高一一道不等式填空题求一个清晰的阶梯思路 求高中不等式题目及答案
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x=8y/(y-2)\uff1d8+16/(y-2)
x+y=y+16/(y-2)+8
=(y-2)+16/(y-2) +10
\u22658+10=18
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(1)\u8bc1\u660e\uff1aniA \uff1cmiA \uff1b(2)\u8bc1\u660e\uff1a(1+m)n\uff1e(1+n)m8.(\u2605\u2605\u2605\u2605\u2605)\u82e5a\uff1e0\uff0cb\uff1e0\uff0ca3+b3=2\uff0c\u6c42\u8bc1\uff1aa+b\u22642\uff0cab\u22641. \u53c2\u8003\u7b54\u6848 \u8bc1\u6cd5\u4e00\uff1a(\u5206\u6790\u7efc\u5408\u6cd5\uff09\u6b32\u8bc1\u539f\u5f0f\uff0c\u5373\u8bc14(ab)2+4(a2+b2)\uff0d25ab+4\u22650\uff0c\u5373\u8bc14(ab)2\uff0d33(ab)+8\u22650\uff0c\u5373\u8bc1ab\u2264 \u6216ab\u22658.\u2235a\uff1e0\uff0cb\uff1e0\uff0ca+b=1\uff0c\u2234ab\u22658\u4e0d\u53ef\u80fd\u6210\u7acb\u22351=a+b\u22652 \uff0c\u2234ab\u2264 \uff0c\u4ece\u800c\u5f97\u8bc1.\u8bc1\u6cd5\u4e8c\uff1a(\u5747\u503c\u4ee3\u6362\u6cd5)\u8bbea= +t1\uff0cb= +t2.\u2235a+b=1\uff0ca\uff1e0\uff0cb\uff1e0\uff0c\u2234t1+t2=0\uff0c|t1|\uff1c \uff0c|t2|\uff1c \u663e\u7136\u5f53\u4e14\u4ec5\u5f53t=0\uff0c\u5373a=b= \u65f6\uff0c\u7b49\u53f7\u6210\u7acb.\u8bc1\u6cd5\u4e09\uff1a(\u6bd4\u8f83\u6cd5\uff09\u2235a+b=1\uff0ca\uff1e0\uff0cb\uff1e0\uff0c\u2234a+b\u22652 \uff0c\u2234ab\u2264 \u8bc1\u6cd5\u56db\uff1a(\u7efc\u5408\u6cd5)\u2235a+b=1\uff0c a\uff1e0\uff0cb\uff1e0\uff0c\u2234a+b\u22652 \uff0c\u2234ab\u2264 .\u8bc1\u6cd5\u4e94\uff1a(\u4e09\u89d2\u4ee3\u6362\u6cd5\uff09\u2235 a\uff1e0\uff0cb\uff1e0\uff0ca+b=1\uff0c\u6545\u4ee4a=sin2\u03b1\uff0cb=cos2\u03b1\uff0c\u03b1\u2208(0\uff0c )2 \u4e00\u30011.\u89e3\u6790\uff1a\u4ee4 =cos2\u03b8\uff0c =sin2\u03b8\uff0c\u5219x=asec2\u03b8\uff0cy=bcsc2\u03b8\uff0c\u2234x+y=asec2\u03b8+bcsc2\u03b8=a+b+atan2\u03b8+bcot2\u03b8\u2265a+b+2 .\u7b54\u6848\uff1aa+b+2 2.\u89e3\u6790\uff1a\u75310\u2264|a\uff0dd|\uff1c|b\uff0dc| (a\uff0dd)2\uff1c(b\uff0dc)2 (a+b)2\uff0d4ad\uff1c(b+c)2\uff0d4bc�\u2235a+d=b+c\uff0c\u2234\uff0d4ad\uff1c\uff0d4bc\uff0c\u6545ad\uff1ebc.\u7b54\u6848\uff1aad\uff1ebc3.\u89e3\u6790\uff1a\u628ap\u3001q\u770b\u6210\u53d8\u91cf\uff0c\u5219m\uff1cp\uff1cn\uff0cm\uff1cq\uff1cn.