x三次方-3x 2=0
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X1,X2为方程x²+3x+1=0的两根那么x1²+3x1+1=0x1²=-3x1-1x1(-3x1-1)+8x2+20=-3x1²-x1+8x2+20=-3(
X1.X2是方程:X的平方+3X+1=0的两个实数根则:X1²+3X1+1=0X1²=-3X1-1由韦达定理得:X1+X2=-3X1的三次方+8*X2+20=X1*X1²
x1三次方+8X2+20=x1³+3x1²+x1+8x2+20-3x1²-x1=x1(x1²+3x1+1)+8x2+20-3x1²-x1………………x
x1带入方程得:x1²+3x1+1=0再同乘上x1得:x1³+3x1²+x1=0所以x1³=-3x1²-x1=-3(-3x1-1)-x1=8x1+3所
x1^3+x1^2-3x1=0,x1^3=3x1-x1^2x1+x2=-1(1)x1*x2=-3(2)(1)^2x1^2+x2^2+2x1x2=1(2)代入x1^2+x2^2=74x2^2=27-4x
X1、X2是方程X^2+3X+1=0的两实数根韦达定理得:X1+X2=-3X1X2=1X1^2+3X1+1=0x1^2=-(3x1+1)x1^3+8x2+20=-x1*(3x1+1)+8x2+20=-
(1)由韦达定理,x1+x2=-2/3,x1x2=-2于是,x1^3+x2^3=(x1+x2)(x1²-x1x2+x2²)=-2/3[(x1+x2)²-3x1x2]=-1
x三次方+3x²-10x=X(X^2+3X-10)=X(X+5)(X-2)=0X1=0X2=-5X3=2
X1、X2是方程X^2+3X+1=0的两实数根韦达定理得:X1+X2=-3X1X2=1X1^2+3X1+1=0x1^2=-(3x1+1)x1^3+8x2+20=-x1*(3x1+1)+8x2+20=-
x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)=(x+y+z)[(x-y)^2+(y-z)^2+(z-x)^2]/2≥0x^3+y^3+z^3≥3xyz
由题可得:x1+x2=2;x1*x2=-2;(x2)-2(x2)=2那么:4x1-x1(x2^4-2x2^3)=4x1-x1[x2(x2-2x2)]=4x1-x1(2x2)=4x1-x1(4x2+4)
x²-3x-1=0根据韦达定理得到x1+x2=3x1x2=-1x1^2+x2^2=(x1+x2)^2-2x1x2=9+2=11x1^3+x2^3=(x1+x2)(x1^2-x1x2+x2^2
x^3-3x^2+4=0x^3-2x^2-(x^2-4)=0x^2(x-2)-(x+2)(x-2)=0(x-2)(x^2-x-2)=0(x-2)^2(x+1)=0x=2-1
可以由十字相乘法分解因式为(3x-8)(x+1)=0,解得x1为-1,x2为8/3再问:完整可以吗
由韦达定理得x1+x2=3,x1*x2=3/2则x1²+x2²=(x1+x2)²-2x1*x2=9-3=6x1³+x2³=(x1+x2)(x1
由x2(x的二次方)+x-1=0得出x2(x的二次方)+x=1那么:x3(x的三次方)+2x2(2x的二次方)-3=x3(x的三次方)+x2(x的二次方)+x2(x的二次方)-3=x【x2(x的二次方
-1x1+x2=-3x1^2=-3x1-1x1^3+8x2+20=x1(-3x1-1)+8x2+20=-3x1^2-x1+8x2+20=-3(-3x1-1)-x1+8x2+20=9x1+3-x1+8x
2x^5-5x^4+2x^3-8x^2+3x=2x^5-6x^4+2x^3+x^4-3x^3+x^2+3x^3-9x^2+3x=2x^3(x^2-3x+1)+x^2(x^2-3x+1)+3x(x^2-
x1³+x2³=(x1+x2)(x1²-x1*x2+x2²)=(x1+x2)[(x1+x2)²-3x1*x2]=3×(3²-3×1)=3×6
x三次方+x平方-5x+3=0x^3+x^2-2x-(3x-3)=0x(x^2+x-2)-3(x-1)=0x(x+2)(x-1)-3(x-1)=0(x-1)(x^2+2x-3)=0(x-1)(x+3)