Let a,b and c are rtional numbers which satisfy a-7b+8c=4 and 8a+4b-c=7.Then a^2-b^2+c^2=?
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![Let a,b and c are rtional numbers which satisfy a-7b+8c=4 and 8a+4b-c=7.Then a^2-b^2+c^2=?](/uploads/image/z/3717026-26-6.jpg?t=Let+a%2Cb+and+c+are+rtional+numbers+which+satisfy+a-7b%2B8c%3D4+and+8a%2B4b-c%3D7.Then+a%5E2-b%5E2%2Bc%5E2%3D%3F)
Let a,b and c are rtional numbers which satisfy a-7b+8c=4 and 8a+4b-c=7.Then a^2-b^2+c^2=?
Let a,b and c are rtional numbers which satisfy a-7b+8c=4 and 8a+4b-c=7.Then a^2-b^2+c^2=?
Let a,b and c are rtional numbers which satisfy a-7b+8c=4 and 8a+4b-c=7.Then a^2-b^2+c^2=?
有
a-7b+8c=4 (1)
8a+4b-c=7 (2)
(1)+(2)*8 可得: 65a+25b=60
化简得13a+5b=12 -->a=(12-5b)/13
(1)*8-(2)得-60b+65c=-25
化简得12b-13c=5 --->c=(12b+5)/13
将a,c,代入a²-b²+c²=((12-5b)/13)²+((12b+5)/13)²-b²
=((12-5b)²+(12b+5)²)/13²-b²
=(12²+(5b)²+(12b)²+5²)/13²-b²
=(12²+5²+(12²+5²)b²)/13²-b²
=1+b²-b²
=1
由a-7b+8c=4得a=7b-8c+4 ……(1)代入8a+4b-c=7得b=(13c-5)/12 ……(2)将(2)代入(1)式得a=(13-5c)/12故a^2-b^2+c^2=(13c-5)^2/144-(13-5c)^2/144+c^2=(8c+8)*(18-18c)/144+c^2=1-c^2+c^2=1