\u7b54\u6848\uff1am\uff1cp\uff1cq\uff1cn\u4e8c\u30014.(1)\u8bc1\u6cd5\u4e00\uff1aa2+b2+c2\uff0d = (3a2+3b2+3c2\uff0d1)= \uff3b3a2+3b2+3c2\uff0d(a+b+c)2\uff3d= \uff3b3a2+3b2+3c2\uff0da2\uff0db2\uff0dc2\uff0d2ab\uff0d2ac\uff0d2bc\uff3d= \uff3b(a\uff0db)2+(b\uff0dc)2+(c\uff0da)2\uff3d\u22650 \u2234a2+b2+c2\u2265 \u8bc1\u6cd5\u4e8c\uff1a\u2235(a+b+c)2=a2+b2+c2+2ab+2ac+2bc\u2264a2+b2+c2+a2+b2+a2+c2+b2+c2\u22343(a2+b2+c2)\u2265(a+b+c)2=1 \u2234a2+b2+c2\u2265 \u8bc1\u6cd5\u4e09\uff1a\u2235 \u2234a2+b2+c2\u2265 \u2234a2+b2+c2\u2265 \u8bc1\u6cd5\u56db\uff1a\u8bbea= +\u03b1\uff0cb= +\u03b2\uff0cc= +\u03b3.\u2235a+b+c=1\uff0c\u2234\u03b1+\u03b2+\u03b3=0\u2234a2+b2+c2=( +\u03b1)2+( +\u03b2)2+( +\u03b3)2= + (\u03b1+\u03b2+\u03b3)+\u03b12+\u03b22+\u03b32= +\u03b12+\u03b22+\u03b32\u2265 \u2234a2+b2+c2\u2265 \u2234\u539f\u4e0d\u7b49\u5f0f\u6210\u7acb.\u8bc1\u6cd5\u4e8c\uff1a \u2234 \u2264 \uff1c6\u2234\u539f\u4e0d\u7b49\u5f0f\u6210\u7acb.5.\u8bc1\u6cd5\u4e00\uff1a\u7531x+y+z=1\uff0cx2+y2+z2= \uff0c\u5f97x2+y2+(1\uff0dx\uff0dy)2= \uff0c\u6574\u7406\u6210\u5173\u4e8ey\u7684\u4e00\u5143\u4e8c\u6b21\u65b9\u7a0b\u5f97\uff1a2y2\uff0d2(1\uff0dx)y+2x2\uff0d2x+ =0\uff0c\u2235y\u2208R\uff0c\u6545\u0394\u22650\u22344(1\uff0dx)2\uff0d4\u00d72(2x2\uff0d2x+ )\u22650\uff0c\u5f970\u2264x\u2264 \uff0c\u2234x\u2208\uff3b0\uff0c \uff3d\u540c\u7406\u53ef\u5f97y\uff0cz\u2208\uff3b0\uff0c \uff3d\u8bc1\u6cd5\u4e8c\uff1a\u8bbex= +x\u2032\uff0cy= +y\u2032\uff0cz= +z\u2032\uff0c\u5219x\u2032+y\u2032+z\u2032=0\uff0c\u4e8e\u662f =( +x\u2032)2+( +y\u2032)2+( +z\u2032)2= +x\u20322+y\u20322+z\u20322+ (x\u2032+y\u2032+z\u2032)= +x\u20322+y\u20322+z\u20322\u2265 +x\u20322+ = + x\u20322\u6545x\u20322\u2264 \uff0cx\u2032\u2208\uff3b\uff0d \uff0c \uff3d\uff0cx\u2208\uff3b0\uff0c \uff3d\uff0c\u540c\u7406y\uff0cz\u2208\uff3b0\uff0c \uff3d\u8bc1\u6cd5\u4e09\uff1a\u8bbex\u3001y\u3001z\u4e09\u6570\u4e2d\u82e5\u6709\u8d1f\u6570\uff0c\u4e0d\u59a8\u8bbex\uff1c0\uff0c\u5219x2\uff1e0\uff0c =x2+y2+z2\u2265x2+ \uff1e \uff0c\u77db\u76fe.x\u3001y\u3001z\u4e09\u6570\u4e2d\u82e5\u6709\u6700\u5927\u8005\u5927\u4e8e \uff0c\u4e0d\u59a8\u8bbex\uff1e \uff0c\u5219 =x2+y2+z2\u2265x2+ =x2+ = x2\uff0dx+ = x(x\uff0d )+ \uff1e \uff1b\u77db\u76fe.