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ben duy

ben duy

Đăng ký: 11-03-2012
Offline Đăng nhập: 13-06-2012 - 22:42
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#305793 ${a^3} + {b^3} + {c^3} = 3abc$ Tính $A = (\frac{a}{b} + 1...

Gửi bởi ben duy trong 21-03-2012 - 23:05

\[\begin{array}{l}
1)\,Cho\,\,{a^3} + {b^3} + {c^3} = 3abc \\
Tinh\,A = (\frac{a}{b} + 1)(\frac{b}{c} + 1)(\frac{c}{a} + 1) \\
\end{array}\]
\[\begin{array}{l}
2)\,Cho\,\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0 \\
Tinh\,A = \frac{{yz}}{{{x^2}}} + \frac{{zx}}{{{y^2}}} + \frac{{xy}}{{{z^2}}} \\
3)\,Cho\,{x^2} - 4x + 1 = 0 \\
Tinh\,M = \frac{{{x^4} + {x^2} + 1}}{{{x^2}}} \\
4)\,Cho\,\frac{x}{{{x^2} - x + 1}} = a \\
Tinh\,N = \frac{{{x^2}}}{{{x^4} + {x^2} + 1}} \\
5)\,Cho\,x = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\,,\,y = \frac{{{a^2} - {{(b - c)}^2}}}{{{{(b + c)}^2} - {a^2}}} \\
Tinh\,x + y + xy \\
\end{array}\]