用权方和不等式证明分式不等式

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(637851)[1]:xiR,yiR+(i=1,2,,n),x21y1+x22y2++x2nyn(x1+x2++xn)2y1+y2++yn(1).(1)21,(1).(1).xiR+,yiR+(i=1,2,,n),mR+,xm+11ym1+xm+12ym2++xm+1nymn(x1+x2++xn)m+1(y1+y2++yn)m(2)x1y1=x2y2==xnyn,(2).,([2]),..1a,b0,:aa+3b+b3a+b1.(3)(200351435)(2)aa+3b+b3a+b=(a23)32(a2+3ab)12+(b23)32(b2+3ab)12(a23+b23)32(a2+3ab+b2+3ab)12=(a23+b23)32(a2+b2+6ab)12.(3),(a23+b23)32(a2+b2+6ab)121Z(a23+b23)3a2+b2+6abZa2+b2+3a23b23(a23+b23)a2+b2+6abZa23+b232a13b13.,(3).2a,bR+,|x|12,f(x)=34a2(1+2x)2+4b2(1-2x)2.(200421479)(2)u=[f(x)]3=4(a23)3(1+2x)2+(b23)3(1-2x)24(a23+b23)3(1+2x+1-2x)2=(a23+b23)3,f(x)a23+b23,a231+2x=b231-2xZx=a23-b232(a23+b23).x=a23-b232(a23+b23),f(x)3a2+3b2.3(42IMO2)a,b,c,:aa2+8bc+bb2+8ca+cc2+8ab1.(4)(2)aa2+8bc+bb2+8ca+cc2+8ab=a32(a3+8abc)12+b32(b3+8abc)12+c32(c3+8abc)12(a+b+c)32(a3+b3+c3+24abc)12.,(4),(a+b+c)32(a2+b3+c3+24abc)121Z(a+b+c)3a3+b3+c3+24abc.(5)742006452(a+b+c)3(),(a+b+c)3=a3+b3+c3+aq1bq2cq3,q1,q2,q33q1+q2+q3=3,aq1bq2cq327-3=24,a,b,c2733-3=24,:aq1bq2cq32424a24b24c24=24abc,(a+b+c)3=a3+b3+c3+aq1bq2cq3a3+b3+c3+24abc,(5),(4).1(a+b+c)3(4).28(4):a,b,cR+,8,aa2+bc+bb2+ca+cc2+ab31+.(6)4x,y,zR+,x+y+z=1,u=1x2+1y2+8z2.(2004101504).(2)u=1x2+1y2+8z2=13x2+13y2+23z2(1+1+2)3(x+y+z)2=64,1x=1y=2z,x=y=14,z=12,u=64.x=y=14,z=12,u64.5a,b,nR+,y=a(1+x)n+b(1-x)n,x(-1,1).(20048625).(2)y=a(1+x)n+b(1-x)n=(a2n+2)n2+1(1+x)n2+(b2n+2)n2+1(1-x)n2(a2n+2+b2n+2)n+22(1+x+1-x)n2=1(2)n(a2n+2+b2n+2)n+22.a2n+21+x=b2n+21-x,x=a2n+2-b2n+2a2n+2+b2n+2,,x=a2n+2-b2n+2a2n+2+b2n+2,y1(2)n(a2n+2+b2n+2)n+22.y=asinn+bcosn(a,b,nR+),02(a2n+2+b2n+2)n+22.6(199536IMO2)a,b,c,abc=1,:1a3(b+c)+1b3(c+a)+1c3(a+b)32.(2)1a3(b+c)+1b3(c+a)+1c3(a+b)=(abc)2b3(c+a)+(abc)2c3(a+b)+(abc)2a3(b+c)=(bc)2ab+ac+(ca)2bc+ab+(ab)2ca+cb(bc+ca+ab)22(ab+ac+bc)=bc+ca+ab21233bccaab=32.7(31IMO)a,b,c,dR+,ab+bc+cd+da=1,a3b+c+d+b3c+d+a+c3d+a+b+d3a+b+c13.(2)a3b+c+d+b3c+d+a+c3d+a+b+d3a+b+c=(a2)2ab+ac+ad+(b2)2bc+bd+ab+(c2)2cd+ac+bc+(d2)2ad+bd+cd(a2+b2+c2+d2)22(ab+ac+ad+bc+bd+cd)=(a2+b2+c2+d2)22(1+ac+bd),a2+b2+c2+d2=a2+b22+b2+c22+d2+a22+c2+d22ab+bc+cd+da=1,a2+b2+c2+d2=(a2+c2)+(b2+d2)2ac+2bd,(a2+b2+c2+d2)22(1+ac+bd)a2+b2+c2+d22(1+ac+bd)=132(a2+b2+c2+d2)+(a2+b2+c2+d2)2(1+ac+bd)132+2(ac+bd)2(1+ac+bd)=13,.1..,2005,12..,2002,23.IMO42.(),2004,11842006452

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