一个一元两次方程求解,4(x-1)*(x-1)=9(x-5)*(x-5)
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一个一元两次方程求解,4(x-1)*(x-1)=9(x-5)*(x-5)
一个一元两次方程求解,
4(x-1)*(x-1)=9(x-5)*(x-5)
一个一元两次方程求解,4(x-1)*(x-1)=9(x-5)*(x-5)
4(x-1)*(x-1)=9(x-5)*(x-5)
4(x-1)²=9(x-5)²
4(x²-2x+1)=9(x²-10x+25)
4x²-8x+4=9x²-90x+225
5x²-82x+221=0
(x-13)(x-17/5)=0
∴x=13,x=17/5
(x-5)*(x-5)/(x-1)*(x-1)=4/9
(x-5)/(x-1)= +/- 2/3
当(x-5)/(x-1)= 2/3
3(x-5) = 2(x-1)
x = 15 - 2
x = 13
当(x-5)/(x-1)= - 2/3
3(x-5) = - 2(x...
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(x-5)*(x-5)/(x-1)*(x-1)=4/9
(x-5)/(x-1)= +/- 2/3
当(x-5)/(x-1)= 2/3
3(x-5) = 2(x-1)
x = 15 - 2
x = 13
当(x-5)/(x-1)= - 2/3
3(x-5) = - 2(x-1)
5 x = 15 + 2
x = 17/5
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4(x-1)*(x-1)=9(x-5)*(x-5)
4(x2-2x+1)=9(x2-10x+25)
4x2-8x+4=9x2-90x+225
5x2-82x+221=0
Δ=822-4*5*221=6724-4420=2304>0,方程有两个实根。
X1,2= =
[2(x-1)]²=[3(x-5)]²
2(x-1)=3(x-5)或2(x-1)=-3(x-5)
x=13或x=17/5