Question

a) 3(2-x)+1=4-2x

b) 2(x+4)=3-x

c) 7-3x=x-5

d) 5x-(x-1)=7

Answer 1

a) Ta có: \(3\left(2-x\right)+1=4-2x\)

\(\Leftrightarrow6-3x+1-4+2x=0\)

\(\Leftrightarrow-x+3=0\)

\(\Leftrightarrow-x=-3\)

hay x=3

Vậy: S={3}

b) Ta có: \(2\left(x+4\right)=3-x\)

\(\Leftrightarrow2x+8-3+x=0\)

\(\Leftrightarrow3x+5=0\)

\(\Leftrightarrow3x=-5\)

hay \(x=-\dfrac{5}{3}\)

Vậy: \(S=\left\{-\dfrac{5}{3}\right\}\)

c) Ta có: \(7-3x=x-5\)

\(\Leftrightarrow7-3x-x+5=0\)

\(\Leftrightarrow-4x+12=0\)

\(\Leftrightarrow-4x=-12\)

hay x=3

Vậy: S={3}

d) Ta có: \(5x-\left(x-1\right)=7\)

\(\Leftrightarrow5x-x+1=7\)

\(\Leftrightarrow4x=6\)

hay \(x=\dfrac{3}{2}\)

Vậy: \(S=\left\{\dfrac{3}{2}\right\}\)