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本帖最后由 朱明君 于 2024-1-22 13:47 编辑
一,\(设\ x^a+y^b=z^c{,}\ 其中\ x{,}y{,}z{,}a{,}b{,}c{,}n为正整数,\)
\(则\left( xz^{nb}\right)^a+\left( yz^{na}\right)^b=z^{nab+c}\)
二,\(设\ x^a+y^b=z^c{,}\ 其中\ x{,}y{,}z{,}a{,}b{,}c{,}n{,}为正整数,\)
\(若a是nb的倍数,则\left( xz\right)^a+\left( yz^n\right)^b=z^{a+c}\)
三,\(设\ x^n+y^{n+1}=z^n,其中\ x{,}y{,}z{,}n{,}K为正整数,\)
\(且2^n-1=x=y{,}\ \ 2\left( 2^n-1\right)=z{,}\)
\(则\left( 2^n-1\right)^n+\left( 2^n-1\right)^{n+1}=\left( 2\left( 2^n-1\right)\right)^n\)
\(则\left( xK^{n+1}\right)^n+\left( yK^n\right)^{n+1}=\left( zK^{n+1}\right)^n\)
\(若n=ab,\)
\(则\left( \left( 2^{ab}-1\right)^b\right)^a+\left( 2^{ab\ }-1\right)^{ab+1}=\left( \left( 2\times\left( 2^{ab}-1\right)\right)^b\right)^a\)
\(则\left( \left( 2^{ab}-1\right)^a\right)^b+\left( 2^{ab}-1\right)^{ab+1}=\left( \left( 2\times\left( 2^{ab}-1\right)\right)^a\right)^b\)
\(则\left( \left( 2^{ab}-1\right)^b\right)^a+\left( 2^{ab}-1\right)^{ab+1}=\left( \left( 2\times\left( 2^{ab}-1\right)\right)^a\right)^b\)
\(则\left( \left( 2^{ab\ \ }-1\right)^a\right)^b+\left( 2^{ab}-1\right)^{ab+1}=\left( \left( 2\times\left( 2^{ab}-1\right)\right)^b\right)^a\)
四,\(设n为正整数,\)
\(则\left( 2^n\right)^{n+2}+\left( 2^n\right)^{n+2}=\left( 2\times2^n\right)^{n+1}\)
\(若n+2=ab{,}\ \ n+1=cd{,}\)
\(则\left( \left( 2^n\right)^a\right)^b+\left( \left( 2^n\right)^a\right)^b=\left( \left( 2\times2^n\right)^c\right)^d\)
\(则\left( \left( 2^n\right)^a\right)^b+\left( \left( 2^n\right)^a\right)^b=\left( \left( 2\times2^n\right)^d\right)^c\)
\(则\left( \left( 2^n\right)^b\right)^a+\left( \left( 2^n\right)^b\right)^a=\left( \left( 2\times2^n\right)^c\right)^d\)
\(则\left( \left( 2^n\right)^b\right)^a+\left( \left( 2^n\right)^b\right)^a=\left( \left( 2\times2^n\right)^d\right)^c\)
\(则\left( \left( 2^n\right)^a\right)^b+\left( \left( 2^n\right)^b\right)^a=\left( \left( 2\times2^n\right)^c\right)^d\)
\(则\left( \left( 2^n\right)^a\right)^b+\left( \left( 2^n\right)^b\right)^a=\left( \left( 2\times2^n\right)^d\right)^c\)
五,\(设x,n为正整数,\)
\(则2^{xn}+2^{xn}=2^{xn+1}\)
\(则\left( 2^n\right)^x+\left( 2^n\right)^x=2^{nx+1}\)
\(则\left( 2^x\right)^n+\left( 2^x\right)^n=2^{nx+1}\)
\(则\left( 2^n\right)^x+\left( 2^x\right)^n=2^{nx+1}\)
(第1题)
\(解方程x^{202401}+y^3=z^{202403}\)
\(原方程,1^a+2^3=3^2{,}\ \ 其中a为大于等于1的正整数,\)
\(若a是nb的倍数,则\left( xz\right)^a+\left( yz^n\right)^b=z^{a+c}\)
\(设a=202401,\ 代入公式得\)
\(\left( 1\times3^1\right)^{202401}+\left( 2\times3^{67467}\right)^3=3^{202403}\)
(第2题)
\(解方程x^3+y^4=z^5\)
\(原方程,2^3+1^b+=3^2{,}\ 其中b为大于等于1的正整数,\)
\(则\left( xz^{nb}\right)^a+\left( yz^{na}\right)^b=z^{nab+c}\)
\(设b=4, \ n=4, \ 代入公式得\)
\(\left( 2\times3^{16}\right)^3+\left( 1\times3^{12}\right)^4=\left( 3^{10}\right)^5\)
(第3题)
\(解方程x^n+y^{n+1}=z^n{,}\ \ 其中x,y,z,n为正整数,\)
\(且2^n-1=x=y{,}\ \ 2\left( 2^n-1\right)=z{,}\)
\(一,则\left( 2^n-1\right)^n+\left( 2^n-1\right)^{n+1}=\left( 2\left( 2^n-1\right)\right)^n\)
\(二,则\left( xK^{n+1}\right)^n+\left( yK^n\right)^{n+1}=\left( zK^{n+1}\right)^n\)
\(设n=4{,}\ 代入公式一得\)
\(\left( 2^4-1\right)^4=\left( 2^4-1\right)^5=\left( 2\left( 2^4-1\right)\right)^4\)
\(设n=3{,}\ \ K=4{,}\ \ 代入公式二得\)
\(\left( \left( 2^3-1\right)\times4^4\right)^3+\left( \left( 2^3-1\right)\times4^3\right)^4=\left( 2\left( 2^3-1\right)\times4^4\right)^3\)
(第3题)
\(x^8+y^{15}=z^{17},\)
\(解:原方程是2^3+1^4=3^2,\)
\(则\left( 2\times3^{480}\right)^8+\left( 1\times3^{256}\right)^{15}=\left( 3^{226}\right)^{17}\)
(第4题)
\(x^8+y^{17}=z^{15}\)
解:原方程是\(2^2+2^2=2^3{,}\)
\(则\left( 2\times2^{204}\right)^8+\left( 2\times2^{96}\right)^{17}=\left( 2^{109}\right)^{15}\)
(第5题)
\(x^6+y^8=z^{14}\)
\(解:原方程是2^3+1^4=3^2,\)
\(则\left( 2\times3^{16}\right)^6+\left( 1\times3^{12}\right)^8=\left( 3^7\right)^{14}\)
(第6题)
\(x^9+y^{16}=z^{25}\)
\(解:原方程是2^4+2^4=2^5,\)
\(则\left( 2\times2^{80}\right)^9+\left( 2\times2^{45}\right)^{16}=\left( 2^{29}\right)^{25}\)
(第7题)
\(x^2+y^2=z^5\)
\(解:原方程是5^2+12^2=13^2\)
\(则\left( 5\times13^4\right)^2+\left( 12\times13^4\right)^2=\left( 13^2\right)^5\)
(第8题)
\(a^2+b^2+c^2=d^3\)
\(解:原方程是3^2+3^2+3^2=3^3\)
\(则\left( 3\times3^{12}\right)^2+\left( 3\times3^{12}\right)^2+\left( 3\times3^{12}\right)^2=\left( 3^9\right)^3\)
(第9题)
\(x^5+y^2=z^2\)
\(解:原方程是3^1-1^1=2^1\)
\(则\left( 3^3\right)^5+\left( 1\times3^7\right)^2=\left( 2\times3^7\right)^2\)
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