Cho $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}= 2 ; \frac{a}{x}+\frac{b}{y}+\frac{c}{z}= 2$ .
$(a,b,c,x,y,z \neq 0)$
Tính
$D = (\frac{a}{x})^{2}+(\frac{b}{y})^{2}+ (\frac{c}{z})^{2}$.
Cho $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}= 2 ; \frac{a}{x}+\frac{b}{y}+\frac{c}{z}= 2$ .
$(a,b,c,x,y,z \neq 0)$
Tính
$D = (\frac{a}{x})^{2}+(\frac{b}{y})^{2}+ (\frac{c}{z})^{2}$.
\[{\left( {\frac{a}{x} + \frac{b}{y} + \frac{c}{z}} \right)^2} = 4\]
\[ \Leftrightarrow {\left( {\frac{a}{x}} \right)^2} + {\left( {\frac{b}{y}} \right)^2} + {\left( {\frac{c}{z}} \right)^2} + 2\left( {\frac{{ab}}{{xy}} + \frac{{bc}}{{yz}} + \frac{{ca}}{{zx}}} \right) = 4\]
\[ \Leftrightarrow D + 2\left( {\frac{{abz + bcx + cay}}{{xyz}}} \right) = 4\]
MÀ:\[\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 2 \Rightarrow \frac{{abz + bcx + cay}}{{abc}} = 2 \Rightarrow abz + bcx + cay = 2abc\]
\[ \Rightarrow D + 2\left( {\frac{{2abc}}{{xyz}}} \right) = 4 \Rightarrow D = 4 - \frac{{4abc}}{{xyz}} = \frac{{4\left( {xyz - abc} \right)}}{{xyz}}\]
Mình nghĩ có lẽ bạn cho sai nếu x/a+y/b+z/c=0 thì D=4
\[{\left( {\frac{a}{x} + \frac{b}{y} + \frac{c}{z}} \right)^2} = 4\]
\[ \Leftrightarrow {\left( {\frac{a}{x}} \right)^2} + {\left( {\frac{b}{y}} \right)^2} + {\left( {\frac{c}{z}} \right)^2} + 2\left( {\frac{{ab}}{{xy}} + \frac{{bc}}{{yz}} + \frac{{ca}}{{zx}}} \right) = 4\]
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