如果(f(x),g(x))=1,且f(x)|g(x)h(x),那么f(x)|h(x).这条定理怎么证明?书上的证明是:由(f(x),g(x))=1可知,有u(x),v(x)使u(x)f(x)+v(x)g(x)=1.等式两边乘h(x),得u(x)f(x)h(x)+v(x)g(x)h(x)=h(x),因为f(x)|g(x)h(x),所以f(x)整除等
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![如果(f(x),g(x))=1,且f(x)|g(x)h(x),那么f(x)|h(x).这条定理怎么证明?书上的证明是:由(f(x),g(x))=1可知,有u(x),v(x)使u(x)f(x)+v(x)g(x)=1.等式两边乘h(x),得u(x)f(x)h(x)+v(x)g(x)h(x)=h(x),因为f(x)|g(x)h(x),所以f(x)整除等](/uploads/image/z/8496239-23-9.jpg?t=%E5%A6%82%E6%9E%9C%28f%28x%29%2Cg%28x%29%29%3D1%2C%E4%B8%94f%28x%29%7Cg%28x%29h%28x%29%2C%E9%82%A3%E4%B9%88f%28x%29%7Ch%28x%29.%E8%BF%99%E6%9D%A1%E5%AE%9A%E7%90%86%E6%80%8E%E4%B9%88%E8%AF%81%E6%98%8E%3F%E4%B9%A6%E4%B8%8A%E7%9A%84%E8%AF%81%E6%98%8E%E6%98%AF%3A%E7%94%B1%28f%28x%29%2Cg%28x%29%29%3D1%E5%8F%AF%E7%9F%A5%2C%E6%9C%89u%28x%29%2Cv%28x%29%E4%BD%BFu%28x%29f%28x%29%2Bv%28x%29g%28x%29%3D1.%E7%AD%89%E5%BC%8F%E4%B8%A4%E8%BE%B9%E4%B9%98h%28x%29%2C%E5%BE%97u%28x%29f%28x%29h%28x%29%2Bv%28x%29g%28x%29h%28x%29%3Dh%28x%29%2C%E5%9B%A0%E4%B8%BAf%28x%29%7Cg%28x%29h%28x%29%2C%E6%89%80%E4%BB%A5f%28x%29%E6%95%B4%E9%99%A4%E7%AD%89)
如果(f(x),g(x))=1,且f(x)|g(x)h(x),那么f(x)|h(x).这条定理怎么证明?书上的证明是:由(f(x),g(x))=1可知,有u(x),v(x)使u(x)f(x)+v(x)g(x)=1.等式两边乘h(x),得u(x)f(x)h(x)+v(x)g(x)h(x)=h(x),因为f(x)|g(x)h(x),所以f(x)整除等
如果(f(x),g(x))=1,且f(x)|g(x)h(x),那么f(x)|h(x).这条定理怎么证明?
书上的证明是:
由(f(x),g(x))=1可知,有u(x),v(x)使
u(x)f(x)+v(x)g(x)=1.
等式两边乘h(x),得
u(x)f(x)h(x)+v(x)g(x)h(x)=h(x),
因为f(x)|g(x)h(x),所以f(x)整除等式左端,从而
f(x)|h(x).
为什么"因为f(x)|g(x)h(x),所以f(x)整除等式左端"?
如果(f(x),g(x))=1,且f(x)|g(x)h(x),那么f(x)|h(x).这条定理怎么证明?书上的证明是:由(f(x),g(x))=1可知,有u(x),v(x)使u(x)f(x)+v(x)g(x)=1.等式两边乘h(x),得u(x)f(x)h(x)+v(x)g(x)h(x)=h(x),因为f(x)|g(x)h(x),所以f(x)整除等
等式左端是两项的和:u(x)f(x)h(x)和v(x)g(x)h(x).显然f(x)整除第一项;由条件f(x)整除g(x)h(x)可知f(x)整除v(x)g(x)h(x),即第二项.所以f(x)整除两项的和,即等式左端.
或者:
由f(x)|g(x)h(x),可设g(x)h(x)=w(x)f(x),所以等式左端
=u(x)f(x)h(x)+v(x)g(x)h(x)
=u(x)f(x)h(x)+v(x)w(x)f(x)
=f(x)[u(x)h(x)+v(x)w(x)],
所以被f(x)整除.