Question

用因式分解法解方程，4(y-3)^2-25(y^2-4y+4)=0

用因式分解法解方程，4(y-3)^2-25(y^2-4y+4)=0

我本来想用平方差算的到一半算不来了。。求解。

Answer 1

4(y-3)^2-25(y^2-4y+4) = 0
先整理成A^2-B^2的样子4(y-3)^2-25(y - 2) ^2 = 0
平方差(2y-6+5y-10)(2y-6-5y+10)=0
即(7y-16)(4-3y)=0

y1=16/7 y2=4/3

Answer 2

4(y-3)^2-25(y^2-4y+4)=4(y-3)^2-25(y-2)^2=[2(y-3)+5(y-2)][2(y-3)-5(y-2)]
=(7y-16)(4-3y)=0
y=16/7,4/3

Answer 3

4(y-3)^2-25(y^2-4y+4)=0
4(y-3)^2-5(y-2)^2=0
(2y-6)^2-(5y-10)^2=0
(2y-6+5y-10)(2y-6-5y+10)=0
(7y-16)(4-3y)=0

y1=16/7 y2=4/3

Answer 4

4(y-3)^2-25(y^2-4y+4)=0
4(y-3)^2=25(y^2-4y+4)
[2(y-3)]^2=[5(y-2)]^2
|2y-6|=|5y-10|
剩下自己讨论就可以了..如果不会就追问~

Answer 5

4(y-3)²-25(y²-4y+4)=0
4(y-3)²-25(y-2)²=0
[2(y-3)]²-[5(y-2)]²=0
[2(y-3)+5(y-2)][2(y-3)-5(y-2)]=0
(7y-16)(4-3y)=0
y=16/7或y=4/3