Question

解一元二次方程 <(5t-20)/6>²+ <(10t-40)/9>²=(7-t)²

在线等

Answer 1

先对方程化简方程，左边可以化为 [5（t-4）/6]^2+[10(t-4)/9]^2=(25/36)*(t-4)^2+
(100/81)*(t-4)^2=(625/324)*(t-4)^2=右边的 （7-t）^2 于是可以求结果为25*（t-4）=18*+-(7-t)
此处易得，t=226/43 或者t=-(26/7)

Answer 2

[(5t-20)/6]²+ [(10t-40)/9]²=(7-t)²
(5T-20)^2/36+(10T-40)^2/81=(7-T)^2
9(5T-20)^2+4(10T-40)^2=324(7-T)^2
9(25T^2-200T+400)+4(100T^2-800T+1600)=324(49-14T+T^2)
225T^2-1800T+3600+400T^2-3200T+6400=15876-4536T+324T^2
301T^2+464T-5876=0
应用公式你自己算吧

Answer 3

<(5t-20)/6>²+ <(10t-40)/9>²=(7-t)²
<15(t-4)/18>²+ <20(t-4)/18>²=(7-t)²
(225+400)/18²*(t-4)²=(7-t)²
(25/18)²(t-4)]²=(7-t)²
25/18(t-4)=7-t 或25/18(t-4)=t-7
25/18(t-4)=7-t ==> 25t-100=126-18t ==> 43t=226 ==>t=226/43
25/18(t-4)=t-7==>25t-100=18t-126 ==>7t=-26 ==>t=-26/7