利用根与系数的关系求x2除以x1 x1除以x2
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x1+x2=-3/2x1x2=-2x21+x22=(x1+x2)²+2x1x2=9/4-4=-7/4(x1+1)(x2+1)=x1x2+x1+x2+1=-3/2-2+1=-5/2|x1-x2
(x1+1)(x2+1)=x1x2+x1+x2+1=(-3/2)+(-4/2)+1=-5/2x1^2x2+x1x2^2=(x1x2)(x1+x2)=(-3/2)(-4/2)=3x2/x1+x1/x2=
因为X1和X2是方程的两个根所以X1+X2=2|3X1.X2=-2|3(1)1/X1+1/X2=(X1+X2)|(X1.X2)=-1(2)X1的平方+X2的平方==(X1+X2)^2-2X1.X2=1
根据题意得x1+x2=-2x1x2=-3/2于是|x1-x2|²=(x1+x2)²-4x1x2=(-2)²-4×(-3/2)=4+6=10所以|x1-x2|=√10
根据根与系数的关系可得:x1+x2=-2,x1•x2=−32.(1)(x1-x2)2=x12+x22-2x1x2=x12+x22+2x1x2-4x1x2=(x1+x2)2-4x1x2=(−2)2−4×
二次方程的根为=(1/2a)*(-b±√b²-4ac)因此x1-x2=(1/2a)*(±2√b^2-4ac)=±a√b²-4ac=b²-4ac=1+4=5因此x1-x2=
由韦达定理:x1+x2=6/2=3,x1*x2=3/2,x1^2x2+x1x2^2=x1x2(x1+x2)=(3/2)*3=9/2,(x1+1/x2)(x2+1/x1)=x1x2+2+1/x1x2=3
化简原式为x1x2-3(x1+x2)+9x1x2=1/2x1+x2=-4/2=-2所以原式=31/2
(3)由根与系数的关系可知:x1+x2=3x1*x2=3/2(x1+1/x2)(x2+/x1)=x1*x2+1+1+1/(x1*x2)=3+1+1+2/3;(4)1/(x1)^2【x1²分之
x1+x2=-(-6/2)=3x1*x2=3/2(x1-x2)^2=x1^2-2x1x2+x2^2=x1^2+2x1x2+x2^2-4x1x2=(x1+x2)^2-4x1x2=3^2-4*(3/2)=
解x1,x2是方程两根则有韦达定理有x1+x2=-3/4x1x2=-5/4∴(1)1/x1+1/x2=(x1+x2)/(x1x2)=(-3/4)×(-4/5)=3/5(2)x1²-x1x2+
X1+X2=3/2,X1*X2=-1/2,|X1-X2|=√(X1-X2)^2=√[(X1+X2)^2-4X1X2]=√(9/4+2)=√17/2,∴X1-X2=±√17/2.X2^2/X1+X1^2
x1+x2=3x1x2=3/2∴(1)看不清(2)(x1-x2)²=(x1+x2)²-4x1x2=9-6=3(3)[(x1+1)/x2][(x2+1)/x1]=(x1x2+x1+x
x1+x2=-4/3x1x2=-7/31、x1²+x2²=(x1+x2)²-2x1x2=58/9所以x2/x1+x1/x2=(x1²+x2²)/x1x
把x=x1代入方程得3x1²-2x1-5=0,即3x1²-2x1=5由韦达定理可得:x1+x2=2/3所以3X1²-3X1-X2=3x1²-2x1-x1-x2=
用求根公式就能解出x1和x2,然后你说的这个表达式,有两种情况,一个是大根当x1,一个是小根当x1