(1) 1/(x-4)=4/(x^2-16);(2) 3/(x-1)-(x+2)/[x(x-1)]=0;(3) (2-x)/(x-3)+1/(3-x)=1.
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(1) 1/(x-4)=4/(x^2-16);(2) 3/(x-1)-(x+2)/[x(x-1)]=0;(3) (2-x)/(x-3)+1/(3-x)=1.
(1) 1/(x-4)=4/(x^2-16);
(2) 3/(x-1)-(x+2)/[x(x-1)]=0;
(3) (2-x)/(x-3)+1/(3-x)=1.
(1) 1/(x-4)=4/(x^2-16);(2) 3/(x-1)-(x+2)/[x(x-1)]=0;(3) (2-x)/(x-3)+1/(3-x)=1.
x2-16=4×(x-4)
x2=4×x
最后x不等于=4,否则分母为0
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一个“整式的乘法”的问题请先阅读下列解题过程,再仿做下面的问题.已知X*X + X - 1=0,求X*X*X + 2*X*X + 3的值.解X*X*X + 2X*X +3=X*X*X +X*X -X +X*X +X +3=X{X*X +X -1} +X*X +X -1 +4=0+0+4=4+ x + X*X + X*X*X=0.+ X*X + X*X*X
1/(1-x)+1/(1+x)+2/(1+x*x)+4/(1+x*x*x*x)+8/(1-x*x*x*x*x*x*x*x)怎么做
(x-3/x-2)-(x-2/x-1)=(x-5/x4) -(x-4/x-3)
(x/x-2)+(2/x*2+2x)-(x+6/x*2-4)化简,=x^2/x(x-2)+(2x-4)/x(x^2-4)-(x^2+6x)/x(x^2-4)=x^2/x(x-2)-(x+2)/x(x-2)=(x-2)(x+1)/x(x-2)=(x+1)/x
x(x+1)+(x+2)+(x+3)+(x+4)+(x+5) = 27求x
X+X+X+(X-1)+(X-2)+(X-3)+(X-4)=10(X-3)+(X-4)
x+2/x+1-x+4/x+3=x+6/x+5-x+8/x+7 x=?
解方程(x/x(x+2))+(x/(x+2)(x+4))+.+x/(x+8)(x+10)=1
x^4+x^3+x^2+x+1=0,x^2006+x^2005+x^2004+x^2003+x^2002
写过程x(x-1)+2x(x+1)-3x(2x-5)-x(3x-2)+2x(2-x)=-5x(x-2)-4
1.用公式计算:COSX=1-X*X/2!X*X*X*X/4!-X*X*X*X*X*X/6!.直到最后一项的绝对
(X+4)(x-2)(x-4),其中x=-1
(x^2+x-1)(x^2+x+2)=4
x+2/x+1-x+3/x+2-x+4/x+3+x+5/x+4
(x-2)(x+1)(x+4)(x+7)=19
(x+1)(x+2)(x+3)(x+4)=24
(x-1)(x-2)(x-3)(x-4)=48