用因式分解法解下列一元二次方程。x的平方等于2x。
\u7528\u56e0\u5f0f\u5206\u89e3\u6cd5\u89e3\u4e00\u5143\u4e8c\u6b21\u65b9\u7a0b\uff082x-1\uff09\u7684\u5e73\u65b9=\uff083-x\uff09\u7684\u5e73\u65b9\uff082x-1\uff09²=\uff083-x\uff09²
\uff082x-1\uff09²-\uff083-x\uff09²=0
[\uff082x-1\uff09+\uff083-x\uff09][\uff082x-1\uff09-\uff083-x\uff09]=0
(x+2)(3x-4)=0
x+2=0\u62163x-4=0
x1=-2\uff0cx2=4/3
(x+3)²=2x+6
(x+3)²-(2x+6)=0
(x+3)²-2(x+3)=0
(x+3)(x+3-2)=0
(x+3)(x+1)=0
x+3=0\u6216x+1=0
x1=-3, x2=-1.
x^2-2x=0
x(x-2)=0
x=0或x=2
绛旓細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^2-5X+2X-10=0 X(X-5)+2...
绛旓細锛3锛墄锛坸-2锛+x-2=0 锛坸+1锛(x-2)=0 鎵浠 x=-1 鎴 x=2 锛4锛墄²+3x-10=0 (x+5)(x-2)=0 鎵浠 x=-5 鎴 x=2
绛旓細瑙ｏ細x²-4X+4-3x+6+2=0 x²-7X+12=0 (X-3)(X-4)=0 鈭 X-3=0鎴朮-4=0 X1=3锛孹2=4 (2-3x)+(3x-2)²=0 瑙ｏ細(2-3x)+(2-3x)²=0 (2-3x锕(1+2-3x)=0 (2-3x)(3-3x)=0 鈭 2-3x=0鎴3-3x=0 X1=2锛3锛孹2=1 ...
绛旓細鍥炵瓟锛7銆(y+1)(y-2)=0 y=-1,y=2
绛旓細鐢ㄥ洜寮忓垎瑙ｆ硶瑙ｄ笅鍒楁柟绋锛 1. 3蠂锕櫹囷紞1锕氾紳2锕櫹囷紞1锕 2.锕2蠂-1锕²锛濓箼3-蠂锕²1. 3蠂锕櫹囷紞1锕氾紳2锕櫹囷紞1锕3x(x-1)-2(x-1)=0 (x-1)(3x-2)=0 x-1=0 3x-2=0 鈭磝=1鎴杧=2/3 2.锕2蠂-1锕²锛濓箼3-蠂锕²(2x-1)²-(3-x)...
绛旓細(3x-2)²=(3-2x)²3x-2=3-2x 3x-2=2x-3 x=1 x=-1 2銆2(x-3)²=9-x² 锛2x²-12x+18=9-x²3x²-12x+9=0 x²-4x+3=0 x=1 x=3 3銆(x+2-3)(x+2-1)=0 (x-1)(x+1)=0 x=1 x=-1 ...
绛旓細鍏鍒嗚В鍥犲紡锛氣埆 1/(x³ + 1) dx = 鈭 1/[(x + 1)(x² - x + 1)] dx = 鈭 A/(x + 1) dx + 鈭 (Bx + C)/(x² - x + 1) dx 1 = A(x² - x + 1) + (Bx + C)(x + 1) = Ax² - Ax + A + Bx² + Cx + Bx + ...
绛旓細(x-4)(x-5) = 0 x=4 鎴 x=5 锛5x-1)(2x+4)=3x+6 2锛5x-1)(x+2)=3(x+2)2锛5x-1)=3 10x-2=3 x=1/2 x²-3x-18=0 (x-6)(x+3)=0 x1=6,x2=-3 2y²-5y+2=0 (2y-1)(2y-2)=0 2y-1=0鎴2y-2=0 鈭磝1=1/2,x2=1 x²+5=2鈭...
绛旓細鍥炵瓟锛氶鍏堜袱杈瑰垎鍒睍寮寰楀埌 x2-8x+16=25-20x+4x2 鐒跺悗鍖栫畝 x2-4x+3=0 鏈鍚庡緱鍒 (x-3)(x-1)=0 鎵浠=3鎴1
绛旓細(x-3)(x-3+4x)=0 (x-3)(5x-3)=0 x1=3锛寈2=3/5 (2) x^4-6x²+9=0 (x²-3)²=0 x²-3=0 x1=鈭3锛寈2=-鈭3 锛3锛夊凡鐭(2x-y)=y(y-2x)(xy鈮0),姹倄²+y²/xy鐨勫 x(2x-y)=-y(2x-y)(x+y)(2x-y)=0 鎵浠ワ細2x-y=...
