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已知x>1,y>0,且2/(x%1)+1/y =2,求x+y最小值

解: 2/(x-1)+1/y=2 2/(x-1)=2- 1/y x>1,x-1>0,2/(x-1)>0 2-1/y>0,1/y0,因此y>½ (首先判定y的取值范围) 整理,得x=(4y-1)/(2y-1) (再将x用y表示) x+y=(4y-1)/(2y-1) +y =(4y-2+1)/(2y-1) +y =y+2 +1/(2y-1) =½(2y-1) +1/(2y-1) +5...

x>0,y>0, ∴1=2/x+1/y =(√2)^2/x+1^2/y ≥(√2+1)^2/(x+y) ∴x+y≥3+2√2. 故所求最小值为;3+2√2。

其实题目是关系到“1”的妙用。 对于1/x+9/y=2,两边除以2,则 (1/2)·(1/x+9/y)=1, 所以, x+y=1·(x+y) =(1/2)(1/x+9/y)·(x+y) =(1/2)(10+y/x+9x/y) ≥(1/2)[10+2√(y/x·9x/y)] =(1/2)(10+6) =8, 即1/x+9/y=2且y/x=9x/y, x=2,y=6时, 所求x+y最...

1/x2+1/y2+1/xy≥3 * 3^√ ̄1/x2*y2*xy =3 * 3^√ ̄1/x^3*y^3 x+y=2≥2√ ̄xy 所以√ ̄xy ≤1 所以 3 * 3^√ ̄1/x^3*y^3≤3

x>y>0,x+y≤2, ∴2/(x+3y)+1/(x-y) =(√2)²/(x+3y)+1²/(x-y) ≥(√2+1)²/[(x+3y)+(x-y)] =(3+2√2)/[2(x+y)] ≥(3+2√2)/4. 故(x+3y):√2=(x-y):1且x+y=2, 即x=-1+2√2,y=3-2√2时, 所求最小值为: (3+2√2)/4。

解: 2/x +1/y=1 1/y=1- 2/x=(x-2)/x y=x/(x-2) x>0,y>0,x/(x-2)>0,x>2 x+2y=x+ 2x/(x-2) =x+2(x-2+2)/(x-2) =x +2 +4/(x-2) =(x-2)+ 4/(x-2) +4 x>2,x-2>0,由均值不等式得: (x-2)+ 4/(x-2)≥2√[(x-2)·4/(x-2)]=4 (x-2)+ 4/(x-2) +4≥8 x+2y...

因为:x+y=1,所以有:1/x+2/y=(1/x+2/y)(x+y)=2x/y+y/x+3;因为 x>0,y>0,所以2x/y>0, y/x>0,有2x/y+y/x+3>=3+2乘根号2,当且仅当2x/y=y/x时“=” 号成立(这里用的是平均值不等式),得 y^2=2x^2,又x>0,y>0, 且x+y=1,所以有x=根号2-1的差,y=2-根号...

x>y>0且x+y≤2, 故依均值不等式得 2/(x+3y)+1/(x-y) =1·[2/(x+3y)+1/(x-y)] =[(x+3y)+(x-y)]/4·[2/(x+3y)+1/(x-y)] =3/4+(1/4)·[(x+3y)/(x-y)+2(x-y)/(x+3y)] ≥3/4+(1/4)·2√[(x+3y)/(x-y)·2(x-y)/(x+3y)] =(3+2√2)/4. ∴(x+3y)/(x-y)=2(x-y)/(x+3...

y=(1/x+1/y)(x+2y)=1+2+2y/x+x/y=3+(2y/x+x/y) ≥ 3+2√(2y/x)*(x/y)=3+2√2 当且仅当,2y/x=x/y,x=√2y时,等号成立 最小值=3+2√2 y=1/x+1/y的最小值=3+2√2

(1)∵x-y=3,∴x=y+3,又∵x>2,∴y+3>2,∴y>-1.又∵y<1,∴-1<y<1,…①同理得:2<x<4,…②由①+②得-1+2<y+x<1+4∴x+y的取值范围是1<x+y<5;(2)∵x-y=a,∴x=y+a,又∵x<-1,∴y+a<-1,∴y<-a-1,又∵y>1,∴1<y<-a-1,…①同理得:a+1<x<...

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