“解”字的古今的读音有多少种，分别是什么，请罗列一下，谢谢！

Answer 1

LZ好，用mathematica的精确解命令
Solve[{m*n + p^2 == 7, n*p + m^2 == 8, p*m + n^2 == 9}, {m, n, p}]
与数值解命令
NSolve[{m*n + p^2 == 7, n*p + m^2 == 8, p*m + n^2 == 9}, {m, n, p}]
解答，都显示这个方程组有4组实解和4组复解
4组实解中没有找到正实数解

由于精确解的表达式很麻烦(所给形式已经经过化简).我只给出第一个精确解的m值.从精确解的形式可以看出，想找到具体解题步骤是不太可能的:m=(1/4991)(1/384 (2892 - 4 (43781688 - 5208 Sqrt[56298])^(1/3) -8(3 (1824237 + 217 Sqrt[56298]))^(1/3) - 573/Sqrt[241/16 + 1/24 (43781688 - 5208 Sqrt[56298])^(1/3) + (1824237 + 217 Sqrt[56298])^(1/3)/(4 3^(2/3))])^(3/2) \[Sqrt](1/6 (35/8 - 1/2 \[Sqrt](241/16 + 1/24 (43781688 - 5208 Sqrt[56298])^(1/3) + (1824237 + 217 Sqrt[56298])^(1/3)/(4 3^(2/3))) -1/8 \[Sqrt](1/6 (2892 - 4 (43781688 - 5208 Sqrt[56298])^(1/3) - 8 (3 (1824237 + 217 Sqrt[56298]))^(1/3) - 573/(\[Sqrt](241/16 + 1/24 (43781688 - 5208 Sqrt[56298])^(1/3) + (1824237 + 217 Sqrt[56298])^(1/3)/( 4 3^(2/3)))))))) - 27559/4 \[Sqrt](35/8 - 1/2 \[Sqrt](241/16 + 1/24 (43781688 - 5208 Sqrt[56298])^(1/3) + (1824237 + 217 Sqrt[56298])^(1/3)/(4 3^(2/3))) - 1/8 \[Sqrt](1/ 6 (2892 - 4 (43781688 - 5208 Sqrt[56298])^(1/3) - 8 (3 (1824237 + 217 Sqrt[56298]))^(1/3) - 573/(\[Sqrt](241/16 + 1/24 (43781688 - 5208 Sqrt[56298])^(1/3) + (1824237 + 217 Sqrt[56298])^(1/3)/(4 3^(2/3))))))) + (151463 \[Sqrt](35/8 - 1/2 \[Sqrt](241/16 + 1/24 (43781688 - 5208 Sqrt[56298])^(1/3) + (1824237 + 217 Sqrt[56298])^(1/3)/(4 3^(2/3))) - 1/8 \[Sqrt](1/6 (2892 - 4 (43781688 - 5208 Sqrt[56298])^(1/3) - 8 (3 (1824237 + 217 Sqrt[56298]))^(1/3) - 573/(\[Sqrt](241/16 + 1/24 (43781688 - 5208 Sqrt[56298])^(1/3) + (1824237 + 217 Sqrt[56298])^(1/3)/(4 3^(2/3))))))))/(128 \[Sqrt](241/16 + 1/24 (43781688 - 5208 Sqrt[56298])^(1/3) + (1824237 + 217 Sqrt[56298])^(1/3)/(4 3^(2/3)))) - 9999/16 Sqrt[241/16 + 1/24 (43781688 - 5208 Sqrt[56298])^(1/3) + (1824237 + 217 Sqrt[56298])^(1/3)/(4 3^(2/3))] \[Sqrt](35/8 - 1/2 \[Sqrt](241/16 + 1/24 (43781688 - 5208 Sqrt[56298])^(1/3) + (1824237 + 217 Sqrt[56298])^(1/3)/(4 3^(2/3))) - 1/8 \[Sqrt](1/6 (2892 - 4 (43781688 - 5208 Sqrt[56298])^(1/3) - 8 (3 (1824237 + 217 Sqrt[56298]))^(1/3) - 573/(\[Sqrt](241/16 + 1/24 (43781688 - 5208 Sqrt[56298])^(1/3) + (1824237 + 217 Sqrt[56298])^(1/3)/(4 3^(2/3))))))) - 1/8 (43781688 - 5208 Sqrt[56298])^(1/3) Sqrt[241/16 + 1/24 (43781688 - 5208 Sqrt[56298])^(1/3) + (1824237 + 217 Sqrt[56298])^(1/3)/(4 3^(2/3))] \[Sqrt](35/8 - 1/2 \[Sqrt](241/16 + 1/24 (43781688 - 5208 Sqrt[56298])^(1/3) + (1824237 + 217 Sqrt[56298])^(1/3)/(4 3^(2/3))) - 1/8 \[Sqrt](1/6 (2892 - 4 (43781688 - 5208 Sqrt[56298])^(1/3) - 8 (3 (1824237 + 217 Sqrt[56298]))^(1/3) - 573/(\[Sqrt](241/16 + 1/24 (43781688 - 5208 Sqrt[56298])^(1/3) + (1824237 + 217 Sqrt[56298])^(1/3)/(4 3^(2/3))))))) - 1/4 (3 (1824237 + 217 Sqrt[56298]))^(1/3) Sqrt[241/16 + 1/24 (43781688 - 5208 Sqrt[56298])^(1/3) + (1824237 + 217 Sqrt[56298])^(1/3)/(4 3^(2/3))] \[Sqrt](35/8 - 1/2 \[Sqrt](241/16 + 1/24 (43781688 - 5208 Sqrt[56298])^(1/3) + (1824237 + 217 Sqrt[56298])^(1/3)/(4 3^(2/3))) - 1/8 \[Sqrt](1/6 (2892 - 4 (43781688 - 5208 Sqrt[56298])^(1/3) - 8 (3 (1824237 + 217 Sqrt[56298]))^(1/3) - 573/(\[Sqrt](241/16 + 1/24 (43781688 - 5208 Sqrt[56298])^(1/3) + (1824237 + 217 Sqrt[56298])^(1/3)/(4 3^(2/3))))))) + (241/16 + 1/24 (43781688 - 5208 Sqrt[56298])^(1/3) + (1824237 + 217 Sqrt[56298])^(1/3)/(4 3^(2/3)))^(3/2) \[Sqrt](35/8 - 1/2 \[Sqrt](241/16 + 1/24 (43781688 - 5208 Sqrt[56298])^( 1/3) + (1824237 + 217 