Question

tìm x:     a) \(\left|7x+1\right|-\left|5x+6\right|=0\)

              b)\(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)

Answer 1

a) |7x + 1| - |5x + 6| = 0

Vì |7x - 1| \(\ge\)0\(\forall\)x

     |5x + 6|\(\ge\)0 \(\forall\)x

Do đó : |7x - 1| + |5x + 6| \(\ge\)0\(\forall\)x

Và |7x - 1| + |5x + 6| = 0

<=> 7x - 1 = 0          <=> x = 1/7

và   5x + 6 = 0             và x = -6/5                     (vô lí)

=> x \(\in\varnothing\)

Answer 2

b) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)

<=> \(\frac{3}{2}x+\frac{1}{2}=4x-1\)hoặc\(\frac{3}{2}x+\frac{1}{2}=-4x+1\)

<=>\(\frac{3}{2}x-4x=\frac{-1}{2}-1\)hoặc \(\frac{3}{2}x+4x=\frac{-1}{2}+1\)

<=> \(\frac{-5}{2}x=\frac{-3}{2}\)hoặc \(\frac{11}{2}x=\frac{1}{2}\)

<=>\(x=\frac{-3}{2}:\frac{-5}{2}\)hoặc \(x=\frac{1}{2}:\frac{11}{2}\)

<=> \(x=\frac{3}{5}\)hoặc \(x=\frac{1}{11}\)