b)\(\left(2y+3\right)\left(y+2\right)-\left(y-4\right)\left(2y-1\right)=18\)
⇒\(2y^2+4y+3y+6-2y^2+y+8y+4=18\)
⇒16y+10=18
⇒16y=28
⇒y=\(\dfrac{7}{4}\)
a)\(2y\left(y-1\right)-y\left(-4+2y\right)+4=0\)
⇒\(2y^2-2y+4y-2y+4=0\)
⇒2y+4=0
⇒2y=-4
⇒y=-2
a, 2y2-2y+4y-2y2+4=0
2y+4 =0
2y =-4
y =-2
vậy y=-2
b,2y2+4y+3y+6-( 2y2-y-8y+4)=18
2y2+7y+6-2y2+9y-4=18
16y+2 =18
16y =16
y =1
vậy y=1