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5/x 1 4/y%2

(x-1)^5+5(x-1)^4+10(x-1)3+10(x-1)2+5(x-1)学了二项式,要往二项式上去想(x-1)^5+5(x-1)^4+10(x-1)3+10(x-1)2+5(x-1)+1-1 =[(x-1)+1]^5-1 =x^5 -1 10,10对称 5,5对称 1需要1对称

(1),x≥_1,时,【1】y≥4,x+4+y_2=5, 2(x+1)+y_4=5;x+y=3,2x+y=7;有x=4,y=_1,无解; 【2】,2≤y

所得两式相加得5/x+5/y+5/z=15 得所求为15/5=3

已知x、y∈R+, 故依柯西不等式得 5=x+y+1/x+1/y ≥(x+y)+4/(x+y) →(x+y)^2-5(x+y)+4≤0 →1≤x+y≤4. ∴x=y=2时, 所求最大值为:4。

设1/ x=m 1/y=n 原方程转换为4m^2+25n^2=25,2m+5n=1 m=(1-5n)/2 4*[(1-5n)/2]^2+25n^2=25 25n^2-5n-12=0 (5n-4)(5n+3)=0 n=4/5或n=-3/5 m=-3/2或m=2 n=4/5,m=-3/2 或n=-3/5,m=2 x=-2/3,y=5/4 或x=1/2,y=-5/3

解: (x+y)/3 +(x-y)/2=1 (1) 4(x+y) -5(x-y)=1 (2) (1)×12 4(x+y)+6(x-y)=12 (3) (3)-(2) 11(x-y)=11 x-y=1 (4) 代入(2) 4(x+y)=1+5(x-y)=1+5=6 x+y=3/2 (5) (4)+(5) 2x=5/2 x=5/4 (5)-(4) 2y=1/2 y=1/4 x=5/4 y=1/4

x<5/4 即4x-5<0 y=4x-2+[1/(4x-5)] =4x-5+[1/(4x-5)]+3 令4x-5为t,t<0,原式=t+(1/t)+3 =-[(-t)+(1/-t)]+3 又因为[(-t)+(1/-t)]≥2√[-t×(1/-t)]≈2 得:t+(1/t)≤-2 则原式=t+(1/t)+3≤1 故y=4x-2+[1/(4x-5)]≤1 当t=(1/t)=-1时,...

x0 y=(4x-5)+1/(4x-5)+3 =-[(5-4x)+1/(5-4x)]+3 (5-4x)+1/(5-4x)≥2√[(5-4x)*1/(5-4x)]=2 所以-[(5-4x)+1/(5-4x)]≤-2 y≤-2+3=1 最大值是1

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