Question

Giải các phương trình sau

a,\(\dfrac{6x+5}{2}-\dfrac{10x+3}{4}=2x+\dfrac{2x+1}{2}\)

b,\(\left(x-4\right)\left(x+4\right)-2\left(3x-2\right)=\left(x-4\right)^2\)

Answer 1

a.

\(\dfrac{6x+5}{2}-\dfrac{10x+3}{4}=2x+\dfrac{2x+1}{2}\)

\(\Leftrightarrow\dfrac{12x+10-\left(10x+3\right)}{4}=\dfrac{8x+4x+2}{4}\)

\(\Leftrightarrow12x+10-10x-3=12x+2\)

\(\Leftrightarrow2x+7=12x+2\)

\(\Leftrightarrow-10x=-5\Leftrightarrow x=\dfrac{1}{2}\)

Vậy S = {1/2}

b.

\(\left(x-4\right)\left(x+4\right)-2\left(3x-2\right)=\left(x-4\right)^2\)

\(\Leftrightarrow x^2-16-6x+4=x^2-8x+16\)

\(\Leftrightarrow-6x-12=-8x+16\)

\(\Leftrightarrow2x=28\Leftrightarrow x=14\)

Vậy S = {14}