(x-3)(x²+2x-3)=x+3 求解方程式
2个回答
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移项并化简,得到:
(x-3)(x+3)(x-1) - (x+3) = 0
(x+3) * [(x-3)(x-1) - 1] = 0
(x+3) * [x²-4x+3 - 1] = 0
(x+3) * [x² -4x + 4 - 2] = 0
(x+3) * [(x-2)² - 2] = 0
(x+3) * [(x-2 + √2)(x-2 - √2)] = 0
(x+3) * [x - (2-√2)] * [x - (2+√2)] = 0
那么,这个方程的解共有 3 个:
x+3 = 0,或
x-(2-√2) = 0,或
x - (2+√2) = 0
即
x = -3, 或 x = 2-√2, 或 x = 2+√2
(x-3)(x+3)(x-1) - (x+3) = 0
(x+3) * [(x-3)(x-1) - 1] = 0
(x+3) * [x²-4x+3 - 1] = 0
(x+3) * [x² -4x + 4 - 2] = 0
(x+3) * [(x-2)² - 2] = 0
(x+3) * [(x-2 + √2)(x-2 - √2)] = 0
(x+3) * [x - (2-√2)] * [x - (2+√2)] = 0
那么,这个方程的解共有 3 个:
x+3 = 0,或
x-(2-√2) = 0,或
x - (2+√2) = 0
即
x = -3, 或 x = 2-√2, 或 x = 2+√2
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