用换元法解方程(1)x+1分之x平方-5x+x(x-5)分之24(x+1)+14=0(2)x+1分之2(x平方+1)+x平方+1分之6(x+1)=7(3)x平方分之x4次方+2x+1+x分之x平方+1=2
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![用换元法解方程(1)x+1分之x平方-5x+x(x-5)分之24(x+1)+14=0(2)x+1分之2(x平方+1)+x平方+1分之6(x+1)=7(3)x平方分之x4次方+2x+1+x分之x平方+1=2](/uploads/image/z/180013-13-3.jpg?t=%E7%94%A8%E6%8D%A2%E5%85%83%E6%B3%95%E8%A7%A3%E6%96%B9%E7%A8%8B%281%29x%2B1%E5%88%86%E4%B9%8Bx%E5%B9%B3%E6%96%B9-5x%2Bx%28x-5%29%E5%88%86%E4%B9%8B24%28x%2B1%29%2B14%3D0%282%29x%2B1%E5%88%86%E4%B9%8B2%28x%E5%B9%B3%E6%96%B9%2B1%EF%BC%89%2Bx%E5%B9%B3%E6%96%B9%2B1%E5%88%86%E4%B9%8B6%EF%BC%88x%2B1%29%3D7%283%29x%E5%B9%B3%E6%96%B9%E5%88%86%E4%B9%8Bx4%E6%AC%A1%E6%96%B9%2B2x%2B1%2Bx%E5%88%86%E4%B9%8Bx%E5%B9%B3%E6%96%B9%2B1%3D2)
用换元法解方程(1)x+1分之x平方-5x+x(x-5)分之24(x+1)+14=0(2)x+1分之2(x平方+1)+x平方+1分之6(x+1)=7(3)x平方分之x4次方+2x+1+x分之x平方+1=2
用换元法解方程(1)x+1分之x平方-5x+x(x-5)分之24(x+1)+14=0(2)x+1分之2(x平方+1)+x平方+1分之6(x+1)=7
(3)x平方分之x4次方+2x+1+x分之x平方+1=2
用换元法解方程(1)x+1分之x平方-5x+x(x-5)分之24(x+1)+14=0(2)x+1分之2(x平方+1)+x平方+1分之6(x+1)=7(3)x平方分之x4次方+2x+1+x分之x平方+1=2
(1)令(x^2-5x)/(x+1)=t,则(x+1)/(x^2-5x)=1/t,所以原方程换元后为t+24/t+14=0,方程两边同时乘以t得t^2+14t+24=0解得t=-2或t=-12即(x^2-5x)/(x+1)=-2或(x^2-5x)/(x+1)=-12,
解(x^2-5x)/(x+1)=-2得: x=1或x=2
解(x^2-5x)/(x+1)=-12得: x=-3或x=-4
(2)令(x^2+1)/(x+1)=t,(x+1)/(x^2+1)=1/t,所以原方程换元后为2t+6/t=7,方程两边同时乘以t得2t^2+6=7t,解得t=2或t=3/2即(x^2+1)/(x+1)=2或(x^2+1)/(x+1)=3/2
解(x^2+1)/(x+1)=2得: x=1+√2或x=1-√2
解(x^2+1)/(x+1)=3/2得: x=(3+√17)/4或x=(3-√17)/4
(3)(x^4+2x^2+1)/x^2+(x^2+1)/x=(x^2+1)^2/x^2+(x^2+1)/x=[(x^2+1)/x]^2++(x^2+1)/x
令(x^2+1)/x=t,则原方程换元后为t^2+t=2,解得t=-2或t=1即(x^2+1)/x=1或(x^2+1)/x=-2
解(x^2+1)/x=1得:无解
解(x^2+1)/x=-2得:x=-1