\u6545x\u3001y\u3001z\u2208\uff3b0\uff0c \uff3d\u2235\u4e0a\u5f0f\u663e\u7136\u6210\u7acb\uff0c\u2234\u539f\u4e0d\u7b49\u5f0f\u5f97\u8bc1.7.\u8bc1\u660e\uff1a(1)\u5bf9\u4e8e1\uff1ci\u2264m\uff0c\u4e14A =m\u00b7\u2026\u00b7(m\uff0di+1)\uff0c\uff0c\u7531\u4e8em\uff1cn\uff0c\u5bf9\u4e8e\u6574\u6570k=1\uff0c2\uff0c\u2026\uff0ci\uff0d1\uff0c\u6709 \uff0c\u6240\u4ee5 (2)\u7531\u4e8c\u9879\u5f0f\u5b9a\u7406\u6709\uff1a(1+m)n=1+C m+C m2+\u2026+C mn\uff0c(1+n)m=1+C n+C n2+\u2026+C nm\uff0c\u7531(1)\u77e5miA \uff1eniA (1\uff1ci\u2264m \uff0c\u800cC = \u2234miCin\uff1eniCim(1\uff1cm\uff1cn \u2234m0C =n0C =1\uff0cmC =nC =m\u00b7n\uff0cm2C \uff1en2C \uff0c\u2026\uff0cmmC \uff1enmC \uff0cmm+1C \uff1e0\uff0c\u2026\uff0cmnC \uff1e0\uff0c\u22341+C m+C m2+\u2026+C mn\uff1e1+C n+C2mn2+\u2026+C nm\uff0c\u5373(1+m)n\uff1e(1+n)m\u6210\u7acb.8.\u8bc1\u6cd5\u4e00\uff1a\u56e0a\uff1e0\uff0cb\uff1e0\uff0ca3+b3=2\uff0c\u6240\u4ee5(a+b)3\uff0d23=a3+b3+3a2b+3ab2\uff0d8=3a2b+3ab2\uff0d6=3\uff3bab(a+b)\uff0d2\uff3d=3\uff3bab(a+b)\uff0d(a3+b3)\uff3d=\uff0d3(a+b)(a\uff0db)2\u22640.\u5373(a+b)3\u226423\uff0c\u53c8a+b\uff1e0\uff0c\u6240\u4ee5a+b\u22642\uff0c\u56e0\u4e3a2 \u2264a+b\u22642\uff0c\u6240\u4ee5ab\u22641.\u8bc1\u6cd5\u4e8c\uff1a\u8bbea\u3001b\u4e3a\u65b9\u7a0bx2\uff0dmx+n=0\u7684\u4e24\u6839\uff0c\u5219 \uff0c\u56e0\u4e3aa\uff1e0\uff0cb\uff1e0\uff0c\u6240\u4ee5m\uff1e0\uff0cn\uff1e0\uff0c\u4e14\u0394=m2\uff0d4n\u22650 \u2460\u56e0\u4e3a2=a3+b3=(a+b)(a2\uff0dab+b2)=(a+b)\uff3b(a+b)2\uff0d3ab\uff3d=m(m2\uff0d3n)\u6240\u4ee5n= \u2461\u5c06\u2461\u4ee3\u5165\u2460\u5f97m2\uff0d4( )\u22650\uff0c\u5373 \u22650\uff0c\u6240\u4ee5\uff0dm3+8\u22650\uff0c\u5373m\u22642\uff0c\u6240\u4ee5a+b\u22642\uff0c\u75312\u2265m \u5f974\u2265m2\uff0c\u53c8m2\u22654n\uff0c\u6240\u4ee54\u22654n\uff0c\u5373n\u22641\uff0c\u6240\u4ee5ab\u22641.