Đáp án:
$M = \dfrac{3}{4}$
Giải thích các bước giải:
$\dfrac{1}{x}+\dfrac{1}{y} + \dfrac{1}{z}=\dfrac{1}{x + y + z}$
$\to \left(\dfrac{1}{x}+\dfrac{1}{y}\right) + \left(\dfrac{1}{z} - \dfrac{1}{x + y + z}\right) = 0$
$\to \dfrac{x + y}{xy} + \dfrac{x + y}{z(x + y + z)}=0$
$\to (x+y)\left(\dfrac{1}{xy} + \dfrac{1}{z(x + y + z)}\right) = 0$
$\to (x + y)\dfrac{zx + zy + z^2 + xy}{xyz(x + y + z)}=0$
$\to \dfrac{(x + y)(y + z)(z + x)}{xyz(x + y + z)}=0$
$+)\quad x + y = 0$
Ta có: $x^8 - y^8$
$= (x^4-y^4)(x^4 + y^4)$
$= (x-y)(x +y)(x^2 + y^2)(x^4 + y^4)$
$\to x^8 - y^8 = 0$
$\to M = \dfrac{3}{4}$
$+)\quad y + z = 0$
Ta có: $y^9 + z^9$
$= (y^3 + z^3)^3 - 3y^3z^3(y^3 + z^3)$
$y^3 + z^3 = (y + z)^3 - 3yz(y + z)$
$\to y^9 + z^9 = 0$
$\to M = \dfrac{3}{4}$
$+)\quad z + x = 0$
Ta có:
$z^{10} - x^{10}$
$= (z^5 - x^5)(z^5 + x^5)$
$= (z^5 - x^5)[(a^3 + b^3)(a^2 + b^2) - (a + b)]$
$\to z^{10} - x^{10} = 0$
$\to M = \dfrac{3}{4}$
Vậy $M = \dfrac{3}{4}$
