Choose the appropriate method, then the square of the following equation (2x-1) + 3 (2x-1) = 0

Choose the appropriate method, then the square of the following equation (2x-1) + 3 (2x-1) = 0

(2x-1)(2x-1+3)=0
(2x-1)(2x+2)=0
x=1/2,x=-1

There is a suitable method to solve the binary linear equation x (X-2) + X-2 = 0 (x-4) & # 178; = (5-2x) & # 178; X (X-5) = 6 X & # 178; - 15x + 50 = 0

(x + 1) (X-2) = 0, x = - 1 or 2
X-4 = 5-2x or 4-x = 5-2x, x = 3 or 1
(X-6) (x + 1) = 0, x = 6 or - 1
(X-5) (X-10) = 0, x = 5 or 10

10／﹙1＋x＋x²﹚＝6－x－x²

X & # 178; + X + 1 + 10 / (X & # 178; + X + 1) - 7 = 0, let X & # 178; + X + 1 = y  the original equation be reduced to y + Y / 10-7 = 0y & # 178; - 7Y + 10 = 0 (Y-2) (Y-5) = 0  y = 2, y = 5  X & # 178; + X + 1 = 2 x & # 178; + X + 1 = 5  x = (- 1 ± √ 5) / 2 x = (- 1 ± √ 17) / 2