1

$(12x+1)(3x-1)-(6x+2)^{2}=10$

2. $frac{6x-1}{4}-frac{3x+1}{3}=frac{1}{4}$

3. (6-1)(x+1)=20

4. $17-2x+frac{x(3x+4)}{2}=54frac{1}{2}$

5. $frac{x(x-1)}{4}+frac{(x-3)^{2}}{2}=frac{(4-x)^{2}}{3}-frac{1}{3}$

Ответы:

1. (12х+1)(3х-1)-(6х+2)²=10

36 $x^{2}$ +3х-12х-1-36 $x^{2}$ -24х-4-10=0

-33х=15

х= $- frac{5}{11}$

2. $frac{6x-1}{4} - frac{3x+1}{3} = frac{1}{4}$ | *12

18х-3-12х-4-3=0

6х=10

х= $1 frac{2}{3}$

3. (6х-1)(х+1)=20

6х²-х+6х-1-20=0

6х²+5х-21=0

D=5²-4*6*(-21)=529

х₁= $frac{-5+23}{2*6}$ = 1,5

х₂ = $2 frac{1}{3}$

4. $17-2 x+ frac{x(3 x+4)}{2} = 54 frac{1}{2}$ | * 2

34-4х+х(3х+4)=109

34-4х+3х²+4х-109=0

3х²=75

х²=25

х=+5,-5

5. $frac{x( x-1)}{4} + frac{(x-3)^{2}}{2} = frac{(4-x)^2}{3} - frac{1}{3}$ | *12

3х^2-3x+6x^2-36[+54=64-4x^2-32-4

13x^2-7-6=0

D=(-7)^2-4*13*(-6)=361

X1= $frac{7+19}{2*13} = 1$

X2= $frac{7-19}{2*13} = - frac{6}{13}$