Given a = 2x & # 178; + 3ax-2x-1, B = - X & # 178; + AX-1, then the value of 3A + 6B does not contain x term, and the value of a is obtained 3A+6B =3(A+2B) =3[2x²+3ax-2x-1+2(-x²+ax-1)] =15ax-6x-9 The value of ∵ 3A + 6B does not contain the term X ∴15a-6=0 ∴a=2/5 That's right, but why does 15ax-6x-9 change to 15a-6 = 0!

Given a = 2x & # 178; + 3ax-2x-1, B = - X & # 178; + AX-1, then the value of 3A + 6B does not contain x term, and the value of a is obtained 3A+6B =3(A+2B) =3[2x²+3ax-2x-1+2(-x²+ax-1)] =15ax-6x-9 The value of ∵ 3A + 6B does not contain the term X ∴15a-6=0 ∴a=2/5 That's right, but why does 15ax-6x-9 change to 15a-6 = 0!

3A+6B
=3(A+2B)
=3[2x²+3ax-2x-1+2(-x²+ax-1)]
=15ax-6x-9 - here
=(15a-6)x-9
It has nothing to do with x value
The coefficient before x is 0
∴15a-6=0
∴a=2/5

Given a = 2x & # 178; + 3ax-2x-1, B = - X & # 178; + AX-1, and the value of 3A + 6B does not contain x term, find the value of A

3A+6B
=6x²+9ax-6x-3-6x²+6ax-6
=(15a-6)x-9
Without x, the coefficient is 0
15a-6=0
a=3/5

Solve the equation: three quarters x (x-one-third) = 0

Three quarters x (x-one-third) = 0
So x-one-third = 0
X = one third