已知方程x的平方减去6x加q等于0-搜答案-智慧作业帮

1式都化为2004,则2004^2x^2-2004^2x+x-1=0因式分解(2004^2x+1)(x-1)=0则p=12式分解为(2003x-1)(x-1)=0q=1/2003p-q=2002/20

x^2-x-1=0,x^2-x=1.(1)又有x^2=x+1,x^3=x^2+x.(2).-x^3+2x^2+2009=-x^2-x+2x^2+2009[(2)代入]=x^2-x+2009=1+200

x的平方减去3x加1等于0求2x的平方减去6x加3x²-3x+1=0两边同乘以2得：2x²-6x+2=0两侧同加上1得：2x²-6x+3=1所以：2x²-6x+

3x^2+2xy+y^2-6x-6y+9=0(x+y)^2-6(x+y)+9=0(x+y-3)^2=0x+y=3

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∵x^2-3x+1=0显然x不等于0∴x-3+1/x=0∴x+1/x=3∴x^2+1/x^2+2=9∴x^2+1/x^2=7∴原式=1/(x^2+1+1/x^2)=1/(7+1)=1/8

1)方程x²-6x+5=0的两个实数根x1=1,x2=5,m

x^4：表示x的4次方2x^4－x³－7x²＋x＋2=0(2x^4－8x²)－(x³－4x)＋(x²－3x＋2)=02x²(x²－

答案是0两式相减可得(P-Q）X=P-Q所以X=1将X带入任意一式得P+Q=0

x²+y²-2x-4y+5=0,可化为(x²-2x+1)+(y²-4y+4)=0,即(x-1)²+(y-2)²=0,因此x-1=0,y-2=

答案是(x-6)(x-1)=0x=6或x=1

x2-3x+m=1m0即13>4m,即m

根号x=根号a-根号1/a所以等式两边平方有x=a+1/a-2X²+4X=X²+4X+4-4=(X+2)²-4=(a+1/a)²-4=a²+1/a&#

x^2-6x+2=0x^2-6x+9-7=0(x-3)^2=7x-3=正负根号7x=3加减根号7

原式=x(x-9)/x(x+3)-(x+3)(x-3)/(x-3)²=(x-9)/(x+3)-(x+3)/(x-3)=(x²-12x+27-x²-6x-9)/(x+3)(

解x^2-6x+q=0(x^2-6x+9)=9-q(x-3)^2=9-q配成(x-p)^2=7∴q=2,p=3∴x^2-6x+q=2(x^2-6x+9)=(11-q)(x-3)^2=11-q=9∴即(

x-1/x=3(x-1/x)²=3²x²-2+1/x²=9x²+1/x²=11

x^2-2x-3=(x+1)(x-3)2x^2-x-1=(2x+1)(x-1)x^2+3x-4=(x-1)(x+4)3x^2-8x-3=(3x+1)(x-3)

(2x^2+my-12)-(nx^2-3y+6)=(2-n)x^2+(m+3)y-18,因为不含有x和y,所以2-n=0,m+3=0,n=2,m=-3,所以m+n+mn=-3+2+(-3)x2=-7

x²+px+q=0(x+p/2)²+q-p²/4=0(x+p/2)²=p²/4-qx+p/2=±√(p²/4-q)x=√(p²/4