This is very difficult math question!
来源:百度知道 编辑:UC知道 时间:2024/06/07 03:58:39
若方程2/3-2(x+1/2005)=1.
求13+24(x+1/2005)的值。
求13+24(x+1/2005)的值。
设y=x+1/2005
2/3-2(x+1/2005)=1
2/3-2y=1
y=-1/6
13+24(x+1/2005)
=13+24y
=13-24*1/6
=13-4
=9
2/3-1=2x+2/2005
2x=2/3-1-2/2005
x=1/3-1/2-1/2005
13+24[(1/3-1/2-1/2005)+1/2005]
=13+24(-1/6)
=13-4
=9
方法一: 因为2/3-(-1/3)=1
所以2(x+1/2005)=-1/3即(x+1/2005)=-1/6(等式的基本性质)
所以原式=13+24*(-1/6)=13-4=9 (因果表示法)
方法2: 把(x+1/2005)看成一个整体设为a
则原式=2/3-2a=1 a=-1/6
则13+24(x+1/2005)=13+24a=13+(-4)=9 (局部观察法)
方法3:
设y=x+1/2005
2/3-2(x+1/2005)=1
2/3-2y=1
y=-1/6
13+24(x+1/2005)
=13+24y
=13-24*1/6
=13-4
=9 (列方程法)
因为2/3-(-1/3)=1
所以2(x+1/2005)=-1/3即(x+1/2005)=-1/6(等式的基本性质)
所以原式=13+24*(-1/6)=13-4=9
把(x+1/2005)看成一个整体设为a
则原式=2/3-2a=1 a=-1/6
则13+24(x+1/2005)=13+24a=13+(-4)=9
so easy
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