Question

Pfv gente me ajudem nessa

Answer 1

Para resolver nós precisaremos tirar o MMC (minimo multiplo comum) em alguns casos e também fazer a distributiva...

A)

$-3(3x-42)=2(7x-52) \\ -9x + 126 = 14x -104 \\ -9x - 14x = -104-126 \\ -23x = -230 * (-1) \\ 23x = 230 \\ x = \frac{230}{23} \\ \boxed{x = 10}$

B)

$\frac{x}{2}+\frac{1-x}{5}=\frac{1}{2} \rightarrow MMC(2, \ 5) = 10 \downarrow \\ \frac{5x}{10}+\frac{2*(1-x)}{10} = \frac{5}{10} \rightarrow Como \ existe \ denominador \ nos \ dois \ lados \ n\acute{o}s \ podemos \ cancelar. \\ 5x + 2-2x=5 \\ 3x = 5 - 2 \\ 3x = 3 \\ x = \frac{3}{3} \\ \boxed{x=1}$

C)

$\frac{x+3}{2}+\frac{x+2}{3}=\frac{-1}{2} \rightarrow MMC(2, \ 3) = 6 \downarrow \\ \frac{3*(x+3)}{6}+\frac{2*(x+2)}{6} = \frac{3*(-1)}{6} \\ 3x + 9 + 2x + 4 = -3 \\ 5x = -3-9-4 \\ 5x = -16 \\ \boxed{x = \frac{-16}{5}}$

D)

$\frac{3+x}{2}-(1-x)=\frac{x-1}{4} \rightarrow MMC(2, \ 4 ) = 4 \downarrow \\ \frac{2*(3+x)}{4}-\frac{4*(1-x)}{4} = \frac{x-1}{4} \\ 6+2x-4+4x=x-1 \\ 6x-x = -1+4-6 \\ 5x = -3 \\ x = \boxed{-\frac{3}{5}}$

E)

$\frac{3x-1}{2}-\frac{4x+2}{4}-\frac{2x-4}{3}=\frac{x-5}{6} \rightarrow MMC(2,4,3,6) = 12 \downarrow \\ \frac{6*(3x-1)}{12}-\frac{3*(4x+2)}{12}-\frac{4*(2x-4)}{12}=\frac{2*(x-5)}{12} \\ 18x-6-12x-6-8x+16=2x-10 \\ 18x-12x-8x-2x = -10-16+6+6 \\-4x = -14 * (-1) \\ 4x = 14 \\ x = \frac{14}{4} \\ \boxed{x = \frac{7}{2}}$

F)

$\frac{2(x-1)}{3}+\frac{3(1+x)}{2}=\frac{1}{2}-\frac{x-1}{3} \rightarrow MMC(2, \ 3) = 6 \downarrow \\ \frac{2x-2}{3}+\frac{3+3x}{2}=\frac{1}{2}-\frac{x-1}{3} \\ \frac{2*(2x-2)}{6} + \frac{3*(3+3x)}{6}=\frac{3}{6}-\frac{2*(x-1)}{6} \\ 4x - 4 + 9 + 9x = 3 - 2x + 2 \\ 13x+2x = 3+2+4-9 \\ 15x = 0 \\ \boxed{x = 0}$