Question

Giải hệ phương trình sau : \(\left\{{}\begin{matrix}x+y+z=a\\x^2+y^2+z^2=b^2\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{c}\end{matrix}\right.\)

Answer 1

\(x+y+z=a\)

\(\Leftrightarrow\left(x+y+z\right)^2=a^2\)

\(\Leftrightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)=a^2\)

\(\Leftrightarrow b^2+2\left(xy+yz+zx\right)=a^2\)

\(\Leftrightarrow xy+yz+zx=\frac{a^2-b^2}{2}\)

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{c}\)

\(\Leftrightarrow\frac{xy+yz+zx}{xyz}=\frac{1}{c}\Leftrightarrow xyz=\left(xy+yz+zx\right)c=\frac{a^2-b^2}{2}.c\)

\(x^2+y^2+z^2=b^2\)

\(\Leftrightarrow x^2+\left(y+z\right)^2-2yz=b^2\)

\(\Leftrightarrow x^2+\left(a-x\right)^2-2\left[\frac{\left(a^2-b^2\right)c}{2x}\right]=b^2\)

\(\Leftrightarrow x^2+a^2-2ax+x^2-\frac{\left(a^2-b^2\right)c}{x}=b^2\)

\(\Leftrightarrow2x^3-2ax^2+\left(a^2-b^2\right)x-\left(a^2-b^2\right)c=0\)

\(x,y,z\) là nghiệm của phương trình trên.

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