$\dfrac1{AM^2}+\dfrac1{AN^2}=\dfrac4{3AB^2}$
$\Leftrightarrow \dfrac{AM^2}{AN^2}+1=\dfrac{4AM^2}{3AB^2} (1)$
Mà $AB//DN \Rightarrow \dfrac{AM}{AN}=\dfrac{BM}{BC}=\dfrac{BM}{AB}$
$(1)\Leftrightarrow \dfrac{BM^2}{AB^2}+1=\dfrac{4AM^2}{AB^2} (*)$
Kẻ $AH$ vuông góc với $BC$
Mà $ABCD$ là hình bình hành có $\widehat{BAD}=120^o \Rightarrow \widehat{ABC}=60^o$
$\Rightarrow \widehat{BAH}=90^o-\widehat{ABC}=30^o=2\widehat{MAB}$
$\Rightarrow AM$ là phân giác $\widehat{BAH}$
$\Rightarrow \dfrac{MB}{MH}=\dfrac{AB}{AH}=\dfrac1{\cos _\widehat{BAH}}=\dfrac1{\cos _{30^o}}=\dfrac2{\sqrt3}$
$\Rightarrow \dfrac {MB}{BH}=\dfrac2{2+\sqrt 3} \Rightarrow MB=\dfrac2{2+\sqrt 3}BH$
Mà $\dfrac {BH}{AB}=\sin_{30^o}=\dfrac 12 \Rightarrow BH=\dfrac12 AB (2)$
$\Rightarrow MB=\dfrac1{2+\sqrt 3}AB (3)$
$\Rightarrow MB^2=\dfrac1{7+4\sqrt3}AB^2$
$\Leftrightarrow \dfrac{MB^2}{AB^2}=\dfrac1{7+4\sqrt3} (**)$
$(2;(3) \Rightarrow MH=BH-MB=\dfrac 12 AB-\dfrac1{2+\sqrt 3}AB=\dfrac {\sqrt3}{4+2\sqrt3}AB$
$\Rightarrow AM^2=MH^2+AH^2=(\dfrac {\sqrt3}{4+2\sqrt3}AB)^2+(\dfrac{\sqrt3}2AB)^2= \dfrac 34\big(1+\dfrac1{7+4\sqrt3}\big)AB^2$
$\Leftrightarrow \dfrac{4AM^2}{3AB^2}=1+\dfrac1{7+4\sqrt3} (***)$
$(*);(**);(***) \Rightarrow \dfrac1{AM^2}+\dfrac1{AN^2}=\dfrac4{3AB^2} \Leftrightarrow 1+\dfrac1{7+4\sqrt3}=1+\dfrac1{7+4\sqrt3}$ luôn đúng
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