已知a,b是两个连续的正整数,且满足a^2-b^2=2013,求a,b的值a^2-b^2=2007(a+b)(a-b)=1*3*3*223(是怎么变过来的)因为a,b都是正整数,且a+b>a-b.所以有:a+b=2007,a-b=1,解得:a=1004,b=1003a+b=3*223,a-b=3,解得:a=336,b=333a+b
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![已知a,b是两个连续的正整数,且满足a^2-b^2=2013,求a,b的值a^2-b^2=2007(a+b)(a-b)=1*3*3*223(是怎么变过来的)因为a,b都是正整数,且a+b>a-b.所以有:a+b=2007,a-b=1,解得:a=1004,b=1003a+b=3*223,a-b=3,解得:a=336,b=333a+b](/uploads/image/z/5403933-45-3.jpg?t=%E5%B7%B2%E7%9F%A5a%2Cb%E6%98%AF%E4%B8%A4%E4%B8%AA%E8%BF%9E%E7%BB%AD%E7%9A%84%E6%AD%A3%E6%95%B4%E6%95%B0%2C%E4%B8%94%E6%BB%A1%E8%B6%B3a%5E2-b%5E2%3D2013%2C%E6%B1%82a%2Cb%E7%9A%84%E5%80%BCa%5E2-b%5E2%3D2007%28a%2Bb%29%28a-b%29%3D1%2A3%2A3%2A223%28%E6%98%AF%E6%80%8E%E4%B9%88%E5%8F%98%E8%BF%87%E6%9D%A5%E7%9A%84%29%E5%9B%A0%E4%B8%BAa%2Cb%E9%83%BD%E6%98%AF%E6%AD%A3%E6%95%B4%E6%95%B0%2C%E4%B8%94a%2Bb%3Ea-b.%E6%89%80%E4%BB%A5%E6%9C%89%EF%BC%9Aa%2Bb%3D2007%2Ca-b%3D1%2C%E8%A7%A3%E5%BE%97%EF%BC%9Aa%3D1004%2Cb%3D1003a%2Bb%3D3%2A223%2Ca-b%3D3%2C%E8%A7%A3%E5%BE%97%EF%BC%9Aa%3D336%2Cb%3D333a%2Bb)
已知a,b是两个连续的正整数,且满足a^2-b^2=2013,求a,b的值a^2-b^2=2007(a+b)(a-b)=1*3*3*223(是怎么变过来的)因为a,b都是正整数,且a+b>a-b.所以有:a+b=2007,a-b=1,解得:a=1004,b=1003a+b=3*223,a-b=3,解得:a=336,b=333a+b
已知a,b是两个连续的正整数,且满足a^2-b^2=2013,求a,b的值
a^2-b^2=2007
(a+b)(a-b)=1*3*3*223(是怎么变过来的)
因为a,b都是正整数,且a+b>a-b.
所以有:
a+b=2007,a-b=1,解得:a=1004,b=1003
a+b=3*223,a-b=3,解得:a=336,b=333
a+b=223,a-b=9,解得:a=116,b=107
已知a,b是两个连续的正整数,且满足a^2-b^2=2013,求a,b的值a^2-b^2=2007(a+b)(a-b)=1*3*3*223(是怎么变过来的)因为a,b都是正整数,且a+b>a-b.所以有:a+b=2007,a-b=1,解得:a=1004,b=1003a+b=3*223,a-b=3,解得:a=336,b=333a+b
a^2-b^2=(a+b)(a-b),这是平方差公式,因式分解常用的.
2007=1*3*3*223,因为2007的数字和=9,所以2007可以除3,一试便知.
同理:
a^2-b^2=2013
(a+b)(a-b)=1*3*671=1*3*11*61=11*183=33*61
因为a,b都是正整数,且a+b>a-b.
所以有:
a+b=2013,a-b=1,解得:a=1007,b=1006
a+b=671,a-b=3,解得:a=337,b=334
a+b=183,a-b=11,解得:a=97,b=86
a+b=61,a-b=33,解得:a=47,b=14