Question

Cho \(\)x, y, z \(\in\) R thỏa mãn \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\).

Hãy tính giá trị của biểu thức: M = \(\dfrac{3}{4}+\left(x^8-y^8\right)\left(y^9+z^9\right)\left(z^{10}-x^{10}\right)\)

Answer 1

Từ \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\Rightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}-\dfrac{1}{x+y+z}=0\)

\(\Rightarrow\dfrac{x+y}{xy}+\dfrac{x+y+z-z}{z\left(x+y+z\right)}=0\)

\(\Rightarrow\left(x+y\right)\left(\dfrac{1}{xy}+\dfrac{1}{z\left(x+y+z\right)}\right)=0\)

\(\Rightarrow\left(x+y\right)\left(\dfrac{zx+zy+z^2+xy}{xyz\left(x+y+z\right)}\right)=0\)

\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)

Ta có: x⁸ - y⁸ = (x + y)(x - y)(x² + y²)(x⁴ + y⁴)

y⁹ + z⁹ = (y + z)(y⁸ - y⁷z + y⁶z² - ... + z⁸)

z¹⁰ - x¹⁰ = (z + x)(z⁴ - z³x + z²x² - zx³ + z⁴)(z⁵ - x⁵)

Vậy M = \(\dfrac{3}{4}\) + (x + y)(y + z)(z + x) = \(\dfrac{3}{4}\)