Toán 8 VIOLYMPIC

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P

phamvananh9

[TEX][/TEX]
Bài 1:

A=$2.(x^2 - y^2).(x^4 + x^2y^2 + y^4) - 6.(x^4 + y^4)$

=$ 4.x^4 + 4.x^2y^2 +4y^4 - 6.x^4 - 6.y^4$

=$ -2.( x^2 - y^2)^2 = -2.2^2 = -8$
 
E

eye_smile

2.$169=(3x+2y)^2 \le (3^2+2^2)(x^2+y^2)=13(x^2+y^2)$

\Rightarrow $x^2+y^2 \ge 13$
 
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