因式分解2x^2－5x-3
2x^2-5x-3 \u56e0\u5f0f\u5206\u89e3\u5341\u5b57\u76f8\u4e58\u6cd5
\uff082x+1\uff09(x-3)
\uff082x-1)(x+3)=0
2x-1=0 x=1/2
x+3=0
x=-3
应该是(2x+1)(x-3)
=(2x+1)(x-3)
=2x²-6x+x-3
=2x(x-3)+(x-3)
=(x-3)(2x+1)
(x-3)(2x+1)
绛旓細2x^2-5x-3 =2x^2-6x+x-3 =2x(x-3)+(x-3)=(x-3)(2x+1)
绛旓細2x鐨勫钩鏂-5x-3鐢ㄥ崄瀛楃浉涔樻硶鍥犲紡鍒嗚В 瑙ｏ細2 1 脳 1 -3 鎵浠 鍘熷紡=(2x+1)(x-3)
绛旓細2x`2-5x-3 x -3 2x 1 鍙互寰楀嚭 (x-3)(2x+1) 鍦ㄧ嚎鏁欎綘
绛旓細鍥犲紡鍒嗚В娉曪細2x^2-5x-3 =0 (x-3) (2 x+1) =0 x=-1/2 鎴 x=3 鍏紡娉曪細a=2锛宐=-5锛宑=-3 鈻=b^2-4ac =49 浠ｅ叆鍏紡x=(-b卤鈭∆)/2a 鎵浠=-1/2 鎴 x=3
绛旓細绛旀涓(2x+1)(x-3)
绛旓細2x^2-5x-3=0 2x^2+x-6x-3=0 x(2x+1)-3(2x+1)=0 (2x+1)(x-3)=0 x1=3 x2=-1/2
绛旓細锛2x+1)(x-3)=0 2 1 1 -3 杩欐牱鍥犲紡鍒嗚Вx^2鐨勭郴鏁板氨鏄乏杈逛袱浣嶆暟鐩镐箻鎵寰楋紝甯搁噺灏辨槸鍙宠竟涓や綅鏁扮浉涔橈紝x鐨勭郴鏁板氨鏄氦鍙夌浉涔樺啀鐩稿姞锛屽嵆2*锛-3锛+1*1=-5 鍙互杩欐牱鐪 2x 1 x -3 杩欐牱骞宠鍐欏氨鐨勶紙2x+1)(x-3)=0
绛旓細2x²-5x-3=0 寰椼2x+1銆戙恱-3銆=0 x=-1/2鎴3 婊℃剰璇烽噰绾崇瓟妗堚濄
绛旓細鎵浠2x^2-5x-3=(x-3)(2x+1)(x-3) 甯屾湜瀵逛綘鏈夊府鍔.,1,鎶婂師寮鍒嗚В鎴 2 -1 浜ゅ弶鐩镐箻 锛岀劧鍚庣浉鍔犵粨鏋滀负璐5绛変簬浜屾椤圭郴鏁 鎵浠 缁撴灉涓猴紙2x-1)(x-3)!!! 鐧惧害鐧剧涓婃湁锛 杈撳叆鍗佸瓧鐩镐箻娉!1 3,2,=锛2X+1锛夛紙X-3锛,1,2x^2-5x-3 =(2x+1)(x-3),0,鐢ㄥ崄瀛楃浉涔樻硶鍒嗚В涓嬪垪澶氶」...
绛旓細瑙ｏ細鍥犲紡鍒嗚В 锛2x+1锛夛紙x-3锛=0 瑙ｅ緱x=-1/2鎴杧=3 ~濡傛灉浣犺鍙垜鐨勫洖绛旓紝璇峰強鏃剁偣鍑汇愰噰绾充负婊℃剰鍥炵瓟銆戞寜閽畘~鎵嬫満鎻愰棶鐨勬湅鍙嬪湪瀹㈡埛绔彸涓婅璇勪环鐐广愭弧鎰忋戝嵆鍙倊浣犵殑閲囩撼鏄垜鍓嶈繘鐨勫姩鍔泘~O(鈭鈭)O锛屼簰鐩稿府鍔╋紝绁濆叡鍚岃繘姝ワ紒
