x1-x二的绝对值等于
韦达公式的运用x1-x2的绝对值一元二次方程中代表两个根在数轴上的距离，因为（X1+X2）^2-4X1*X2=X1^2+X2^2-2X1*X2=(x1-x2)^2所以就用|(x1-x2)|=(（X1+X2）^2-4X1*X2)^1/
2 再结合韦达定理来求出x1-x2的绝对值
绛旓細|x1-x2|=鈭(x1-x2)²=鈭歔(x1+x2)²-4x1x2]=鈭(b²/a²-4c/a)=[鈭(b²-4ac)]/|a| 鏍规嵁涓鍏冧簩娆℃柟绋媋x²+bx+c=0(a鈮0)x1+x2=-b/a x1*x2=c/a 鎵浠 |x1-x2|=鈭(x1-x2)²=鈭歔(x1+x2)²-4x1x2]=鈭(b²/...
绛旓細x1鍑x2鐨勭粷瀵瑰=[鈭(b²-4ac)]/|a| |x1-x2|=鈭(x1-x2)²=鈭歔(x1+x2)²-4x1x2]=鈭(b²/a²-4c/a)=[鈭(b²-4ac)]/|a| 鏍规嵁涓鍏冧簩娆℃柟绋媋x²+bx+c=0(a鈮0)x1+x2=-b/a x1*x2=c/a 鎵浠 |x1-x2|=鈭(x1-x2)²=...
绛旓細|x1-x2|銆傛牴鎹浉鍏宠祫鏂欐煡璇紝x1-x浜岀殑缁濆鍊肩瓑浜巪x1-x2|锛岀粷瀵瑰兼槸鎸囦竴涓暟鍦ㄦ暟杞翠笂鎵瀵瑰簲鐐瑰埌鍘熺偣鐨勮窛绂伙紝鐢ㄢ渱|鈥濇潵琛ㄧず銆
绛旓細x1鍑x2鐨勭粷瀵瑰鍏紡锛殀x1-x2|=鈭(x1-x2)2=鈭歔(x1+x2)2-4x1x2]=鈭(b2/a2-4c/a)=[鈭(b2-4ac)]/|a|銆傛垨鑰呬辅x1-x2涓2=锛坸1-x2)2=(x1+x2)2-4x1x2銆備竴鍏冧簩娆℃柟绋嬶細ax2+bx+c=0(a鈮0)锛寈1+x2=-b/a锛寈1*x2=c/a銆
绛旓細浣犳槸鎯抽棶x1-x浜岀殑缁濆鍊兼槸浠涔堝悧锛焫1-x浜岀殑缁濆鍊兼槸|x1-x2|=鈭(x1-x2)?=鈭歔(x1+x2)?-4x1x2]=鈭(b?/a?-4c/a)=[鈭(b?-4ac)]/|a|銆傜粷瀵瑰兼槸鎸囦竴涓暟鍦ㄦ暟杞翠笂鎵瀵瑰簲鐐瑰埌鍘熺偣鐨勮窛绂伙紝鐢ㄢ渱|鈥濇潵琛ㄧず銆
绛旓細(x1涓x2)鐨勭粷瀵瑰 鎴戞潵绛 1涓洖绛 #鐑# 浣犲彂鏈嬪弸鍦堜細浣跨敤閮ㄥ垎浜哄彲瑙佸姛鑳藉悧? 鍖垮悕鐢ㄦ埛 2014-07-16 灞曞紑鍏ㄩ儴 鏇村杩介棶杩界瓟 杩界瓟 鑲畾瀵 璇烽噰绾 宸茶禐杩 宸茶俯杩< 浣犲杩欎釜鍥炵瓟鐨勮瘎浠锋槸? 璇勮 鏀惰捣 涓轰綘鎺ㄨ崘: 鐗瑰埆鎺ㄨ崘 鎽搁奔涓轰粈涔堜篃浼氱劍铏? 涓鍒嗛挓鐨凪BTI鎬ф牸娴嬭瘯闈犺氨鍚? 涓婃捣涓轰粈涔堟湁...
绛旓細x1+x2=-b/a x1x2=c/a 鎵浠(x1-x2)²=(x1+x2)²-4x1x2 =b²/a²-4c/a =(b²-4ac)/a²鎵浠x1-x2|=鈭(b²-4ac)/|a|
绛旓細闊﹁揪鍏紡鐨勮繍鐢x1-x2鐨勭粷瀵瑰涓鍏冧簩娆℃柟绋嬩腑浠ｈ〃涓や釜鏍瑰湪鏁拌酱涓婄殑璺濈锛屽洜涓猴紙X1+X2锛塣2-4X1*X2=X1^2+X2^2-2X1*X2=(x1-x2)^ 2鎵浠ュ氨鐢▅(x1-x2)|=(锛圶1+X2锛塣2-4X1*X2)^1/ 2 鍐嶇粨鍚堥煢杈惧畾鐞嗘潵姹傚嚭x1-x2鐨勭粷瀵瑰 ...
绛旓細x1锛峹2鐨勭粷瀵瑰锛岀敤缁濆鍊肩鍙疯〃绀哄氨鏄瘄x1-x2| 缁濆鍊硷細缁濆鍊兼槸鎸囦竴涓暟鍦ㄦ暟杞翠笂鎵瀵瑰簲鐐瑰埌鍘熺偣鐨勮窛绂诲彨鍋氳繖涓暟鐨勭粷瀵瑰硷紝缁濆鍊肩敤鈥渱 |鈥濇潵琛ㄧず銆倈b-a|鎴東a-b|琛ㄧず鏁拌酱涓婅〃绀篴鐨勭偣鍜岃〃绀篵鐨勭偣鐨勮窛绂汇
绛旓細鐢变笉绛夊紡1寰梮灏忎簬4鍒嗕箣63锛涚敱鏂圭▼缁1銆2寰x1绛変簬璐2鍒嗕箣5锛x2绛変簬4鍒嗕箣11锛屼袱涓В鍧囨弧瓒充笉绛夊紡锛屾晠閮芥垚绔嬶紒
