cộng trừ phân thức

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tranvanhung7997

1.Cmr: nếu 1/a + 1/b + 1/c = 2 và a+b+c=abc thì 1/a^2 + 1/b^2 + 1/c^2 = 2
2.Cho 1/x + 1/y + 1/z = 0. Tính yz/x^2 + zx/y^2 + xy/z^2

\[ Bài 1: Ta có: \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\]
\[=(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2-2(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca})\]
\[=(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2-2(\frac{a+b+c}{abc})\]
\[=2^2-2.1\] (Vì \[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2 \] và \[a+b+c=abc)\]
\[=2\]
 
T

tranvanhung7997

\[Bài 2: Ta có: \frac{yz}{x^2}+\frac{zx}{y^2}+\frac{xy}{z^2}\]
\[=xyz.( \frac{1}{x^3}+ \frac{1}{y^3}+ \frac{1}{z^3})\]
\[=xyz.[(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}).(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}-\frac{1}{xy}-\frac{1}{yz}-\frac{1}{zx})+3.\frac{1}{x}.\frac{1}{y}.\frac{1}{z}]\]
\[=xyz.\frac{3}{xyz}\] \[(vì \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0)\]
\[=3\]
 
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