若分式x的平方-2x 1分之x 2有意义,则x应满足
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上边的方程可以写成x^2=6-3x,这个方程有两个根:x1和x2.所以x1^2=6-3*x1,x2^2=6-3*x2所以让求的式子变成了:x2/(6-3*x1)+x1/(6-3*x2)然后通分:(15
4x的平方+根号3x=14x的平方+根号3x-1=0若两个根分别是X1,X2,则x1+x2=-√3/4,x1*x2=-1/4X1分之X2+X2分之X1=(x2²+x1²)/(x1x
x²-4x+2=0由韦达定理得:x1+x2=4,x1·x2=2∴(1)x1+x2+3x1x2=4+3*2=10(2)x2/x1+x1/x2=(x2²+x1²)/x1x2=
5x²+x-5=0两根x1,x2,由韦达定理得x1+x2=-1/5x1x2=-5/5=-1x1²+x2²=(x1+x2)²-2x1x2=(-1/5)²
△=4(k+1)²-4(k²-1)≥0解得:k≥-1根据韦达定理x1+x2=-2(k+1)x1*x2=k²-1x1²+x2²=(x1+x2)²
X1+X2=-B/A=2X1*X2=C/A=1/2求得X1=1+根号2或者X1=1-根号2从而求出X2的值X1/X2+X2/X1=(X1*X1+X2*X2)/(X1X2)=6
∵x²+6x+3=0∴x1+x2=-6x1x2=3x1/x2+x2/x1=(x1+x2)²-2x1x2/x1x2=10
x1+x2=4x1x2=1/2原式=(x1+x2)²÷(x1+x2)/x1x2=x1x2(x1+x2)=2
1/x1+1/x2=(x1+x2)/x1x2伟达定理x1+x2=-b/ax1x2=c/a1-2
∵x1,x2是一元二次方程x的平方x2-2x-1=0的两个根∴a=1,b=-2根据韦达定理,X1+X2=-b/a∴X1+X2=2
x-x+3=0所以x1+x2=1,x1x2=3因此(1)(X1+2)(X2+2)=x1x2+2(x1+x2)+4=3+2x1+4=9(2)(X1-X2)=(x1+x2)-4x1x2=1-4x3=-11
设x1,x2是方程2x平方+4x-3=0的两个根,则x1+x2=-2x1·x2=-3/2∴x1平方+x2平方=(x1+x2)²-2x1·x2=(-2)²-2×(-3/2)=4+3=
1.1/x1+1/x2=(x1+x2)/x1x2=2x1+x2=-b/a=mx1x2=c/a=-4-m/4=2m=-82.这两句话对不对?为什么?对.因为德尔塔等于0代表有两个相等的实根,一般不说一个
由题知,x1,x2是方程4x²-4mx+m+2=0的两个实数根,判别式⊿=(4m)²-4*4*(m+2)=16[m²-m-2]≥0,即m≤-1或m≥2所以,由韦达定理x1
2x平方-5x-1=0X平方-5/2X-1/2=0X平方-5/2X=1/2X平方-5/2X+5/4的平方=1/2+5/4的平方(X-5/4)平方=33/16X-5/4=正负根号33/4X=正负根号33
x1+x2=-3/2x1x2=-21/x1+1/x2=(x1+x2)/x1x2=(-3/2)/(-2)=3/4x1²+x2²=(x1+x2)²-2x1x2=(-3/2)&
X1+X2=-6/2=-3X1*X2=-3/21/X1+1/X2=(X1+X2)/(X1X2)=-3/(-3/2)=2
x1+x2=4x1x2=-1(x1+x2)^2/(1/x1+1/x2)=(x1+x2)^2*x1x2/(x1+x2)=x1x2*(x1+x2)=-4
x1+x2=3/2,x1*x2=1/2所以(x1+x2)^2=x1^2+2x1x2+x2^2所以x1^2+x2^2=(x1+x2)^2-2x1x2=9/4-1=5/4
2x²+5x-3=0(2X-1)(X+3)=0所以有X1=1/2X2=-3或者X1=-3X2=1/2则|x1-x2|=3.51/x1²+1/x2²=4+1/9=37/9