Sqrt[56298])^(1/3)/(4 3^(2/3))) - 1/8 \[Sqrt](1/6 (2892 - 4 (43781688 - 5208 Sqrt[56298])^(1/3) - 8 (3 (1824237 + 217 Sqrt[56298]))^(1/3) - 573/(\[Sqrt](241/16 + 1/24 (43781688 - 5208 Sqrt[56298])^(1/3) + (1824237 + 217 Sqrt[56298])^(1/3)/(4 3^(2/3))))))) + (1/(32 Sqrt[6]))(43781688 - 5208 Sqrt[56298])^(1/3) \[Sqrt](2892 - 4 (43781688 - 5208 Sqrt[56298])^(1/3) - 8 (3 (1824237 + 217 Sqrt[56298]))^(1/3) - 573/Sqrt[241/16 + 1/24 (43781688 - 5208 Sqrt[56298])^(1/3) + (1824237 + 217 Sqrt[56298])^(1/3)/(4 3^(2/3))]) \[Sqrt](35/8 - 1/2 \[Sqrt](241/16 + 1/24 (43781688 - 5208 Sqrt[56298])^(1/3) + (1824237 + 217 Sqrt[56298])^(1/3)/(4 3^(2/3))) - 1/8 \[Sqrt](1/6 (2892 - 4 (43781688 - 5208 Sqrt[56298])^(1/3) - 8 (3 (1824237 + 217 Sqrt[56298]))^(1/3) - 573/(\[Sqrt](241/16 + 1/24 (43781688 - 5208 Sqrt[56298])^(1/3) + (1824237 + 217 Sqrt[56298])^(1/3)/(4 3^(2/3))))))) + (1/(16 Sqrt[2] 3^(1/6)))(1824237 + 217 Sqrt[56298])^(1/3) \[Sqrt](2892 - 4 (43781688 - 5208 Sqrt[56298])^(1/3) - 8 (3 (1824237 + 217 Sqrt[56298]))^(1/3) - 573/Sqrt[241/16 + 1/24 (43781688 - 5208 Sqrt[56298])^(1/3) + (1824237 + 217 Sqrt[56298])^(1/3)/(4 3^(2/3))]) \[Sqrt](35/8 - 1/2 \[Sqrt](241/16 + 1/24 (43781688 - 5208 Sqrt[56298])^( 1/3) + (1824237 + 217 Sqrt[56298])^(1/3)/(4 3^(2/3))) - 1/8 \[Sqrt](1/6 (2892 - 4 (43781688 - 5208 Sqrt[56298])^(1/3) - 8 (3 (1824237 + 217 Sqrt[56298]))^(1/3) - 573/(\[Sqrt](241/16 + 1/24 (43781688 - 5208 Sqrt[56298])^( 1/3) + (1824237 + 217 Sqrt[56298])^(1/3)/(4 3^(2/3))))))) - 43 \[Sqrt](2/3 (241/16 + 1/24 (43781688 - 5208 Sqrt[56298])^(1/3) + (1824237 + 217 Sqrt[56298])^(1/3)/(4 3^(2/3)))) \[Sqrt](2892 - 4 (43781688 - 5208 Sqrt[56298])^(1/3) - 8 (3 (1824237 + 217 Sqrt[56298]))^(1/3) - 573/Sqrt[241/16 + 1/24 (43781688 - 5208 Sqrt[56298])^(1/3) + (1824237 + 217 Sqrt[56298])^(1/3)/(4 3^(2/3))]) \[Sqrt](35/8 - 1/2 \[Sqrt](241/16 + 1/24 (43781688 - 5208 Sqrt[56298])^(1/3) + (1824237 + 217 Sqrt[56298])^(1/3)/(4 3^(2/3))) - 1/8 \[Sqrt](1/6 (2892 - 4 (43781688 - 5208 Sqrt[56298])^(1/3) - 8 (3 (1824237 + 217 Sqrt[56298]))^(1/3) - 573/(\[Sqrt](241/16 + 1/24 (43781688 - 5208 Sqrt[56298])^( 1/3) + (1824237 + 217 Sqrt[56298])^(1/3)/(4 3^(2/3))))))) - 35/8 \[Sqrt](3/2 (241/16 + 1/24 (43781688 - 5208 Sqrt[56298])^( 1/3) + (1824237 + 217 Sqrt[56298])^(1/3)/( 4 3^(2/3)))) \[Sqrt](2892 - 4 (43781688 - 5208 Sqrt[56298])^(1/3) - 8 (3 (1824237 + 217 Sqrt[56298]))^(1/3) - 573/Sqrt[241/16 + 1/24 (43781688 - 5208 Sqrt[56298])^(1/3) + (1824237 + 217 Sqrt[56298])^(1/3)/(4 3^(2/3))]) \[Sqrt](35/8 - 1/2 \[Sqrt](241/16 + 1/24 (43781688 - 5208 Sqrt[56298])^(1/3) + (1824237 + 217 Sqrt[56298])^(1/3)/(4 3^(2/3))) - 1/8 \[Sqrt](1/6 (2892 - 4 (43781688 - 5208 Sqrt[56298])^(1/3) - 8 (3 (1824237 + 217 Sqrt[56298]))^(1/3) - 573/(\[Sqrt](241/16 + 1/24 (43781688 - 5208 Sqrt[56298])^(1/3) + (1824237 + 217 Sqrt[56298])^(1/3)/( 1787/32 \[Sqrt](3/2 (2892 - 4 (43781688 - 5208 Sqrt[56298])^(1/3) - 8 (3 (1824237 + 217 Sqrt[56298]))^(1/3) - 573/Sqrt[241/16 + 1/24 (43781688 - 5208 Sqrt[56298])^(1/3) + (1824237 + 217 Sqrt[56298])^(1/3)/(4 3^(2/3))])) \[Sqrt](35/8 - 1/2 \[Sqrt](241/16 + 1/24 (43781688 - 5208 Sqrt[56298])^(1/3) + (1824237 + 217 Sqrt[56298])^(1/3)/(4 3^(2/3))) - 1/8 \[Sqrt](1/6 (2892 - 4 (43781688 - 5208 Sqrt[56298])^(1/3) - 8 (3 (1824237 + 217 Sqrt[56298]))^(1/3) - 573/(\[Sqrt](241/16 + 1/24 (43781688 - 5208 Sqrt[56298])^(1/3) + (1824237 + 217 Sqrt[56298])^(1/3)/(4 3^(2/3))))))))