\u8bc1\u6cd5\u4e09\uff1a\u56e0a\uff1e0\uff0cb\uff1e0\uff0ca3+b3=2\uff0c\u6240\u4ee52=a3+b3=(a+b)(a2+b2\uff0dab)\u2265(a+b)(2ab\uff0dab)=ab(a+b)\u4e8e\u662f\u67096\u22653ab(a+b)\uff0c\u4ece\u800c8\u22653ab(a+b)+2=3a2b+3ab2+a3+b3=�(a+b)3\uff0c\u6240\u4ee5a+b\u22642\uff0c(\u4e0b\u7565\uff09\u8bc1\u6cd5\u56db\uff1a\u56e0\u4e3a \u22650\uff0c\u6240\u4ee5\u5bf9\u4efb\u610f\u975e\u8d1f\u5b9e\u6570a\u3001b\uff0c\u6709 \u2265 \u56e0\u4e3aa\uff1e0\uff0cb\uff1e0\uff0ca3+b3=2\uff0c\u6240\u4ee51= \u2265 \uff0c\u2234 \u22641\uff0c\u5373a+b\u22642\uff0c(\u4ee5\u4e0b\u7565\uff09\u8bc1\u6cd5\u4e94\uff1a\u5047\u8bbea+b\uff1e2\uff0c\u5219a3+b3=(a+b)(a2\uff0dab+b2)=(a+b)\uff3b(a+b)2\uff0d3ab\uff3d\uff1e(a+b)ab\uff1e2ab\uff0c\u6240\u4ee5ab\uff1c1\uff0c\u53c8a3+b3=(a+b)\uff3ba2\uff0dab+b2\uff3d=(a+b)\uff3b(a+b)2\uff0d3ab\uff3d\uff1e2(22\uff0d3ab)\u56e0\u4e3aa3+b3=2\uff0c\u6240\u4ee52\uff1e2(4\uff0d3ab)\uff0c\u56e0\u6b64ab\uff1e1\uff0c\u524d\u540e\u77db\u76fe\uff0c\u6545a+b\u22642(\u4ee5\u4e0b\u7565)
你好，
对于分式不等式，解法并不复杂，但是一定要有正确的解题思路和顺序。
总体来说，我想说两点：分类讨论的思想以及一些特定的处理方式。
先谈比较固定的一类问题，就是变量在分母上的不等式，是分式不等式的一种形式。记住一个大前提：不能直接将分母乘出来或者随意等号左右两边移项，除非正负可以判断。
当然 ，最完整的步骤是首先通过式子本身拿出最明了而又最重要的取值范围，就是分母不为0，由此算出第一个限制条件，下面的解答中这条不是主要说的，不单独计算这一点，但要记住。
具体看题目，比如第3题，这道题目的分子给的很巧可以利用“十字相乘”处理成
(x+2)(2x-1)，结合分母3x+5，注意！！这时不可以将分母直接乘到右边，而是结合分子分母考虑：要使式子大于0，相当于(x+2)(2x-1)和(3x+5)的乘积大于0，或(x+2)(2x-1)小于0且(3x+5)小于0，很自然，可以在数轴上取出-2,1/2，-5/3，三个数形成了四个区间段，分别判断每个式子正负，就能定出范围，最后结果是：{x|-2<x<-5/或x>1/2}
再如第4题，结合第四题谈一下上面提到的固定处理方法：只考虑一边1/（x-1）<3,对于这种类型的式子，正确的做法是把右边的3移到左边进行通分：化成(4-3x)/(x-1)<0,然后采用刚才的方法，考察(4-3x)和(x-1)乘积小于0的结果。同样地，另半边不等式也这样处理，自己考虑一下。
再看选择题，这道题就很好地诠释了分类讨论思想的重要性：
这样考虑，讨论x的范围：若x>0，把分母上的x乘到左边，得到x平方大于1，解得x大于1或x小于-1，所以x大于1；若x<0，同样处理，得到x平方小于1，x大于-1小于1，所以x大于-1小于0.综上，-1<x<0或x>1。
几道题目答案可能有问题，可以再跟老师协商，不过要读好题目，填空题的要求问的是集合，不要写成区间的形式。
总之注意上面这些解题顺序和思想，再多加练习，不等式还是很明了的。
希望这些会帮助你，下面是一张关于分式不等式的配图。
三类分式不等式。
【1】图片来自网络
绛旓細(x+2)(2x-1)锛岀粨鍚堝垎姣3x+5锛屾敞鎰忥紒锛佽繖鏃朵笉鍙互灏嗗垎姣嶇洿鎺ヤ箻鍒板彸杈癸紝鑰屾槸缁撳悎鍒嗗瓙鍒嗘瘝鑰冭檻锛氳浣垮紡瀛愬ぇ浜0锛岀浉褰撲簬(x+2)(2x-1)鍜(3x+5)鐨勪箻绉ぇ浜0锛屾垨(x+2)(2x-1)灏忎簬0涓(3x+5)灏忎簬0锛屽緢鑷劧锛屽彲浠ュ湪鏁拌酱涓婂彇鍑-2,1/2锛-5/3锛屼笁涓暟褰㈡垚浜嗗洓涓尯闂存锛屽垎鍒垽鏂瘡涓紡瀛愭璐燂紝...
绛旓細2^a + 2^b = 2^(a+b) ...锛1锛2^a + 2^b + 2^c = 2^(a+b+c) ...锛2锛夋牴鎹潎鍊涓嶇瓑寮寰 2^a + 2^b 鈮 2鈭歔2^(a+b)]鍙2^a + 2^b = 2^(a+b)鈭 2^(a+b) 鈮 2 * 鈭歔2^(a+b)]鍗2^(a+b) 鈮 2 * 2^[(a+b)/2] = 2^[(a+b+2)/2...
