rlz87tm45t的部落格

標題:

有幾題數唔式

發問:

2y-5/3y-6 - 1/2 = 5y-4/2y-4 1-3x/1+3x + 3x+1/3x-1 = 12/(1+3x)(1-3x) 2x+5/5x+3 - 2x+1/5x+2 =0 4x+3/9 = 8x+37/18 - 5x - 7x/12 -29 求未知數, 謝

最佳解答:

• S.1 Maths Question-s
• 問1題數學題~_0
• 求救呀 大埔有乜巴士去西九龍中心

• S.1 Maths Question-s
• 問1題數學題~_0
• 求救呀 大埔有乜巴士去西九龍中心

 

此文章來自奇摩知識+如有不便請留言告知

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

其他解答:

so clear