初三数学用因式分解法解下列方程
\u3010\u521d\u4e09\u6570\u5b66\u3011\u7528\u56e0\u5f0f\u5206\u89e3\u6cd5\u89e3\u4e0b\u5217\u65b9\u7a0b x^2=5x\u89e3\uff1ax^2=5x
x^2-5x=0
x(x-5)=0\u2014\u2014\u63d0\u53d6\u516c\u56e0\u5f0f\u6cd5
x=0\u6216x=5
\u6240\u4ee5\uff0c\u65b9\u7a0b\u7684\u89e3\u4e3a\uff1ax=0\u6216x=5\u3002
3x²-12x=-12
\u89e3\uff1ax²-4x+4 =0
\ufe59 X-2\ufe5a²=0
\u2234X1=X2=2
1.(x-2)(x-2)=0 ∴x=22.(x-6)(x+6)=0∴x=6或x=-63.(x-1)(3x-2)=0∴x=1或x=2/34.(x-4)(3x+2)=0∴x=4或x=-2/35.(3x-2)(2x+1)=0∴x=2/3或x=-1/26.(3x-3)(x-3)=0∴x=1或x=3
绛旓細1. 3锛坸^2-4x+4)=0 (x-2)^2=0, x=2,鎴栬厁=-22 4锛坸^2-36)=0 x=6鎴栬厁=-63 锛3x-2锛夛紙x-1锛=0 3x-2=0锛屾垨鑰厁-1=0 閭ｄ箞x=2/3锛屾垨鑰厁=14 锛2x-1+3-x)(2x-1-3+x)=0 (2x-2)(3x-4)=0 x=1,鎴栬厁=4/35 (3x-2)(2x...
绛旓細锛1锛墄²+10x+9=0 瑙ｏ細鍘鏂圭▼鍥犲紡鍒嗚В寰楋細(x+1)(x+9)=0 瑙ｅ緱x=-1鎴杧=-9 锛2锛墄²-x-7/4=0 瑙ｏ細鍘熸柟绋嬮厤鏂瑰緱锛歺²-x+ 1/4 -2=0 鍗(x- 1/2)²=2 x-1/2=鏍瑰彿2鎴杧-1/2=-鏍瑰彿2 鎵浠ュ師鏂圭▼鐨勮В涓猴細x=1/2 + 鏍瑰彿2 鎴杧=1/2 - 鏍瑰彿2 锛...
绛旓細鈭 x1=-2锛寈2=-12 鎵浠ワ紝鍘鏂圭▼鐨勮В涓簒1=-2锛寈2=-12 甯屾湜浣犺兘閲囩撼銆
绛旓細鍥炵瓟锛2銆戙 瑙:鍖栫畝寰, 3x²+2x-8=0 (3x-4)(x+2)=0 瑙ｅ緱,x1=4/3;x2= -2; 3銆戙 瑙:銆(x+2)-4銆²=0 銆恱-2銆²=0 瑙ｅ緱,x=2
绛旓細x^2-4x+4=0 (x-2)^2=0 x1=x2=2 3. 3x(x-1)=2(x-1)3x(x-1)-2(x-1)=0 (x-1)(3x-2)=0 x1=1, x2=2/3 4. (2x-1)^2=(3-x)^2 (2x-1)^2-(x-3)^2=0 [(2x-1)+(x-3)][(2x-1)-(x-3)]=0 (3x-4)(x+2)=0 x1=4/3, x2=-2 ...
绛旓細²=0 x=2 (7) (2x-1)^2= - 8 x鏃犺В (8) (x-2)^2=2x(x-2)x²-4x+4-2x²+4x=0 -x²+4=0 x²-4=0 (x+2)(x-2)=0 x1=-2,x2=2 2銆瑙ｆ柟绋 x^2-2鈭3 x= - 3 x^2-2鈭3 x +3=0 (x-鈭3)²=0 x=鈭3 ...
绛旓細5x(x-3)=2(3-x)5x(x-3)+2(x-3)=0 (x-3)(5x+2)=0 x-3=0鎴5x+2=0 x1=3;x2=-2/5 2. 9锛坸-2锛²=4x²[ 3锕檟-2锕歖²-( 2x)²=0 [ 3锕檟-2锕+2x][3锕檟-2锕-2x]=0 (5x-6)(x-6)=0 5x-6=0鎴杧-6=0 x1=6/5;x2=6 3. ...
绛旓細杩欐牱
绛旓細瑙:鐢遍鎰忥紝6x+15=4x²+10x 4x²+4x-15=0 (2x-3)(2x+5)=0 瑙ｅ緱锛歺=3/2 鎴 -5/2 鐢遍鎰忥紝3x²+2x-8=0 锛3x-4锛夛紙x+2锛=0 瑙ｅ緱锛歺=4/3 鎴 -2
绛旓細杩欏簲璇ユ槸鐢ㄩ厤鏂规硶锛岃屼笉鏄鍥犲紡鍒嗚В娉锛侊紒锛7銆亁^2+4x=2 ==> x^2+4x+4=2+4 ==> (x+2)^2=6 ==> x+2=卤鈭6 ==> x=-2卤鈭6 8銆(x-3)^2+4(x-3)-9=0 ==> (x-3)^2+4(x-4)+4=13 ==> [(x-3)+2]^2=13 ==> (x-1)^2=13 ==> x-1=卤鈭13 ==>...
