1除以(x^2 x-2)对x求微分,范围是-1

由x^2-2x-3=0得：x=3或x=-1；[(x^2-3)/(x-1)-2]÷[1/(x-1)]=[(x^2-3)/(x-1)-2](x-1)=x^2-3-2(x-1)=x^2-2x-1当x=3时,

[x的平方-x分之x+1-x的平方-2x+1分之x]除以x分之1=[(x+1)/x(x-1)-x/(x-1)²]×x=(x+1)/(x-1)-x²/(x-1)²=(x&#

((x+2)/(x^2-2x)-(x-1)/(x^2-4x+4))/((x^2-16)/(x^2+4x))=((x+2)(x-2)-(x-1)x)/((x-2)^2*x)/((x-4)/x)=((x-

X除以（X-2）+（X-9）除以（X-7)=(X+1)除以（X-1）+（X-8)除以（X-6）(x-2+2/x-2)+(x-7-2/x-7)=(x-1+2/x-1)+(x-6-2/x-6)1+(2/x

x^2+2√5x+2=0x^2+2√5x+(√5)^2-3=0(x+√5)^2=3x+√5=±√3x1=-√5+√3,x2=-√5-√38x(2x+1)=1516x^2+8x=15x^2+1/2x=1

4x三次除以(-2x)平方-(2x平方-x)除以(1/2x)=4x^3/4x^2-(2x^2-x)/(1/2x)=x-2x(2x-1)/x=x-2(2x-1)=x-4x+2=-3x+2

x/(x²-x+1)=7x²-x+1=x/7x²+1=8x/7平方x^4+2x²+1=64x²/49两边减去x²x^4+x²+1=

原式={（x+1)/x}÷{x-(1+x²)/(2x)}={（x+1)/x}÷{（2x²-1-x²)/2x}={（x+1)/x}*{2x/(x²-1)}={(x

两边乘(x-1)(x+2)3mx-3m-4x+1=2nx+4n(3m-4-2n)x=3m-1+4n所以3m-4-2n=03m-1+4n=0相减6n+3=0n=-1/2m=(2n+4)/3=5/3所以m

解因为(2x-3)/[(x-1)(x+2)]=a/(x-1)+b/(x+2)=[a(x+2)+b(x-1)]/[(x-1)(x+2)]所以2x-3=a(x+2)+b(x-1)=（a+b）x+（2a-b

这是什么题啊,我考MPA的复习资料里也有这题.你要是今年考的话,现在才开始复习综合能力的数学会不会有些晚了啊.咱俩一样啊,我也刚刚开始复习.不过你要是明年考的话现在开始更要把握好时间了.我现在就后悔以

∵[(2x-1)/(x-3)*(x+5)]=[A/(x-3)]+[B/(x+5)]∴(2x-1)/[x-3)(x+5)]=[(A+B)x+(5A-3B)]/[(x-3)(x+5)]∴A+B=2,5A-

x/(x-1)+1=(2x-1)/x[x²+x(x-1)]/x(x-1)=(2x-1)(x-1)/x(x-1)当x(x-1)≠02x²-x=(2x-1)(x-1)x=x-1无解

∵x^2+x+1=(x+1/2)^2+3/4＞0则3x^2+2x+2≥M(x^2+x+1)即(3-M)x^2+(2-M)x+(2-M)≥0,∵对任意M恒成立,∴3-M＞0,且△=(2-M)^2-4(3

4x/(2x-1)-(x+1)/(x-2)=1先通分[4x(x-2)-(x+1)(2x-1)]/[(2x-1)(x-2)]=14x2-8x-(2x2+x-1)=2x2-5x+2x=-1/4其中x2代表

x²+1=4x两边平方x^4+2x²+1=16x²两边减去11x²x^4-9x²+1=5x²所以x²÷(x^4-9x²+

(x-21)除以(x-56)=2x-21=2(x-56)x-21=2x-1122x-x=112-21x=91在右上角点击【评价】,然后就可以选择【满意,问题已经完美解决】了.

题目不对再问：是滴发错了3x-4除以（x-1)（x-2）=a除以x-1+b除以x-2再答：右边通分=[a(x-2)+b(x-1)]/(x-1)(x-2)=[(a+b)x-(2a+b)]/(x-1)(x

X²+x-2分之1+x除以x+2分之x²-1,=（x+1）/(x+2)(x-1)÷（x+1）（x-1）/(x+2)=1/(x-1)²=1/(1/2-1)²=4

x=10