用因式分解法解下列方程；【2X+1]的平方=X的平方，求X的ŀ
用因式分解法解方程(2x+1)²=x²
(2x+1)²-x²=0
[(2x+1)+x][(2x+1)-x]=0
(3x+1)(x+1)=0
3x+1=0或者x+1=0
x=-1/3或者x=-1
x=-1
绛旓細宸茬粡鏄鍥犲紡 2x(x+5)=0 x=0 鎴 x+5=0 x=-5
绛旓細锛2x-1锛²=2x-1 锛2x-1锛²-锛2x-1锛=0 锛2x-1锛夛紙2x-1-1锛=0 锛2x-1锛夛紙2x-2锛=0 2x-1=0鎴2x-2=0 x=1/2锛2鍒嗕箣1锛夋垨x=1 鏈嬪弸锛岃閲囩撼姝ｇ‘绛旀锛屼綘浠彧鎻愰棶锛屼笉閲囩撼姝ｇ‘绛旀锛屽洖绛旈兘娌℃湁鍔诧紒锛侊紒鏈嬪弸锛岃銆愰噰绾崇瓟妗堛戯紝鎮ㄧ殑閲囩撼鏄垜绛旈鐨勫姩鍔涳紝濡傛灉娌℃湁鏄庣櫧...
绛旓細(2x-1)^2=(3-x)^2 (2x-1)^2-(3-x)^2=0 (2x-1+3-x)(2x-1-3+x)=0 (x+2)(3x-4)=0 x1=-2 x2=4/3
绛旓細1. X锛2X锛1锛夛紳3锛2X+1)X锛2X锛1锛-3锛2X+1)=0 (2X+1)(X-3)=0 X=-1/2,X=3 2. (X-2)^2=2-X (X-2)^2+(X-2)=0 (X-2)(X+1)=0 X=2,X=-1 3. X(2X-1)=3(1-2X)X(2X-1)+3(2X-1)=0 (2X-1)(X+3)=0 X=1/2,X=-3 4. X^2-3X-10=0 X^...
绛旓細瑙ｏ細宸﹁竟灞曞紑寰楋細2x�0�5-3x+1=3绉婚」寰楋細2x�0�5-3x-2=0鍒嗚В鍥犲紡寰:(x-2)(2x-1)=0鎵浠-2=0鎴2x-1=0鎵浠1=-0.5 x2=2
绛旓細1锛歽(y+5)=-25/4 y²+5y+25/4=0 (y+5/2)²=0 y=-5/2 2:3(x-3)=(x-3)^2;(x-3)(3-x+3)=0 (x-3)(6-x)=0 x=3 x=6 3:(2x+3)^2=24x 4x²+12x+9-24x=0 4x²-12x+9=0 (2x-3)²=0 x=3/2 4:1.2y-0.04=9y^2 9y...
绛旓細2x^2-3x-1=0 閿欎簡鍚 鏄笉鏄2x^2-3x+1=0锛(2x-1)(x-1)=0 x=1/2,x=1 濡傛灉娌￠敊 鍒檟=(3卤鈭17)/4 x^2-4x-12=0 (x-6)(x+2)=0 x=6,x=-2 (x-7)(x-8)=72 x^2-15x+56-72=0 x^2-15x-16=0 (x-16)(x+1)=0 x=16,x=-1 2x^2-3x-2=0 (2x+1)(x...
绛旓細(2x-1)²-(3-x)²=0 (2x-1+3-x)(2x-1-3+x)=0 (x+2)(3x-4)=0 x+2=0鎴3x-4=0 x1=-2锛寈2=4/3
绛旓細1.杩愮敤骞虫柟宸叕寮忓彲浠ヨ繘琛姹傝В锛
绛旓細瑙ｆ硶1鍥犲紡鍒嗚В 鐢2x(x+1)=3(x+1)寰2x(x+1)-3(x+1)=0 鍗筹紙x+1锛夛紙2x-3锛=0 瑙ｅ緱x=-1鎴杧=3/2 娉2鍏紡娉 鐢2x(x+1)=3(x+1)寰2x^2+2x=3x+3 鍗2x^2-x-3=0 鍏段=锛-1锛塣2-4*2*(-3)=25 鏁厁=锛1+鈭25锛/2脳2=3/2鎴杧=锛1-鈭25锛/2脳2=-1 閰嶆柟娉3...
