Question

Giải gấp giúp mình với ạ

Answer 1

a: \(\dfrac{x}{3}-\dfrac{2x+1}{6}=\dfrac{x}{6}-x\)

=>\(\dfrac{2x-2x-1}{6}=\dfrac{x}{6}-\dfrac{6x}{6}\)

=>-5x=-1

=>\(x=\dfrac{1}{5}\)

b: \(\dfrac{x+2}{5}-\dfrac{1}{2}x=\dfrac{1-2x}{4}+\dfrac{1}{4}\)

=>\(\dfrac{x+2}{5}-\dfrac{x}{2}=\dfrac{-2x+1+1}{4}=\dfrac{-2x+2}{4}=\dfrac{-x+1}{2}\)

=>\(\dfrac{x+2}{5}=\dfrac{-x+1}{2}+\dfrac{x}{2}=\dfrac{-x+1+x}{2}=\dfrac{1}{2}\)

=>2x+4=5

=>2x=1

=>\(x=\dfrac{1}{2}\)

c: \(\dfrac{x-1}{2}+\dfrac{x-1}{4}=1-\dfrac{2\left(x-1\right)}{3}\)

=>\(\dfrac{2\left(x-1\right)+\left(x-1\right)}{4}=\dfrac{3-2\left(x-1\right)}{3}\)

=>\(\dfrac{3x-3}{4}=\dfrac{3-2x+2}{3}=\dfrac{-2x+5}{3}\)

=>3(3x-3)=4(-2x+5)

=>9x-9=-8x+20

=>17x=29

=>\(x=\dfrac{29}{17}\)

d: \(\dfrac{x+1}{3}+\dfrac{3\left(2x+1\right)}{4}=\dfrac{2x+3\left(x+1\right)}{6}+\dfrac{12x+7}{12}\)

=>\(\dfrac{4\left(x+1\right)}{12}+\dfrac{9\left(2x+1\right)}{12}=\dfrac{2\left(2x+3x+3\right)}{12}+\dfrac{12x+7}{12}\)

=>4x+4+18x+9=2(5x+3)+12x+7

=>22x+13=10x+6+12x+7

=>22x=22x

=>0x=0(luôn đúng)