绝对值x1-x2等于什么 x1的绝对值=x2,为什么x1等于正负x2
x1-x2\u7684\u7edd\u5bf9\u503c\u4e00\u5143\u4e8c\u6b21\u65b9\u7a0b\u4e2d\u4ee3\u8868\u4ec0\u4e48\uff1f\u610f\u4e49\u662f\u4ec0\u4e48\uff1f\u7528\u6765\u6c42\u4ec0\u4e48\u7684\uff1f \u4e3a\u4ec0\u4e48\u7b49\u4e8ex1+x2\u7684\u5e73\u65b9\u97e6\u8fbe\u516c\u5f0f\u7684\u8fd0\u7528
x1-x2\u7684\u7edd\u5bf9\u503c\u4e00\u5143\u4e8c\u6b21\u65b9\u7a0b\u4e2d\u4ee3\u8868\u4e24\u4e2a\u6839\u5728\u6570\u8f74\u4e0a\u7684\u8ddd\u79bb\uff0c\u56e0\u4e3a\uff08X1+X2\uff09^2-4X1*X2=X1^2+X2^2-2X1*X2=(x1-x2)^2
\u6240\u4ee5\u5c31\u7528|(x1-x2)|=(\uff08X1+X2\uff09^2-4X1*X2)^1/2 \u518d\u7ed3\u5408\u97e6\u8fbe\u5b9a\u7406\u6765\u6c42\u51fax1-x2\u7684\u7edd\u5bf9\u503c
\u56e0\u4e3a\u7edd\u5bf9\u503c\u662f\u6b63\u7684
\u4f46\u662fX1\u5982\u679c\u662f\u8d1f\u7684\u8bdd\u53ea\u8981\u52a0\u7edd\u5bf9\u503c\u5c31\u53d8\u6b63\u7684\u4e86
\u6240\u4ee5\u6709+x1\u548c-x2
x1x2=c/a
所以(x1-x2)²=(x1+x2)²-4x1x2
=b²/a²-4c/a
=(b²-4ac)/a²
所以|x1-x2|=√(b²-4ac)/|a|
x1+x2=-b/a
x1x2=c/a
(x1+x2)^2=b^2/a^2
(x1-x2)^2=b^2/a^2-4c/a=b^2-4ac/a^2
绝对值x1-x2=√b^2-4ac/a的绝对值
即绝对值a分之√△
绝对值X1-X2 就等于X1-X2平方开根号.就等于(X1平方+X2平方+两倍X1乘以X2)开跟号.然后用韦达定理就OK啦
用韦达
定理做
绛旓細瑙:鈭祒1锛寈2婊¤冻缁濆鍊锛X1-X2锛=鈭5,鈭(x1-x2)^2=5,鈭祒1.x2骞冲潎鏁版槸x=(x1+x2)/2.鈭存柟宸=[(x1-x)^2+(x2-x)^2]/2 =[(x1-x2)^2/4+(x2-x1)^2/4]/2 =[2(x1-x2)^2/4]/2 =(x1-x2)^2/4 =5/4
绛旓細x1鈥攛2鐨缁濆鍊肩瓑浜鏍瑰彿涓嬶紙x1+x2锛夌殑骞虫柟鈥4x1x2鐒跺悗甯﹀叆鏃㈠彲銆
绛旓細鍥犱负 锛坸1+x2锛夌殑骞虫柟-4x1x2=(x1鐨勫钩鏂+x2鐨勫钩鏂+2x1x2)-4x1x2=x1鐨勫钩鏂+x2鐨勫钩鏂-2x1x2=(x1-x2)鐨勫钩鏂
绛旓細鍥犱负x1锛寈2鐨勫ぇ灏忓叧绯讳笉鏄庣‘锛屾墍浠x1锛峹2|锛6 涓よ竟骞虫柟寰 x1²锛2x1x2+x2²锛36 (x1+x2)²锛4x1x2锛36 x1+x2涓巟1x2鍙牴鎹牴涓庣郴鏁扮殑鍏崇郴寰楀埌
绛旓細闊﹁揪鍏紡鐨勮繍鐢 x1-x2鐨缁濆鍊涓鍏冧簩娆℃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-x2|=鈭氾紙x1+x2锛塣2-4x1x2 鐒跺悗澶ф鏄埄鐢ㄤ簩鍏冧竴娆℃柟绋嬫牴涓庣郴鏁扮殑鍏崇郴锛屽彲浠ュ緱 x1+x2=D x1x2=F 甯﹀叆鍗冲彲銆
绛旓細瀵逛簬ax2+bx+c锛x1-x2鐨缁濆鍊肩瓑浜鏍瑰彿涓媥1-x2鐨勫钩鏂癸紝鑰寈1-x2鐨勫钩鏂瑰睍寮鏄痻12+x22-2x1x2,涔熷氨鏄(x1+x2)2-4x1x2,鑰寈1+x2鍜寈1x2鍙敤闊﹁揪瀹氱悊琛ㄧず锛屽寲绠鍚庡緱瑙
绛旓細|x1-x2|=鈭(x1-x2)^2=鈭(x1^2-2x1x2+x2^2)=鈭(x1^2+2x1x2+x2^2-4x1x2)=鈭歔(x1+x2)^2-4x1x2]=鈭歔(-b/a)^2-4c/a]=鈭歔(b^2-4ac)/a^2]=鈭氣柍/|a|
绛旓細涓鍏冧簩娆℃柟绋媋x²+bx+c=0(a鈮0)x1+x2=-b/a x1*x2=c/a |x1-x2|=鈭(x1-x2)²=鈭歔(x1+x2)²-4x1x2]=鈭(b²/a²-4c/a)=[鈭(b²-4ac)]/|a|
绛旓細杩欏睘浜庨煢杈惧畾鐞嗙殑鎷撳睍銆 棣栧厛鍙互鐢ㄤ簬姹倈x1-x2| 鍏舵 鍦ㄨВ鏋愬嚑浣曚腑 甯稿簲鐢 灏ゅ叾鏄湪鐩寸嚎鍜屾洸绾跨浉浜ゆ眰浜ょ偣闂寸嚎娈甸暱搴︽椂銆
