Question

решите через дискриминант пожалуйста :(

1. x^2+3=3-x

2. 5x+2=2-2x^2

3. x^2 - 6x= 4x-25

4. $3x^{2}$+9=12x-$x^{2}$

5. x(x+2)=3

6. x(x-5)=-4

7. x(2x+1)=3x+4

8. $\frac{x}{2x+3} = \frac{1}{x}$

9. $\frac{x}{10-3x} = \frac{1}{x}$

10. 4 + $\frac{21}{x}$ = X

11. 6 + $\frac{7}{x}$ = X

12. $\frac{5}{x+3} + \frac{4}{x} =3$

Answer 1

Ответ:

1. x^2 + x = 0

D = 1 - 4 * 1 * 0 = 1 - 0 = 1

2. 2x^2 + 5x = 0

D = 25 - 4 * 2 * 0 = 25

3. x^2 - 10x + 25 = 0

D = 100 - 4 * 1 * 25 = 0

4. 4x^2 - 12x + 9 = 0

D = 144 - 4 * 4 * 9 = 0

5. x^2 + 2x - 3 = 0

D = 4 - 4 * 1 * ( - 3 ) = 16 = 4^2

6. x^2 - 5x + 4 = 0

D = 25 - 4 * 1 * 4 = 9 = 3^2

7. 2x^2 - 2x - 4 = 0

x^2 - x - 2 = 0

D = 1 - 4 * 1 * ( - 2 ) = 9 = 3^2

8. x^2 = 2x + 3

x^2 - 2x - 3 = 0

D = 4 - 4 * 1  * ( - 3 ) = 16 = 4^2

9. x^2 = 10 - 3x

x^2 + 3x - 10 = 0

D = 9 - 4 * 1 * ( - 10 ) = 49 = 7^2

10. 4x + 21 = x^2

x^2 - 4x - 21 = 0

D = 16 - 4 * 1 * ( - 21 ) = 100 = 10^2

11. 6x + 7 = x^2

x^2 - 6x - 7 = 0

D = 36 - 4 * 1 * ( - 7 ) = 64 = 8^2

12. $\frac{5x+ 4x + 12}{x ( x + 3 )}-3 = 0$

$\frac{5x+4x+12-3x^{2}-9x }{x(x+3)}=0$

$\frac{-3x^{2}+12 }{x(x+3)}=0$

$\frac{3x^{2}-12 }{x(x+3)}=0$

3x^2 - 12 = 0; + [ x ≠ 0; x ≠ -3]

D = 0 - 4 * 3 * ( - 12 ) = 0 + 144 = 144 = 12^2

Объяснение: