Question

[Nguyễn Hải An]

giúp em vs

rảnh em hậu tạ sau ạ !!!

Answer 1

[Lightning Farron]

Bài 4:

Áp dụng BĐT AM-GM ta có:

\(\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}\ge2\sqrt{\dfrac{a^2}{b^2}\cdot\dfrac{b^2}{a^2}}=2\)

\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}\cdot\dfrac{b}{a}}=2\Rightarrow3\left(\dfrac{a}{b}+\dfrac{b}{c}\right)\ge6\)

\(\Rightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}-3\left(\dfrac{a}{b}+\dfrac{b}{a}\right)\ge2-6=-4\)

\(\Rightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}-3\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+4\ge-4+4=0\) (đúng)

Answer 2

[Nguyệt Nguyệt]

Hình vẽ không chính xác lắm thông cảm

a) Vì OM song song với AB nên \(\dfrac{OM}{AB}=\dfrac{OD}{BD}\)

Vì OM song song với CD nên \(\dfrac{OM}{CD}=\dfrac{OA}{AC}\)

Vì AB song song với CD nên \(\dfrac{OA}{AC}=\dfrac{OB}{BD}\) nên \(\dfrac{OM}{CD}=\dfrac{OB}{BD}\)

Do đó \(\dfrac{OM}{AB}+\dfrac{OM}{CD}=\dfrac{OD}{BD}+\dfrac{OA}{AC}=\dfrac{OD}{BD}+\dfrac{OB}{BD}=1\)

Hay \(OM\left(\dfrac{1}{AB}+\dfrac{1}{CD}\right)=1\) suy ra \(\dfrac{1}{AB}+\dfrac{1}{CD}=\dfrac{1}{OM}\)

Lại có ON song song với CD nên \(\dfrac{ON}{CD}=\dfrac{OB}{BD}\) mà \(\dfrac{OB}{BD}=\dfrac{OM}{CD}\) nên \(\dfrac{ON}{CD}=\dfrac{OM}{CD}\) hay OM = ON = \(\dfrac{1}{2}\)MN

Suy ra \(\dfrac{1}{AB}+\dfrac{1}{CD}=\dfrac{1}{OM}=\dfrac{1}{\dfrac{1}{2}MN}=\dfrac{2}{MN}\)

b) Dễ chứng minh S_ADC = S_BDC

Mà S_ADC = S_AOD+S_OCD và S_BDC = S_BOC+S_OCD

Suy ra S_AOD = S_BOC

Lại có \(\dfrac{S_{AOD}}{S_{AOB}}=\dfrac{OD}{OB}\) và \(\dfrac{S_{OCD}}{S_{BOC}}=\dfrac{OD}{OB}\)

Nên \(\dfrac{S_{AOD}}{S_{AOB}}=\dfrac{S_{OCD}}{S_{BOC}}\) \(\Leftrightarrow\) \(S_{AOD}.S_{BOC}=S_{AOB}.S_{OCD}\)

Hay \(S_{AOD}=S_{BOC}=\sqrt{S_{AOB}.S_{OCD}}=\sqrt{a^2.b^2}=ab\)

Khi đó \(S_{ABCD}=S_{AOD}+S_{BOC}+S_{AOB}+S_{OCD}=ab+ab+a^2+b^2=a^2+b^2+2ab=\left(a+b\right)^2\)

A B C D O M N