恆等式
求下列恆等式中常數A及B的值.
1.2x^2+Ax -3=2(x+1)^2+B
公式的代入法
1.1/y = 1/x + c [x]
2.x = 1-y/1+y [y]
3.y =c + a-x/x
數學問題...
2007-11-27 6:34 am
回答 (2)
2007-11-27 6:50 am
✔ 最佳答案
1)RHS
=2(x^2+2x+1)+B
=2x^2+4x+(2+B)
LHS
=2x^2+Ax -3
∴A=4 ,B=-5
1.1/y = 1/x + c [x]
x=y+cxy
x-cxy=y
x(1-cy)=y
x=y/(1-cy)
2.x = 1-y/1+y [y]
x+xy=1-y
xy+y=1-x
y(x+1)=1-x
y=(1-x)/(1+x)
3.y =c + a-x/x
xy=cx+a-x
xy-cx+x=a
x(y-c+1)=a
x=a/(1+y-c)
2007-11-27 7:08 am
1.
2x^2+Ax -3=2(x+1)^2+B
右方
=2(x^2+2x+1)+B
=2x^2+4x+(2+B)
所以 2x^2+Ax-3 = 2x^2+4x+(2+B)
A=4
2+B=-3
B=-5
∴A=4 ,B=-5
轉主項?
1.1/y = 1/x + c [x]
x=y+cxy
x-cxy=y
x(1-cy)=y
x=y/(1-cy)
2.x = 1-y/1+y [y]
x+xy=1-y
xy+y=1-x
y(x+1)=1-x
y=(1-x)/(1+x)
3.y =c + a-x/x
xy=cx+a-x
xy-cx+x=a
x(y-c+1)=a
x=a/(1+y-c)
2x^2+Ax -3=2(x+1)^2+B
右方
=2(x^2+2x+1)+B
=2x^2+4x+(2+B)
所以 2x^2+Ax-3 = 2x^2+4x+(2+B)
A=4
2+B=-3
B=-5
∴A=4 ,B=-5
轉主項?
1.1/y = 1/x + c [x]
x=y+cxy
x-cxy=y
x(1-cy)=y
x=y/(1-cy)
2.x = 1-y/1+y [y]
x+xy=1-y
xy+y=1-x
y(x+1)=1-x
y=(1-x)/(1+x)
3.y =c + a-x/x
xy=cx+a-x
xy-cx+x=a
x(y-c+1)=a
x=a/(1+y-c)
收錄日期: 2021-04-13 14:35:02
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https://hk.answers.yahoo.com/question/index?qid=20071126000051KK04501