有幾題數唔式

我估你打數本有少少問題少左好多()
你係唔係想問e個?
(y-5)/(3y-6)-1/2=(5y-4)/(2y-4)
首先左邊要通分數
2(y-5)/(6(y-1)) - (3(y-1))/6(y-1)=(5y-4)/(2y-4)
(2y-10 - 3y + 1) /(6(y-1))=(5y-4)/(2y-4)
(-y-9) / (6(y-1)) = (5y-4) / (2y-4)
(-y-9) x (2y-4 ) = (5y-4) x 6(y-1)
(-y-9) x (y-2) = (5y-4) x 3(y-1)
-y^2 - 7y + 18 = 15y^2 -27y + 12
16y^2 - 20y -6 = 0
8y^2 - 10y - 3 = 0
(2y -3 )(4y +1) = 0
y= 3/2 or y=-1/4


(1-3x)/(1+3x) + (3x+1)/(3x-1) = 12/((1+3x)(1-3x))
同樣通分母
((1-3x)(3x-1))/((1+3x)(3x-1)) + ((3x+1)(1+3x))/((1+3x)(3x-1)) = 12/((1+3x)(1-3x))
((3x+1)^2 - (3x-1)^2) / ((1+ 3x)(3x-1) = 12/((1+3x)(1-3x))
由於a^2 - b^2 = (a+b)(a-b) 同時消去分母
(3x+1+3x-1)(3x+1-3x+1) = -12
(6x)(2) = -12
x = -1

(2x+5)/(5x+3) - (2x+1)/(5x+2) =0
左右兩邊同時乖(5x + 3)(5x +2)
(2x+5)(5x+2) - (2x+1)(5x +3) = 0
10x^2 +4x + 25x + 10 - ( 10x^2 + 6x + 5x +3) = 0
18x = 7
x = 7/18

4x+3/9 = 8x+37/18 - 5x - 7x/12 -29←完全睇唔明sorry

so clear