命题(1)x 1^x的最小值是2,(2)x2 2^根号下x2 1的最小值是2
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x1,x2是x²+(2-M)x+(1+M)=0的两个根x1+x2=M-2x1x2=1+Mx1²+x2²>=2x1x2=2(1+M)当且仅当x1=x2时,有最小值.即根的判
(x1+5x2)(x2+5x1)=x1x2+5x1^2+5x2^2+25x1x2=26x1x2+5(x1^2+x2^2)=5(x1+x2)^2+16x1x2=(2m)^2+16(m-1)=4m^2+1
X1,X2是方程X^2-2aX+a+b=0两实数根x1+x2=2ax1*x2=a+b且△=(-2a)^2-4(a+b)≥0a^2≥a+b=x1*x2(X1-1)^2+(X2-1)^2=(x1^2-2x
根据韦达定理x1+x2=-b/a=-2/mx1x2=c/a=m/m=1∴x1²+x2²=(x1+x2)²-2x1x2=4/m²-2≥-2所以最小值为-2
由题意,y=(x1-2x2)(2x1-x2)=2x1²-x1x2-4x1x2+2x2²=2(x1+x2)²-9x1x2因为x1,x2是x²-kx+k-1=0的实
x^2-2mx+m+2=0△=4m^2-4(m+2)≥0m^2-m-2≥0(m-2)(m+1)≥0m≥2,m≤-1x1^2+x2^2=(x1+x2)^2-2x1x2=(2m)^2-2*2=4m^2-4
∵x1、x2是关于x的方程x2-kx+k-1=0的两个实数根,∴x1+x2=k,x1x2=k-1,∴y=(x1-2x2)(2x1-x2)=2x12-x1x2-2x1x2+2x22=2x12-3x1x2
x^2-2kx+1=k^2x^2-2kx+1-k^2=0x1+x2=-(b/a)=-(-2k)=2kx1x2=c/a=1-k^2x1^2+x2^2=(x1+x2)^2-2x1x2=(2k)^2-2(1
x1,x2是方程x²-(2m-1)x+(m²+2m-4)=0的两个实数根所以x1+x2=2m-1,x1x2=m²+2m-4Δ=(2m-1)²-4(m²
当x^2+2x-1>x+1,解得x1时,f(x)=x^2+2x-1.那么当-2
由⊿=(-2m)²-4(1-m²)=8m²-4≥0,得m²≥1/2.又x1+x2=2mx1x2=1-m²则x1²+x2²=(x1+
方程x^2+2mx+2m+3=的两实根是x1,x2那么Δ=4m²-4(2m+3)≥0即m²-2m-3≥0解得m≤-1或m≥3又根据韦达定理:x1+x2=-2m,x1x2=2m+3∴
△=(2-m)^2-4(1+m)=4-4m+m^2-4-4m=m^2-8m≥0m(m-8)≥0m≤0或m≥8x1^2+x2^2=(x1+x2)^2-2x1x2=(m-2)^2-2(1+m)=m^2-4
方程x^2-2mx+m+2=0求解得出x1=m-√(m^2-m-2)x2=m+√(m^2-m-2)代入方程(x1)^2+(x2)^2得出(x1)^2+(x2)^2=4m^2-2m-4方程x^2-2mx
用维达定理(X2)+(X1)=(-a分之b)=(-1分之-2)=2(X1)*(X2)=(a分之c)=(-1分之m-3)所以(X2)+(X1)最小是2
首先方程有根Δ=4a²-4(a+b)≥0即a+b≤a²X1+X2=2ax1x2=a+b(x1-1)²+(x2-1)²=x1²+x2²-2(x
根据韦达定理,x1+x2=2a,x1*x2=6,x1^2+x2^2=(x1+x2)^2-2x1*x2=4a^2-12,a=0,时最小值为-12.
Δ=4k²-4(1-k²)=8k²-4≥0;k²≥1/2;∴k≥√2/2或k≤-√2/2;x1+x2=2k;x1x2=1-k²;x1²+x2
x1+x2=2k,x1*x2=1-k^2有两个实根4k^2-4(1-k^2)>=08k^2-4>=0k^2>=1/2x1^2+x2^2=(x1+x2)^2-2x1x2=4k^2-2(1-k^2)=6k
楼上两位都错啦首先要满足判别式为非负数才OK嘛!所以,m^2>=1/2,于是,x1^2+x2^2=(x1+x2)^2-2x1x2=6m^2-2>=1故当m=正负二分之根号二时,有最小值1