[tex]\begin{pmatrix} \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \end{pmatrix}:\begin{pmatrix} \frac{1}{x+y+z} \end{pmatrix}=1[/tex] .Tính B=[tex]\begin{pmatrix} x^{29}+y^{29} \end{pmatrix}\begin{pmatrix} y^{11}+z^{11} \end{pmatrix}\begin{pmatrix} z^{2013}+x^{2013} \end{pmatrix}[/tex]
ĐKXĐ:
[tex]\begin{pmatrix} \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z} \end{pmatrix}:\begin{pmatrix} \dfrac{1}{x+y+z} \end{pmatrix}=1\\\Leftrightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\\\Leftrightarrow \frac{xy+yz+zx}{xyz}=\frac{1}{x+y+z}\\\Leftrightarrow xyz=(xy+yz+zx)(x+y+z)\\\Leftrightarrow (xy+yz+zx)(x+y+z)-xyz=0\\\Leftrightarrow xy(x+y+z) -xyz +(yz+zx)(x+y+z)=0\\\Leftrightarrow xy(x+y+z -z) + z(x+y)(x+y+z)=0\\\Leftrightarrow xy(x+y) +z(x+y)(x+y+z)=0\\\Leftrightarrow (x+y)[xy+z(x+y+z)]=0\\\Leftrightarrow (x+y)[xy+zx+z(y+z)]=0\\\Leftrightarrow (x+y)[x(y+z)+z(y+z)]=0\\\Leftrightarrow (x+y)(y+z)(x+z)=0[/tex]
Ta có [tex]B=\begin{pmatrix} x^{29}+y^{29} \end{pmatrix}\begin{pmatrix} y^{11}+z^{11} \end{pmatrix}\begin{pmatrix} z^{2013}+x^{2013} \end{pmatrix}\vdots (x+y)(y+z)(z+x)[/tex]
Nên [TEX]B=0[/TEX]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
