x的平方加2x减3等于0,求解,求步骤 x的平方加2x减三等于零用配方法解
x\u7684\u5e73\u65b9\u51cf2x\u52a03\u7b49\u4e8e\u96f6\u7528\u914d\u65b9\u6cd5\uff0c\u600e\u4e48\u505a\uff1f\u8fd9\u9898\u6570\u5b57\u6709\u95ee\u9898\u2026\uff0c\u53ef\u80fd\u662f\u51cf3\uff0c\u52a0\u679c\u662f\u52a03\uff0c\u6b64\u9898\u6ca1\u6709\u5b9e\u6570\u6839
x\uff3e2-2x+3\uff1d0
x\uff3e2-2x+1+2\uff1d0
\uff08x-1\uff09\uff3e2+2\uff1d0
\u56e0\u4e3a\uff08x-1\uff09\uff3e2\u22650\uff0c\u6240\u4ee5\uff08x-1\uff09\uff3e2+2\u22652\uff0c\u4e0d\u53ef\u80fd\u7b49\u4e8e\u96f6
\u89c1\u4e0b\u56fe
x²+2x-3=0
解:原不等式可化为
(x+3)(x-1)=0
x1=-3,或x2=1
绛旓細x²+2x+3=0 x²+2x+1+2=0 x²+2x+1=-2 (x+1)²=-2,鏃犲疄鏁拌В.濡傛灉鍘熼涓锛歺²+2x-3=0 x²+2x+1-4=0 x²+2x+1=4 (x+1)²=4 x+1=卤2 x1=2-1=1锛泋2=-2-1=-3锛...
绛旓細x鐨勫钩鏂鍑2x鍑3绛変簬0 x=3锛x=-1 2x鍑14x鍔5 = 5-12x 绛変簬璐31锛屾垨鑰呰礋7
绛旓細x鐨勫钩鏂鍑2x鍑3绛変簬0 (x-3)(x+1)=0 x1=3 x2=-1
绛旓細X绛変簬-1鎴3 X鐨勫钩鏂鐨勫父鏁伴」鏄1锛2X鐨勫父鏁伴」鏄2锛屽父鏁伴」涓3锛b鏄2X鐨勫父鏁伴」 a鏄疿骞虫柟鐨勫父鏁伴」 c涓哄父鏁伴」 鎵浠ワ紝鍙敤b鐨勫钩鏂鍑忓幓4鍊嶇殑ac 鍙緱锛宐鐨勫钩鏂光4ac 绛変簬16 鍥犱负锛孹绛変簬-b鍔犲噺4ac闄や互2a 灏哸bc 浠e叆鍙緱X绛変簬-1鎴3
绛旓細鍥炵瓟锛歺^2-2x-3=0 (x+1)(x-3)=0 x+1=0 x-3=0 x1=-1 x2=3
绛旓細瑙o細X²-2X-3=0 (X+1)(X-3)=0 X+1=0鎴朮-3=0 X=-1鎴朮=3 鎵浠ュ師鏂圭▼鐨勮В涓篨1=-1,X2=3
绛旓細x^2-2x-3=0 鍙樺舰寰楋細(x-3)(x+1)=0锛瑙e緱锛歺=3鎴杧=-1
绛旓細x²-2x-3=0 (x+1)(x-3)=0 x=-1鎴杧=3
绛旓細x²-2x-3=0 (x+1)(x-3)=0 x1=-1锛寈2=3 (x1)²x2+(x2)²x1 =(-1)²*3+3²*(-1)=1*3-9*1 =-6
绛旓細-10 x^2+2x+3=0 3x^2+6x+9=0 3x^2+6x=-9 3x^2+6x-1=-10