已知X1=-1是一元二次方程X的平方 mx-5=0的一个跟
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![已知X1=-1是一元二次方程X的平方 mx-5=0的一个跟](/uploads/image/f/4227253-61-3.jpg?t=%E5%B7%B2%E7%9F%A5X1%3D-1%E6%98%AF%E4%B8%80%E5%85%83%E4%BA%8C%E6%AC%A1%E6%96%B9%E7%A8%8BX%E7%9A%84%E5%B9%B3%E6%96%B9+mx-5%3D0%E7%9A%84%E4%B8%80%E4%B8%AA%E8%B7%9F)
x1三次方+8X2+20=x1³+3x1²+x1+8x2+20-3x1²-x1=x1(x1²+3x1+1)+8x2+20-3x1²-x1………………x
x=x1则x1²+3x1+1=0所以x1²=-3x1-1且x1+x2=-3所以原式=-3x1-1-3x2+20=-3(x1+x2)+19=-3*(-3)+19=28
∵一元二次方程x2-4x+1=0的两个实数根是x1、x2,∴x1+x2=4,x1•x2=1,∴(x1+x2)2÷(1x1+1x2)=42÷x1+x2x1x2=42÷4=4.
1/x1+1/x2=1则x1+x2=x1*x2由根与系数间关系x1+x2=2k+3,x1*x2=k^2所以2k+3=k^2即k^2-2k-3=0所以k=3或k=-1
2x1²+4x2²-6x2+2011=2x1²+2x2²+2x2²-6x2+2011=2(x1²+x2²)+2(x2²-
x1,x2是一元二次方程x²-4x+1=0的两个实数根∴x1+x2=4,x1x2=1⒈(X1+X2)²=4²=16⒉1/x1+1/x2=(x1+x2)/x1x2=4⒊(x
x1+x2=4x1x2=1所以原式=(x1+x2)²÷[(x1+x2)/x1x2]=x1x2(x1+x2)²/(x1+x2)=x1x2(x1+x2)=1×4=4
由韦达定理x1+x2=-(-3)/1=3x1*x2=(-1)/1=-1所以x1x2^2+x1^2x2=x1x2(x1+x2)=(-1)*3=-3
x²-3x-1=0的两个根分别是x1,x2则x1+x2=3,x1*x2=-1x1²x2+x1x2²=x1x2(x1+x2)=-3如果不懂,请Hi我,
你好x1²x2+x1x2²=x1x2(x1+x2)=(-1)*3=-3韦达定理x1x2=c/ax1+x2=-b/aa、b、c分别是方程ax²+bx+c=0的系数,这里x&
x1+x2=—b/a,x1乘x2=c/a先把式子代入x1乘x2+2(x1+x2)>0得(1-3m)/2+2>0解得m<5/3由于一元二次方程2x^2-2x+1-3m=0有实数根所以判别式≥0,4-4*
x1+x2=2x1x2=m-1x1²+x1x2=x1(x1+x2)=2x1=1x1=1/2x2=3/2x1x2=m-1=3/4m=7/4
整理得,x1*x2(x1+x2)可知,x1+x2=4,x1*x2=1/2,代入得2
x1+x2=4x1x2=1所以1/x1+1/x2=(x1+x2)/x1x2所以原式=(x1+x2)²*x1x2/(x1+x2)=x1x2(x1+x2)=1*4=4
(1)根据题意得△=(-2)2-4×2×(m+1)≥0,解得m≤-12;(2)根据题意得x1+x2=1,x1x2=m+12,∵7+4x1x2>x12+x22,∴7+4x1x2>(x1+x2)2-2x1
x1、x2是一元二次方程2x^2+3x-1=0的两根所以由韦达定理得x1+x2=-3/2x1x2=-1/2以x1+x2,x1x2为根的方程两根之和=(x1+x2)+x1x2=-3/2+(-1/2)=-
∵x1,x2是一元二次方程4kx2-4kx+k+1=0的两个实数根,∴x1+x2=1,x1x2=k+14k,∴x1x2+x2x1-2=x12+x22x1x2-2=(x1+x2)2−2x1x2x1x2-
(x1-1)(x2-1)=x1*x2-(x1+x2)+1因为x1+x2=-1/ax1*x2=1/a代入x1*x2-(x1+x2)+1整理得2/a+1
x^2-4x+1=0x^2-4x+4=3(x-2)^2=3x1,x2=2±根号3|x1-x2|=2根号3