用牛顿迭代法求解非线性方程组
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clearclcFvwuzux1i
put请输入初值epsi
put请输入误差精度bjacobia
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gthx1bsubsbvix1iFsubsFvix1ie
dbevalbFevalFx2x1bFi1whilemaxabsx2x1epsx1x2Fwuzubjacobia
Fvfori11le
gthx1bsubsbvix1iFsubsFvix1ie
dbevalbFevalFx2x1bFii1ifi1000breake
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difi1000disp
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迭代法很可能不收敛elsedisp方程的根是：xx2E
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fwuzumfu
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Fvwuzusymsxyf1x210xy28f2xy2x10y8Ff1f2vxyE
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