绛旓細鍥炵瓟锛氭牴鎹熀鏈涓嶇瓑寮,寰梮+鈮4,褰撲笖浠呭綋x=2鏃,鍑芥暟f(x)=x+4/x鐨勬渶灏忓间负m=4.鐢辨寰楀埌鍑芥暟鍦▁鈭圼1,3]鏃跺厛鍑忓悗澧,姣旇緝f(1)涓巉(3)鐨勫ぇ灏忓緱鍒板嚱鏁扮殑鏈澶у糓=5, 鍗斥埖x鈭圼1,3]涓烘鏁 鈭磝+4/x鈮2=4 褰撲笖浠呭綋x=2鏃,鍑芥暟f(x)=x+4/x鐨勬渶灏忓间负m=4, 鐢辨鍙緱鍑芥暟鍦(1,2)...
绛旓細14,16 15, -1 -1 16,涓嶇瓑寮鏃犺В 17,30.25
绛旓細鎵浠3=2a²+(1+b²)鈮2鈭歔2a²脳(1+b²)]鍗斥垰a²(1+b²)=a鈭(1+b²)鈮3/2鈭2=3鈭2/4 鎵浠鈭(1+b²)鐨勬渶澶у兼槸3鈭2/4銆13.宸茬煡5/x+3/y=2锛岋紙x>0锛寉>0锛夛紝鍒檟y鐨勬渶灏忓兼槸 瑙ｏ細鈭5/x+3/y=2鈮2鈭(15/xy)鍗1鈮15...
绛旓細瑙ｆ瀽锛 鐢遍鎰忚涓鏀挗绗旂殑浠峰间负x鍏冿紝涓鏀搮绗旂殑浠峰间负y鍏冿紝閭ｄ箞鍙垪涓嶇瓑寮缁勫緱锛歿 6x+3y>24 (1){ 4x+5y<22 (2)(1)脳2鍙緱锛12x+6y>48 锛3锛(2)寮忎袱杈瑰悓涔樹互-3锛屽彲寰楋細-12x-15y>-66 閭ｄ箞锛12x+6y+(-12x-15y)>48+(-66)鍗-9y>-18 鎵浠ワ細y<2 鑰 (2)寮忓彲鍖栦负...
绛旓細1,鍥犱负鏄涓嶇瓑寮鏄眰灏忎簬闆剁殑锛屾墍浠ョ嚎搴旇浠庢渶宸 绔偣鐨勪笅鏂瑰紑濮嬬┛锛屼緷娆＄┛杩囨瘡涓偣锛岀┛濂戒箣鍚庡氨寰堢洿瑙傜殑鐪嬪嚭锛屽湪X杞翠笅鏂圭殑鍖哄煙锛屽氨 鏄笉绛夊紡鐨勮В锛屽嵆锛堣礋鏃犵┓锛孉)鍜岋紙0,1)2.鍚岀悊锛屽綋0<A<1鏃讹紝涓嶇瓑寮忕殑瑙ｆ槸锛堣礋鏃犵┓锛0锛夊拰锛圓,1)3,褰揂>1鏃讹紝涓嶇瓑寮忕殑瑙ｆ槸锛堣礋鏃犵┓锛0锛夊拰锛1锛孉)...
绛旓細涓嶆垚绔嬶紝鏃犺В锛涘綋x-1<0鏃讹紝ax-a+1>0锛屽嵆 ax>a-1 璁総=锛坅-1锛/a=1-1/a 褰揳>0鏃讹紝x>t锛 0<t<1锛屽垯 1-1/a <x< 1锛涘綋a<0鏃讹紝x<t锛宼>1锛寈<1锛涘綋a=0鏃讹紝1>0锛屾垚绔嬶紱鍥犱负鏄濉┖棰锛岄鐩笂娌℃湁璇存槑a鐨勫彇鍊艰寖鍥达紝鍒欐妸瀹冭涓烘槸浠绘剰瀹炴暟锛屾墍浠 1-1/a <x< 1 ...
绛旓細鏁皔=sin^2x+4/(sin^2x)锛5/4脳锛1锛峜os2x锛/2 锛5/8脳(1-cos2x)鍥犱负锛-1鈮os2x鈮1 鎵浠ワ紝0鈮鈮5/4 鎵浠ワ紝y鐨勫煎煙涓篬0锛5/4]
绛旓細瑙ｉ泦涓篟锛屽垯m搴斿皬浜0锛屼笖鏂圭▼mx^2+2(m+1)x+9m+4=0濮嬬粓鏈夎В 4(m+1)^2-4m(9m+4)>=0 4m^2+8m+4-36m^2-16m>=0 32m^2+8m-4<=0 8m^2+2m-1<=0 (2m+1)(4m-1)<=0 2m+1>=0,4m-1<=0,m>=-1/2,m<1/4,-1/2<=m<0 2m+1<=0,4m-1>=0,鏃犺В 鎵浠-1/2<...
