+8=2 0.3x=2-8 0.3x=-6 x=-6/0.3 x = - 20

2) 4-x=1+4x - x-4x=1-4 - 5x=-3 x=-3 / (-5) x = 0.6

3) 7-2 (x+3) = 9-6x - 2x+6x=9-7+6 4x=8 x=8/4 x = 2

4) 4 (x-0.5) - 2 (x+0.3) = - 2.6 4x-2x = - 2.6+2+0.6 2x = 0 x=0/2 x = 0

5) 0.4x-6 = - 12 0.4x = - 12+6 0.4x = - 6 x = - 6/0.4 x = - 15

6) x+6=5+4x x-4x=5-6 - 3x = - 1 x = - 1 / (-3) x = 1/3

7) 13-3 (x+1) = 4-5x - 3x + 5x = 4-13+3 2x = - 6 x = - 6/2 x = - 3

8) 0.2 (3x-5) - 0.3 (x-1) = - 0.7 0.6x - 0.3x = - 0.7+1-0.3 0.3x = 0 x = 0/0.3 x = 0

1) x^2+x/4-3-7x/20=0,32) 1/2x^2-3x-9+3/x^2-x-6=x/2x^2+7x+6

1) x^2+x/4-3-7x/20=0,3

2) 1/2x^2-3x-9+3/x^2-x-6=x/2x^2+7x+6