toán 9 tìm min

T

transformers123

Bài 1:
$D=\dfrac{1}{16x}+\dfrac{1}{4y}+\dfrac{1}{z} =\dfrac{1}{16x}+\dfrac{4}{16y}+\dfrac{16}{16z} \ge \dfrac{(1+2+4)^2}{16(x+y+z)} = \dfrac{49}{16}$
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