已知x x x 1=1 3,求分式的值.
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1)由1/x-1/y=5,得(y-x)/xy=5,所以y-x=5xy,即x-y=-5xy(3x+5xy-3y)/(x-3xy-y)=[3(x-y)+5xy]/[(x-y)-3xy]=(-15xy+5x
∵a-1/a=6∴﹙a-1/a﹚²=6²a²+1/a²=38a²+1/a²+1=38+1=39.
x=0:10;a=[2,1,1];b=[1,0,1,1,9];y=polyval(a,x)./polyval(b,x)
x+1/x=3(x+1/x)^2=3^2=9x^2+1/x^2+2=9x^2+1/x^2=7
由已知1,分母为0,b=6,由已知2,a=-5.则6a+9b=6*6+9*(-5)=-9
y²-3y+1=0两边除以y得y-3+1/y=0y+1/y=3两边平方得y²+2+1/y²=9∴y²+1/y²=7y²+1+1/y²
2x=-yy=-2xxy/(x^2-y^2)=x*(-2x)/[x^2-(-2x)^2]=-2x^2/[x^2+4x^2]=-2x^2/5x^2=-2/5
x+x分之一=3两边平方:x²+1/x²+2=9x²+1/x²=9-2=7
两边乘x(x+1)3(x+1)+x(kx+3)=2x(x+1)有增根则分母等于0x(x+1)=0x=0,x=-1x=0代入3(x+1)+x(kx+3)=2x(x+1)3=0,不成立x=-1代入3(x+
由1/x-1/y=2001得:y-x/xy=2001所以:xy/y-x=1/2001xy/x-y=-1/2001x-xy-y/(x-y)=x-y-xy/(x-y)=1-xy/(x-y)原式=1+1/2
1/a+1/b=5/(a+b)(a+b)²=5aba²+b²=3ab(a²+b²)/ab=3a/b+b/a=3.
依据题意:当x=4时,分母2x+b=0∴8+b=0b=-8当x=-3时,分子x-a=0∴-3-a=0a=-3∴(a+b)/(a-2b)=(-3-8)/(-3+16)=-11/13【很高兴为你解决以上问
1/(a+1)-(a+3)/(a^2-1)*(a^2-2a+1)/(a^2+4a+3)=1/(a+1)-(a+3)/(a+1)(a-1)*(a-1)^2/(a+3)(a+1)=1/(a+1)-(a-1
X=-2时,分式无意义,所以a-2b+3×(-2)=0,即a-2b=6x=1时,分式的值为0,所以1+a+b=0,即a+b=-1a、b的两个方程联立解得:a=4/3,b=-7/3
2的m次方=2的(-4)次方∴m=-4(1/3)的n次方=273的(-n)次方=3³-n=3∴n=-3∴n^m=(-3)^(-4)=1/81
1.最简单的方法是用数代(填空题的话)ab互为倒数设a=2则b=1/22/3+1/3=12.移项x+7/x+6-x+6/x+5=x+10/x+9-x+9/x+8-1/X^2+11x+30=-1/X^2
y=4x带入3x-y=3x-4x=-xx+y=x+4x=5x(x+y)/(3x-y)=5x/(-x)=-5
原式=[(x-y)/x]/[(x²-2xy+y²)/x]=(x-y)/(x-y)²因为x-y=2011-2012=-1所以原式=-1/(-1)²=-1
x+1/x=3(x+1/x)^2=3^2=9x^2+1/x^2+2=9x^2+1/x^2=7
同学,你这道题是不是题干有问题啊?分子上这一项你确定是ab²?这样的话就会出现b的3次方,那就求不出分式的值了,所以下面把ab²改为ab了.2a=3b,得a=3b/2分子:9b