已知(x方+2x分之x-2-x方+4x+4分之x-1)除以x+2分之x-4
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将x^2+y^2+5/4=2x+y化为将x^2-2x+1+y^2-y+1/4=0即(x-1)^2+(y-1/2)^2=0上面的等式成立,则有x-1=0且y-1/2=0即x=1,y=1/2所以xy/(x
(x+1)/x²-2x²/(x+1)=1设(x+1)/x²=t,则方程变为:t-2/t=1,即:t²-t-2=0,即(t-2)(t+1)=0∴t1=2t2=-1
已知x的二次方+3x+1=0,x+3+1/x=0x+1/x=-3(x+1/x)^2=(-3)^2x^2+2+1/x^2=9x^2+1/x^2=7x的平分+x的平方分之一+2x+x分之2=7+2*(-3
x/(x²-x+1)=33x²-3x+3=x3x²+3=4x平方9x^4+18x²+9=16x²两边减去9x²9(x^4+x²+1
【俊狼猎英】团度为您解答~原式=[(x^2-y^2)*2(x-y)^2]/[xy(xy-y^2)]先约掉一个(x-y),分子分母同时处以y^3得到2[(y/x)^2-1]*[(y/x)-1]/[(y/
已知x,y满足方程组x方+2y=5,(1)x-y=-1(2)(1)+(2)*2x²+2x-3=0解得x1=1x2=-3代入(2)y1=2y2=-21.x1=1y1=2时x+y分之x的三次方-
(X-2)分之【(X-2)三次方-(X-1)方+1】=【(X-2)三次方-(X-1+1)(x-1-1)】/(X-2)=【(X-2)三次方-X(x-2)】/(X-2)=(x-2)^2-1=x^2-4x+
我操,你这样鬼能看懂啊?分子是啥分母是啥,那些加、除都在什么位置上啊?
X+1/X=2(X+1/X)²=X²+2+(1/X)²=4X²+(1/X)²=2(X²+(1/X)²)²=X^4+2+(
题目表达不清楚
2-x分之x-x方+4x+4分之4-x方=2-x/x-x²+4x+4/4-x²=2-1-x²+4x+1-x²=2+4x-2x²
2x/(x^2-1)-1/(x+1)=2x/(x^2-1)-(x-1)/(x^2-1)=(2x-x+1)/(x^2-1)=(x+1)/(x^2-1)=1/(x-1)=1/(根号3+1-1)=1/(根号
x分之4再问:有过程吗他那个是x-1分之1-x+1分之1的再答:[1/(x-1)-1/(x+1)]/[x/(2x方-2)]=[2/(x方-1)]*[2(x方-1)/x]=(2*2)/x=x/4
把每个式子因式分解x的方+5x+4=(x+1)(x+4)x方-5x+6=(x-2)(x-3)………………
f(X)等于1+2X二次方分之X二次方f(a)=a²/(1+2a²)
(x-1)/(x^2+3x+2)+6/(2+x-x^2)-(10-x)/(4-x^2)=(x-1)/(x+1)(x+2)-6/(x+1)(x-2)-(x-10)/(x+2)(x-2)=[(x-1)(x
x/(x²+2x+1),(x-1)/(x²+x),1/(x²-1)式1:x/(x+1)²,式2:(x-1)/[x(x+1)],式3:1/[(x+1)(x-1)]
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