Đáp án:
Ta có :
`(1/x + 1/y + 1/z)/[(1/(x + y + z)) = 1`
`=> 1/x + 1/y + 1/z = 1/(x + y + z)`
`=> 1/x + 1/y + 1/z - 1/(x + y + z) = 0`
`=>( 1/x + 1/y) + [1/z - 1/(x + y + z)] = 0`
`=> y/(xy) + x/(xy) + (x + y + z)/(z(x + y + z)) - z/(z(x + y + z)) = 0`
`=> (x + y)/(xy) + (x + y)/(z(x + y + z)) = 0`
`=> [(x + y)z(x + y + z) + (x + y)xy]/[xyz(x + y + z)] = 0`
`=> (x + y)z(x + y + z) + (x + y)xy = 0`
`=> (x + y)[z(x + y + z) + xy] = 0`
`=> (x + y)(zx + zy + z^2 + xy) = 0`
`=> (x + y)[(zx + z^2) + (zy + xy)]= 0`
`=> (x + y)[z(x + z) + y(z + x)] = 0`
`=> (x + y)(x + z)(y + z) = 0`
th1 : `x + y = 0 =>y = -x`
`=> B =(x^{21}+y^{21})(y^{11}+z^{11})(z^{2017}+x^{2017})`
`= [x^{21} - x^{21}].(y^{11}+z^{11})(z^{2017}+x^{2017})`
`= 0. (y^{11}+z^{11})(z^{2017}+x^{2017})`
`= 0`
th2 : `x + z = 0 => x = -z`
`=> B =(x^{21}+y^{21})(y^{11}+z^{11})(z^{2017}+x^{2017})`
` = (x^{21}+y^{21})(y^{11}+z^{11})(z^{2017} - z^{2017})`
`= (x^{21}+y^{21})(y^{11}+z^{11}) . 0`
`= 0`
th3 : `y + z = 0 => z = -y`
làm tương tự `=> B = 0`
Giải thích các bước giải:
