(2x减1)的平方等于(3减x)的平方 用因式法解 过程
\u62ec\u53f72x\u51cf\u4e00\u62ec\u53f7\u7684\u5e73\u65b9\u7b49\u4e8e\u62ec\u53f73-x\u62ec\u53f7\u7684\u5e73\u65b9\u7528\u56e0\u5f0f\u5206\u89e3\u6cd5\u7b97(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
x=-2,\u6216x=4/3
\uff082x-1)\u7684\u5e73\u65b9=2\uff083-x)\u7684\u5e73\u65b9
(2x\uff0d1)²\uff0d2(3\uff0dx)²=0
4x²\uff0d4x\uff0b1\uff0d18\uff0d2x²\uff0b12x=0
2x²\uff0b8x\uff0d17=0
x²\uff0b4x\uff0b4=25\uff0f2(\u4e24\u8fb9\u540c\u9664\u4ee52\uff0c\u518d\u79fb\u9879)
(x\uff0b2)²=(5\u221a2\uff0f2\uff09²
x=\ufe632\u00b15\u221a2\uff0f2
\u6240\u4ee5x1=\ufe632\uff0b5\u221a2\uff0f2
x2=\ufe632\uff0d5\u221a2\uff0f2
\u4e0d\u61c2\u7684\u8fd8\u53ef\u4ee5\u95ee\uff01\u6ee1\u610f\u8bf7\u53ca\u65f6\u91c7\u7eb3\uff01 O(\u2229_\u2229)O
(2x-1)^2 - (3-x)^2=0
(2x-1+3-x)(2x-1-3+x)=0
(x+2)(3x-4)=0
x+2=0或3x-4=0
解得x=-2或x=4/3
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绛旓細濡傚浘鎵绀
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绛旓細(x-1)² = 3x-3 绉婚」鎻愬彇鍏洜寮忥紝鍙互绠鍖栬绠楄繃绋嬶紝鍏蜂綋姝ラ濡備笅 (x-1)² - 锛3x-3锛 = 0 (x-1)² - 3(x-1) = 0 (x-1)(x-1-3) = 0 (x-1)(x-4) = 0 涓や釜鏁扮殑涔樼Н绛変簬0锛屽彧闇瑕佸叾涓竴涓暟涓洪浂婊¤冻鏉′欢 x-1 = 0 鎴 x-4 = 0 x=1 鎴 x ...
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