解方程组:(√2-1)x+y=3,x+(1-√2)y=2
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解方程组:(√2-1)x+y=3,x+(1-√2)y=2
rt,解方程组,我解蒙了.不得不来求助.
rt,解方程组,我解蒙了.不得不来求助.
答:
(√2-1)x+y=3
两边同乘以(√2+1)有:
(√2+1)(√2-1)x+(√2+1)y=3(√2+1)
x+(1+√2)y=3√2+3
x+(1-√2)y=2
上两式相减得:(1+√2-1+√2)y=3√2+1
2√2y=3√2+1
y=3/2+√2/4
代入x=2+(√2-1)y得:
x=2+(√2-1)(3/2+√2/4)
=2+3√2/2+1/2-3/2-√2/4
=1+5√2/4
所以:
x=1+5√2/4,y=3/2+√2/4
(√2-1)x+y=3
两边同乘以(√2+1)有:
(√2+1)(√2-1)x+(√2+1)y=3(√2+1)
x+(1+√2)y=3√2+3
x+(1-√2)y=2
上两式相减得:(1+√2-1+√2)y=3√2+1
2√2y=3√2+1
y=3/2+√2/4
代入x=2+(√2-1)y得:
x=2+(√2-1)(3/2+√2/4)
=2+3√2/2+1/2-3/2-√2/4
=1+5√2/4
所以:
x=1+5√2/4,y=3/2+√2/4
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