Question

Giải phương trình sau:

\(\dfrac{2x-1}{5}-\dfrac{x-2}{3}=\dfrac{x+7}{15}\)

\(\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)

Answer 1

a, Ta có : \(\dfrac{2x-1}{5}-\dfrac{x-2}{3}=\dfrac{x+7}{15}\)

=> \(\dfrac{3\left(2x-1\right)-5\left(x-2\right)}{15}=\dfrac{x+7}{15}\)

=> \(\dfrac{6x-3-5x+10}{15}=\dfrac{x+7}{15}\)

=>\(\dfrac{x+7}{15}=\dfrac{x+7}{15}\)

Vậy phương trình thỏa mãn với mọi x

b, Ta có :\(\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)

=>\(\dfrac{5x+2-2\left(8x+1\right)}{6}=\dfrac{4x+2-25}{5}\)

=>\(\dfrac{5x+2-16x+2}{6}=\dfrac{4x-23}{5}\)

=>\(\dfrac{-11x+4}{6}=\dfrac{4x-23}{5}\)

=> 6(4x-23)= 5(-11x+4) => 24x-138=-55x+20 => 79x =158 =x=2

Vậy x=2