Bonjour,
[tex]A = 3(2x-1) - 5x=3x-1\\\\\\3(2x-1)-5x=3x-1\\\\6x-3-5x=3x-1\\\\x-3=3x-1\\\\x-x-3=3x-x-1\\\\-3=2x-1\\\\-3+1=2x-1+1\\\\-2=2x\\\\\frac{-2}{2}=\frac{2x}{2}\\\\-1=x[/tex]
~
[tex]B = 3x+2-4(x+1) = 3(x+2)-2(5x+1)\\\\\\3x+2-4(x+1) = 3(x+2)-2(5x+1)\\\\3x+2-4x-4=3x+6-10x-2\\\\-x-2=-7x+4\\\\-x+x-2=-7x+x+4\\\\-2=-6x+4\\\\-2-4 = -6x+4-4\\\\-6=-6x\\\\\frac{-6}{-6} =\frac{-6x}{-6}\\\\1 = x[/tex]
~
[tex]C = \frac{x-1}{5}+\frac{x+1}{3}=2 \\\\\\\frac{x-1}{5}+\frac{x+1}{3}=2 \\\\15(\frac{x-1}{5}+\frac{x+1}{3})=2*15\\\\\frac{15x-15}{5}+\frac{15x+15}{3}=30\\\\\frac{3(15x-15)}{5*3}+\frac{5(15x+15)}{3*5}=30\\\\\frac{45x-45}{15}+\frac{75x+75}{15}=30\\\\\frac{45x-45+75x+75}{15}=30\\\\\frac{120x+30}{15}=30\\\\\frac{15(8x+2)}{15}=30\\\\8x+2=30\\\\8x+2-2=30-2\\\\8x=28\\\\\frac{8x}{8} = \frac{28}{8}\\\\x = \frac{7}{2}[/tex]
~
[tex]D = \frac{3-2x}{6}+\frac{3+x}{8}=\frac{3-4x}{4}+x\\\\\\\frac{3-2x}{6}+\frac{3+x}{8}=\frac{3-4x}{4}+x \\\\24( \frac{3-2x}{6}+\frac{3+x}{8})=24(\frac{3-4x}{4}+x)\\\\24( \frac{4(3-2x)}{6*4}+\frac{3(3+x)}{8*3})=24(\frac{3-4x}{4}+x)\\\\24( \frac{4(3-2x)+3(3+x)}{24})=24(\frac{3-4x}{4}+x)\\\\24( \frac{12-8x+9+3x}{24})=24(\frac{3-4x}{4}+x)\\\\24( \frac{12-8x+9+3x}{24})=24(\frac{3-4x}{4}+x)\\\\12-8x+9+3x=24(\frac{3-4x}{4}+x)\\\\21-5x=24(\frac{3-4x}{4}+x)\\\\21-5x=24(\frac{3-4x}{4}+\frac{4x}{4} )[/tex]
[tex]21-5x=24(\frac{3-4x+4x}{4})\\\\21-5x=24(\frac{3}{4})\\\\21-5x=\frac{72}{4} \\\\21-5x=18\\\\21-21-5x=18-21\\\\-5x=-3\\\\\frac{-5x}{-5}=\frac{-3}{-5}\\\\x= \frac{3}{5}[/tex]
J’espère t’avoir aidé.
Si tu as des questions n’hésites pas à me les demander.
Bonne journée et bonne continuation.
