已知x²+y²=1,求证:x/(1+y)-y/(1+x)=2(x-y)/1+x+y注意运用有关比例的知识(比如和比性质之类的)
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/01 07:40:59
xSJ@A$FߑAfXl uvRR.Ib ;;$j6_)e&n'N'eFЌG m+Zmɹxhy6y;ח'[^g .ǧhJX5AٌF32ȡB%е!߹00,0rc]jD%ԽJQ}=肂aŚ2@:cuQ#T4E F< +̢HHQHaYHDՃaT2>
已知x²+y²=1,求证:x/(1+y)-y/(1+x)=2(x-y)/1+x+y注意运用有关比例的知识(比如和比性质之类的)
已知x²+y²=1,求证:x/(1+y)-y/(1+x)=2(x-y)/1+x+y
注意运用有关比例的知识(比如和比性质之类的)
已知x²+y²=1,求证:x/(1+y)-y/(1+x)=2(x-y)/1+x+y注意运用有关比例的知识(比如和比性质之类的)
∵x^2+y^2=1, ∴x^2=1-y^2=(1+y)(1-y), ∴x/(1+y)=(1-y)/x,
∴由等比定理,有:x/(1+y)=(1+x-y)/(1+x+y).······①
∵x^2+y^2=1, ∴y^2=1-x^2=(1+x)(1-x), ∴y/(1+x)=(1-x)/y,
∴由等比定理,有:y/(1+x)=(1-x+y)/(1+x+y).······②
①-②,得:
x/(1+y)-y/(1+x)
=[(1+x-y)-(1-x+y)]/(1+x+y)
=2(x-y)/(1+x+y).
∴ x/(1+y)-y/(1+x)=2(x-y)/(1+x+y).
∵(X+Y+1)²=X²+Y²+1+2X+2Y+2XY=2+2X+2Y+2XY=2(1+X+Y+XY)
∴(X+Y+1)/(1+X+Y+XY)=2/(X+Y+1)
∴(X-Y)(X+Y+1)/(1+X+Y+XY)=2(X-Y)/(X+Y+1)
∴(X+X²-Y-Y²)/(1+X)(1+Y)=2(X-Y)/(X+Y+1)
∴[X(1+X)-Y(1-Y)]/(1+X)(1+Y)=2(X-Y)/(X+Y+1)
∴X/(1+Y)-Y/(1-X)=2(X-Y)/(X+Y+1)