下面给出方程组的全部数值解，上面的精确解对应下面的第一个数值解.前4组是实解，后四组是复解：
{m=0.0216395,n=-3.00957,p=-2.65803},
{m=-0.0216395,n=3.00957,p=2.65803},
{m=-3.04648,n=0.443303,p=-2.88973},
{m=3.04648,n=-0.443303,p=2.88973},
{m=-2.31541-0.0433529 i,n=-2.57372-0.0322286i,p=-1.02491+0.0908378i},
{m=-2.31541+0.0433529i,n=-2.57372+0.0322286i,p=-1.02491-0.0908378i},
{m=2.31541-0.0433529i,n=2.57372-0.0322286i,p=1.02491+0.0908378i},
{m=2.31541+0.0433529i,n=2.57372+0.0322286i,p=1.02491-0.0908378i}

补充：
把原方程组中的m，n消去以后，得到关于p的八次方程
-1122 p^2 + 738 p^4 - 140 p^6 + 8 p^8 == -529，
把p^2看成变量，上面的方程就变成了四次方程，数学上是可以求解四次方程的解析解，即精确解.求解可得关于p^2的四个解（写成数值形式）
{p^2=1.04218-0.186201i},
{p^2=1.04218+0.186201i},
{p^2=7.06513},
{p^2=8.35051}

再由原方程组中的第一个方程mn+p^2=7，因为要求m，n，p都是正实数，所以p^2就要小于7，而上面关于p^2的四个数值解都不符合要求，所以原方程组没有正实数解.
从上面的推导过程还可以看出，原方程组只有上面已经列出的8组解 9357希望对你有帮助！

参考资料：. saivqru

Answer 2

中古：[kai]（国际音标）

上古：[kree？]

这是学界公认的拟音。

中古一直都读gai，清朝满人统治时变为jiai，清末变为jie。

Answer 3

jie(3):解放
jie（4）：押解
xie(4):是一个姓氏。

Answer 4

楼上说什么呀？怎么不懂！！
解放（第三声）
押解（第四声）

浑身解数（与上读音不同）