求解两道高一数学题

\u4e24\u9053\u9ad8\u4e00\u6570\u5b66\u9898\u6c42\u89e3

\u89e3\uff1a\uff081\uff09\u7531\u9898\u610f\u53ef\u77e5\u5f53\u8be5\u5236\u836f\u5382\u6bcf\u5929\u5e9f\u6c14\u5904\u7406\u91cf\u8ba1\u5212\u4e3a20\u5428\u65f6\uff0c \u6bcf\u5929\u5229\u7528\u8bbe\u5907\u5904\u7406\u5e9f\u6c14\u7684\u7efc\u5408\u6210\u672c\u4e3a(20)402012002000f\u5143\uff0c\u2026\u2026\u20262\u5206 \u8f6c\u5316\u7684\u67d0\u79cd\u5316\u5de5\u4ea7\u54c1\u53ef\u5f97\u5229\u6da680201600\u5143\uff0c \u2026\u2026\u20263\u5206 \u6240\u4ee5\u5de5\u5382\u6bcf\u5929\u9700\u8981\u6295\u5165\u5e9f\u6c14\u5904\u7406\u8d44\u91d1\u4e3a400\u5143\uff0e \u2026\u2026\u20264\u5206
\uff082\uff09\u5e02\u653f\u5e9c\u4e3a\u5904\u7406\u6bcf\u5428\u5e9f\u6c14\u8865\u8d34a\u5143\u5c31\u80fd\u786e\u4fdd\u8be5\u5382\u6bcf\u5929\u7684\u5e9f\u6c14\u5904\u7406\u4e0d\u9700\u8981\u6295\u5165\u8d44\u91d1 \u5f534080x\u65f6\uff0c\u4e0d\u7b49\u5f0f2 80(21005000)0xaxxx\u6052\u6210\u7acb\uff0c \u53732 2(180)50000xax\u5bf9\u4efb\u610f[40,80]x\u6052\u6210\u7acb\uff0c \u2026\u2026\u2026\u2026\u2026\u202613\u5206 \u4ee42 ()2(180)5000gxxax\uff0c\u5219(40)085 (80)0 2gag \u7b54\uff1a\u5e02\u653f\u5e9c\u53ea\u8981\u4e3a\u5904\u7406\u6bcf\u5428\u5e9f\u6c14\u8865\u8d34 85 2 \u5143\u5c31\u80fd\u786e\u4fdd\u8be5\u5382\u6bcf\u5929\u7684\u5e9f\u6c14\u5904\u7406\u4e0d\u9700\u8981\u6295\u5165\u8d44\u91d1\uff0e \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u202616\u5206

\u7b54\uff1a
1\uff09
\u70b9A\uff084,0\uff09\uff0c\u70b9B(x\uff0cy)\u5728\u5706x²+y²=2\u4e0a\uff0cC\u662fAB\u7684\u4e2d\u70b9\uff08a\uff0cb\uff09
\u6240\u4ee5\uff1a
a=(x+4)/2
b=(y+0)/2
\u89e3\u5f97\uff1ax=2a-4\uff0cy=2b
\u4ee3\u5165\u5706\u65b9\u7a0b\uff1a
(2a-4)²+(2b)²=2
(a-2)²+b²=1/2
\u6240\u4ee5\uff1a\u70b9C\u7684\u8f68\u8ff9\u65b9\u7a0b\u4e3a(x-2)²+y²=1/2

2\uff09
\u5706(x+2)²+y²=1\uff0c\u5706\u5fc3(-2,0)\uff0c\u534a\u5f84R=1
\u5706\u5fc3\u5230\u76f4\u7ebf3x+4y+12=0\u7684\u8ddd\u79bb\uff1a
d=|-6+0+12|/\u221a(3²+4²)
=6/5>R=1
\u6240\u4ee5\uff1a\u5706\u4e0e\u76f4\u7ebf\u76f8\u79bb\uff0c\u6ca1\u6709\u4ea4\u70b9
\u6240\u4ee5\uff1a
\u8ddd\u79bb\u6700\u5927\u503c=d+R=6/5+1=11/5
\u8ddd\u79bb\u6700\u5c0f\u503c=d-R=6/5-1=1/5

两道题都可以用作差法解答

（1）设f(x)=ax2+bx+c(a＞0)
则 f(px+qy)=a(px+qy)2+b(px+qy)+c
=ap2x2+aq2y2+2apqxy+bpx+bqy+c

pf(x)+qf(y)=p(ax2+bx+c)+q(ay2+by+c)=apx2+aq2y2+bpx+bqy+c(这里用了p+q=1)
∴f(px+qy)-[pf(x)+qf(y)]
=ap2x2+aq2y2-apx2-aqy2+2apqxy
=ax2p(p-1)+ay2q(q-1)+2apqxy
=-ax2pq-ay2qp+2apqxy（这里把1用p+q代替）
=-apq(x2+y2-2xy)
=-apq(x-y)2＜0
∴f(px+qy)＜[pf(x)+qf(y)]

（2）(b／√a)+(a／√b)-(√a+√b)
=（b／√a-√b）+（a／√b-√a)
=√b／√a（√b-√a）+√a／√b（√a-√b)
=(√b-√a）(√b/√a-√a／√b)
=(√b-√a）(b-a)/√ab
=(√b-√a）2(√b＋√a)/√ab＞0
∴(b／√a)+(a／√b)＞√a+√b

希望能帮到LZ

ap2x2+aq2y2+2apqxy-apx2-aqy2
我来解决后边的的问题，把分给我吧
当然式子中的那个a是可以提出的
然后P2X2-PX2提个PX2即PX2(p-1)
又P+Q=1即PX2(p-1)=-PQX2
同理P2Y2-PY2=-PQY2
现在还有个2pqxy
好了，提个pq出来
再看看
里边是个完全平方，呵呵
结合pq为正数
就这样

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绛旓細绗1棰樻妱閿欎袱澶勶紝z搴旇鏄2锛寈搴旇鏄痑 -a+½<x<a+½瑙ｅ緱a=3/2 绗2棰鎶勯敊涓ゅ锛9搴旇鏄痲 鐢辫В闆嗗緱(x+3)(x-2)<0 鏁寸悊锛屽緱x²+x-6<0 p=1锛宷=-6 p+q=1+(-6)=-5