方程2x² 3x-4=0的两根倒数之和为-搜答案-搜搜考试网

4x^2+2x√(3x^2+x)+x-9=03x^2+x+2x√(3x^2+x)+x^2-9=0[√(3x^2+x)]^2+2x√(3x^2+x)+x^2=9[√(3x^2+x)+x]^2=9√(3x

X²-2X-3=0(X-3)(X+1)=0X-3=0,X+1=0X=3,X=-1

4/(x-2)+(x-1)/[(x-2)(x-3)]-2/(x-3)=0[4(x-3)+(x-1)-2(x-2)]/[(x-2)(x-3)]=03(x-3)/[(x-2)(x-3)]=0不知道解了~~

由题意可知：x≠-1且x≠0而方程x²-3x-4=0可化为：(x-4)(x+1)=0解得：x=4（另x=-1不合题意,舍去）所以：(x²-2x)/(x+1)·(1+1/x)=x(x

#include#includeintmain(){doublex=1,x2;do{x2=x;x-=(2*x*x*x-4*x*x+3*x-6)/(6*x*x-8*x+3);}while(fabs(x-

intf(floatx)返回值,错了.

3x/(x+1)-(x+4)/(x^2+x)=-23x^2-(x+4)=-2(x^2+x)3x^2-x-4=-2x^2-2x5x^2+x-4=0(5x-4)(x+1)=0x1=4/5x2=-1经检验,

2/x^2+x+3/x^2-x-4/x^2-1=0(2/x^2+3/x^2-4/x^2)+x-x-1=01/x^2-1=01/x^2=1x^2=1x=1或-1

4x^2+2x√(3x^2+x)+x-9=0,——》(4x^2+x-9)^2=4x^2*(3x^2+x),——》4x^4+4x^3-71x-18x+81=(x-1)(4x^3+8x^2-6x+81)=

3（x-2）=5x（2-x）3（x-2）-5x（2-x）=03（x-2）+5x(x-2)=0(x-2)(3+5x)=0解得x=2或-3/54x（3x+8）-5（3x+8）=0(3x+8)(4x-5)=

x(x+1)(x²-2x-4)=0x1=0x2=-1x²-2x-4=0x²-2x+1=5(x-1)²=5x3=1+√5x4=1-√5x³-2x+1=0

x^6+x^5(-1-3)+x^4(-4+3+1)+x^3(7+12-1-2)+x^2(-21-4+2)+x(7+8)-14+6x^2-15x+18=0x^6-4x^5+16x^3-17x^2+4=0

因为等比数列：设4个根为：a,aq,aq^2,aq^3a+aq+aq^2+aq^3=4.1a*aq*aq^2+a*aq*aq^3+a*aq^2*aq^3+aq*aq^2aq^3=-56.2a*aq*a

2x^2-4x-3-m=2(x-1)^2-m-5=0所以2(x-1)^2=（m+5）由0

由题意知：X不等于0方程可化为4/(X^2+2X+1)+5/(X^2+2X-5)+3/2=0设X^2+2X=Y则:4/(Y+1)+5/(Y-5)+3/2=0然后通分:Y^2-2Y-15=0所以Y=5O

令：f(x)=x^3-6x^2+9x-4f(x)'=3x^2-12x+9f(x)'=0时,解得x1=1,x2=3limf(x)(x→-∞）=-∞f(1)=0f(3)=-4limf(x)(x→+∞)→=

2x²＋x＋3＝0,2(x²＋2·x·¼＋(¼)²)＝－3＋(1/8).2(x＋¼)²＝－23/8.(x＋¼)²

由题易知x

你好把x^2-x当作整体,(x^2-x)^2-2*4*(x^2-x)+12=0(x^2-x-6)(x^2-x+2)=0因为x^2-x≥-1/4所以x^2-x-6=0(x-3)(x+2)=0x=3或-2