若x^2-3x 1=0,则x x分之一=-搜答案-灵鹊做题网

x1+x2=-(-2)/1=2

x1+x2=-5,x1x2=-31)|x1-x2|^2=(x1+x2)^2-4x1x2=25+12=37|x1-x2|=√372)1/x1^2+1/x2^2=(x1^2+x2^2)/(x1x2)^2=

对称轴x=-1(x1+x2)/2=(1-a)/20

x1,x2是方程2x^2-3x-1=0的两个根,则x1^2+x2^2=(x1+x2)^2-2x1*x2=(3/2)^2+1=13/4

2

x1+x2=m=2方程x^-mx-3=0变为x^2-2x-3=0(x+1)(x-3)=0x=-1或3x1,x2的值为-1或3

x-1）（x-2）=0x=1ORx=2x1＞x2x1=2,x2=1x1-2x=2-1=1

唯达定理：x1+x2=2,x1x2=1/2→x1²+x2²=(x1+x2)²-2x1x2=3→x1/x2+x2/x1=(x1²+x2²)/x1x2=6

由题意x1^2+3x1+1=0x1^2=-1-3x1原式=x1*x1^2+8x2+20=x1(-1-3x1)+8x2+20=-3x1^2-x1+8x2+20=-3(-1-3x1)-x1+8x2+20=

由题意可知Δ=(-2)²-4*2*(3m-1)≥0即4-8(3m-1)≥01-6m+2≥06m≤3解得：m≤1/2又由韦达定理可得：x1+x2=1,x1*x2=(3m-1)/2因为：x1×x

利用x1+X2=-B/A,x1x2=C/A(1)1/x1+1/x2=(x1+x2)/x1x2=-(-6/2)/(3/2)=-2(2)在菱形ABCD中边长是5,所以有OA^2+OB^2=5^2=25OA

答案是与x=9/2时相等过程为当x=X1时,y=2*X1*X1+9（X1）+34,x=X2时,y=2*X2*X2+9(X2)+34因此有2*X1*X1+9（X1）=2*X2*X2+9(X2)得（X1-

已知x1是方程的解,则2x1²-2x1-5=0===>x1²-x1=5/2=2.5又,x1,x2是方程的两个解,则：x1+x2=1,x1x2=-5/2x1³+3x1

先移项3x(x-2)-（x-2）=0把x-2看成一个整体,合并同类项(3x-1)(x-2)=03x-1=0或x-2=0x1=2,x2=1/3

x1.x2是方程2x²-x-3=0的两实根∴x1+x2=1/2x1x2=-3/2∴x1+x2+x1*x2=1/2-3/2=-1

xx+yy+4x-6y+13=0整理得：(x+2)^2+(y-3)^2=0那么只有(x+2)=0(y-3)=0x=-2y=3(x^2-2x)/(x^2+3y^2)=(4+4)/(4+3*9)=8/31

根据韦达定理：x1+x2=63x1+2x2=x1+2x1+2x2=x1+2(x1+x2)=x1+2*6=x1+12=20x1=8代回原方程,得M=-16

韦达定理再问：亲，就是因为没看懂啥意思啊，可以的话，具体过程可以有么？省点木事再答：韦达定理：X1+X2=-5/2,X1X2=-3/2因此|x1-x2|=√(x1+x2)^2-2x1x2=√25/4+

X1+X2=-6/2=-3X1*X2=-3/21/X1+1/X2=(X1+X2)/(X1X2)=-3/(-3/2)=2

x1^2=3-x1,x2^2=3-x2,x1^3-4x2^2+19=3x1-x1^2-4x2^2+19=3x1-x1^2+4x2+7,x1+x2=-1,原式=4+x2-x1^2=4+x2-3+x1=0