Question

[Đào Thị Bạch Cúc]

cho x,y,z thuộc R, thỏa mãn \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\) tính  M=\(\frac{3}{4}+\left(x^2-y^2\right)\cdot\left(y^3+z^3\right)\cdot\left(z^4-x^4\right)\)

Answer 1

[Lã Nguyễn Gia Hy]

Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}=0\)

\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y+z-z}{z\left(x+y+z\right)}=0\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\)

\(\Leftrightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right)=0\Leftrightarrow\left(x+y\right)\left(\frac{zx+z^2+zy+xy}{xyz\left(x+y+z\right)}\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left[z\left(x+z\right)+y\left(x+z\right)\right]=0\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)

\(\Rightarrow\left(x^2-y^2\right)\left(y^3+z^3\right)\left(z^4-x^4\right)=0\).

Vậy  \(M=\frac{3}{4}+\left(x^2-y^2\right)\left(y^3+z^3\right)\left(z^4-x^4\right)=\frac{3}{4}+0=\frac{3}{4}\)

Answer 2

[Đào Thị Bạch Cúc]

thank Gia Hy