求-3x的平方 6X 1的最大值-搜答案-神马作文网

设x^2+y^2=m≥03x平方+2y平方=6x,(x-1)^2+y^2/(3/2)=1,椭圆方程x属于[0,2]3x平方+2y平方=6x,x^2+2m=6x,2m=-x^2+6x,m=-1/2(x-

3x^2+2y^2=6xy^2=-1/2*(3x^2-6x)故x^2+y^2=x^2-1/2*(3x^2-6x)=-1/2*x^2+3x=-1/2(x-3)^2+9/2故最大值9/2无最小值

x²+y²-4x+1=0;(x-2)²+y²=3;所以半径r=√3,圆心：(2,0)1)设k=y/x;y=kx;代入原来方程,整理：（1+k²）x&s

x1+x2=3x1x2=-2007x1的平方加X2的平方=(x1+x2)^2-2x1x2=9+4014=4023

根据韦达定理x1+x2=-3/2,x1x2=-2所以x1²+x2²=(x1+x2)²-2x1x2=(-3/2)²+4=9/4+4=25/4

由韦达定理得x1+x2=-3x1x2=-c²x1²+x2²=(x1+x2)²-2x1x2=9+2c²>=9x1²+x2²的最小值是

ax²+bx+c=0中有：x1+x2=-b/ax1·x2=c/a2X²-9X+6=0中：a=2b=-9c=6x1+x2=-b/a=9/2x1·x2=c/a=3(x1-x2)&sup

(k-2)^2-4(k^2+3k+5)>=0-4

∵x²+6x+3=0∴x1+x2=-6x1x2=3x1/x2+x2/x1=(x1+x2)²-2x1x2/x1x2=10

由韦达定理x1+x2=3x1x2=1x1²+x2²=(x1+x2)²-2x1x2=3²-2*1=7

X的平方-3X+1=0的两个实数根是X1,X2X1+X2=3X1X2=1(X1-X2)^2=(X1+X2)^2-4X1X2=3^2-4=5X1-X2=正负根号5

-2x平方-3x+5=-2(x²+3/2x+9/16)+9/8+5=-2(X+3/4)²+49/8当x=-3/4有最大值49/8

有根,判别式大于等于016m²-40m²+72m+96>=0m²-3m-4

-2x平方-6x+3==-2(x²＋3x－(3/2))=-2(x²＋2·x·(3/2)＋(3/2)²－(3/2)²－(3/2))=-2(x＋(3/2))

3x^2+4x-7=0由韦达到理得：x1+x2=-4/3、x1x2=-7/3.x1^2+x2^2=(x1+x2)^2-2x1x2=16/9+14/3=58/9.1/x1^2+1/x2^2=(x1^2+

根据题意得x1+x2=5/2x1x2=-3于是1/x²1+1/x²2=(x²1+x²2)/x²1*x²2=[(x1+x2)²-2x

X1的平方+X2的平方的和=（x1+x2)的平方-2*x1x2根据韦大定理x1+x2=9/2,x1x2=6/2=3求得结果为73/4

整理得y=(-3x+6x)/2>=0,x(x-2)

首先解x*2-4x+2=0的解,解出x1=根号2+2,x2=2-根号2然后可算x1+x2=根号2+2+2-根号2=4x1x2=（根号2+2）（2-根号2）=4-2=2问题1：x1分之1加x2分之1=x

韦达定理：一元二次方程aX^2+bX+C=0﹙a≠0﹚中,两根X1,X2有如下关系：X1+X2=-b/a,X1·X2=c/a.3x^2-2x-2=0a=3,b=-2,c=-2,-b/a=2/3,c/a