比较代数式(x 1)²与2x的大小.怎么解答
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![比较代数式(x 1)²与2x的大小.怎么解答](/uploads/image/f/5685653-29-3.jpg?t=%E6%AF%94%E8%BE%83%E4%BB%A3%E6%95%B0%E5%BC%8F%28x+1%29%C2%B2%E4%B8%8E2x%E7%9A%84%E5%A4%A7%E5%B0%8F.%E6%80%8E%E4%B9%88%E8%A7%A3%E7%AD%94)
x²+3-3x=(x-3/2)²+3/4>=3/4>0x²+3>3xA>0,B>0,A≠BA³+B³=(A+B)(A²-AB+B²
∵X1、X2是方程X²-4X+2=0的两个根∴X1+X2=4,X1·X2=2X2/X1+X1/X2=X2²/(X1·X2)+X1²/(X1·X2)=(X1²+X
(1)f''(x)=(ln2)^2*2^x>0,故f(x)为下凸函数,根据下凸函数的性质:f(t1x1+t2x2)≤t1f(x1)+t2f(x2),0≤t1,t2≤1,则有[f(x1-1)+f(x2-
两式相减:x²+y²+2-(2x+2y)=x²+y²+1+1-2x-2y=x²-2x+1+y²-2y+1=(x²-2x+1)+(y
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
根据韦达定理:x1+x2=-b/ax1*x2=c/a代入:x1+x2=-5/3x1*x2=-2/3即:x1+x2+x1*x2=(-5/3)+(-2/3)=-7/3
x1^2-4x1+2=0x1^2-3x1=x1-2x1+x2-2=4-2=2
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(3X^2+6)-(2X^2+4X+2)=3X^2+6-2X^2-4X-2=X^2-4X+4=(X-2)^2当X=2时,(X-2)^2=0则3X^2+6=2X^2+4X+2X不等于2时,(X-2)^2
用(x²-1)-(2x-3)=x²-2x+2=(x-1)²+1>0,所以x²-1>2x-3;或者画函数图像,x²-1的抛物线永远在2x-3的直线上方,
x²-2x-4=0由根与系数关系知:x1+x2=2,x1x2=-4x2/x1+x1/x2=(x2²+x1²)/x1x2=[(x1+x2)²-2x1x2]/x1x
已知x1是方程的解,则2x1²-2x1-5=0===>x1²-x1=5/2=2.5又,x1,x2是方程的两个解,则:x1+x2=1,x1x2=-5/2x1³+3x1
[f(x1)+f(x2)]/2=1/2Log(x1*x2)f[(x1+x2)/2]=log[(x1+x2)/2]故前式>=后式
|F(x1)-F(x2)|=|根号下(1+x1^2)-根号下(1+x2^2)|=|(x1^2-x2^2)/(根号下(1+x1^2)+根号下(1+x2^2))|=|(x1-x2)||(x1+x2)/(根
因为(x-3)²-(x-2)(x-4)=x²-6x+9-(x²-6x+8)=9-8=1>0所以(x-3)²>(x-2)(x-4)
(3x2+4)-(2x2+4x)=x2-4x+4=(x-2)2,∵(x-2)2≥0,∴3x2+4≥2x2+4x.
(2X+1)(X-3)-(X-6)(X+1)=2X²-6X+X-3-X²-X+6X+6=X²+3>0(2X+1)(X-3)>(X-6)(X+1)