Ta thấy hàm số này chỉ có cực đại. Và bị chặn 2 đầu. Vậy đầu chặn nào bé hơn chính là min

Vì 4 - 2x² \(\ge0\)

\(-\sqrt{2}\le x\le\sqrt{2}\)

Tại x = \(\sqrt{2}\) thì hàm số = \(2\sqrt{2}\)

Tại x = -\(\sqrt{2}\) thì hàm số = - \(2\sqrt{2}\)

Vậy min là - \(2\sqrt{2}\)tại x = - \(\sqrt{2}\)

A B C O D E F

Ta có : \(\frac{OD}{AD}=\frac{S_{BOC}}{S_{ABC}}\) ; \(\frac{OE}{BE}=\frac{S_{AOC}}{S_{ABC}}\) ; \(\frac{OF}{CF}=\frac{S_{ABO}}{S_{ABC}}\)

\(\Rightarrow\frac{OD}{AD}+\frac{OE}{BE}+\frac{OF}{CF}=\frac{S_{ABC}}{S_{ABC}}=1\)

\(\Rightarrow\left(1-\frac{OD}{AD}\right)+\left(1-\frac{OE}{BE}\right)+\left(1-\frac{OF}{CF}\right)=2\)

\(\Rightarrow\frac{OA}{AD}+\frac{OB}{BE}+\frac{OC}{CF}=2\)

hay \(R\left(\frac{1}{AD}+\frac{1}{BE}+\frac{1}{CF}\right)=2\Rightarrow\frac{1}{AD}+\frac{1}{BE}+\frac{1}{CF}=\frac{2}{R}\) 

mà ta có \(\frac{1}{AD}+\frac{1}{BE}+\frac{1}{CF}\ge\frac{9}{AD+BE+CF}\)

\(\Rightarrow\frac{2}{R}\ge\frac{9}{AD+BE+CF}\)

\(\Rightarrow AD+BE+CF\ge\frac{9R}{2}\)(đpcm)

Khó quá! Em mới học lớp 7

Áp dụng BĐT Cô-si cho 2 số dương ta có:

\(\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab}\left(1\right)\)

\(\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{2}{bc}\left(2\right)\)

\(\frac{1}{c^2}+\frac{1}{a^2}\ge\frac{2}{ac}\left(2\right)\)

Từ (1) ;(2) và (3) suy ra:

\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=\frac{a+b+c}{abc}=6\)

Vậy \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge6\).Dấu "=" xảy ra <=>\(\hept{\begin{cases}a+b+c=6abc\\\frac{1}{a^2}=\frac{1}{b^2}=\frac{1}{c^2}\end{cases}=>a=b=c=\frac{1}{\sqrt{2}}}\)

A = \(x-2\sqrt{xy}+3y-2\sqrt{x}+1\)

\(=\left(\frac{x}{3}-\frac{2\times\sqrt{3}\sqrt{xy}}{\sqrt{3}}+3y\right)+\left(\frac{2x}{3}-\frac{2\times\sqrt{2}\times\sqrt{3}\sqrt{x}}{\sqrt{2}\times\sqrt{3}}+\frac{3}{2}\right)-\frac{1}{2}\)

\(=\left(\frac{\sqrt{x}}{\sqrt{3}}-\sqrt{3y}\right)^2+\left(\sqrt{\frac{2x}{3}}-\sqrt{\frac{3}{2}}\right)^2-\frac{1}{2}\)

\(\ge-\frac{1}{2}\)

\(\Leftrightarrow\left(x-m\right)^2=1-x^2\)

\(\Leftrightarrow2x^2-2mx+m^2-1=0\)

có \(\Delta'=m^2-2\left(m^2-1\right)=2-m^2\)

phương trình có nghiệm duy nhất khi \(\Delta'=0\)<=> 2-m^2=0   <=> m \(\in\left\{\sqrt{2},-\sqrt{2}\right\}\)

vậy...

Cha kho the nhi?

512 

ko chac

Kho the

\(\left(\frac{ID}{AD}+\frac{IE}{BE}+\frac{IF}{CF}\right)\left(\frac{AD}{ID}+\frac{BE}{IE}+\frac{CF}{IF}\right)\ge\left(\sqrt{\frac{ID}{AD}}\sqrt{\frac{AD}{ID}}+\sqrt{\frac{IE}{BE}}\sqrt{\frac{BE}{IE}}+\sqrt{\frac{IF}{CF}}\sqrt{\frac{CF}{IF}}\right)^2\)

\(\Rightarrow\frac{AD}{ID}+\frac{BE}{IE}+\frac{CF}{IF}\ge\left(1+1+1\right)^2\Leftrightarrow\frac{IA+ID}{ID}+\frac{IB+IE}{IE}+\frac{IC+IF}{IF}\ge9\)

\(\Rightarrow\frac{IA}{ID}+\frac{IB}{IE}+\frac{IC}{IF}\ge6\)

Bạn ko hiểu chỗ nào thì hỏi mình nhé!

tôi không biết

Ta có:\(\sqrt{a+1}-\sqrt{a-1}=\frac{\left(\sqrt{a+1}-\sqrt{a-1}\right)\left(\sqrt{a+1}+\sqrt{a-1}\right)}{\sqrt{a+1}+\sqrt{a-1}}\)

\(=\frac{2}{\sqrt{a+1}+\sqrt{a-1}}\)(1)

Lại có:\(\sqrt{a+1}+\sqrt{a-1}< 2\sqrt{a}\)

 \(\Leftrightarrow\left(\sqrt{a+1}+\sqrt{a-1}\right)^2=2a+2\sqrt{a^2-1}< \left(2\sqrt{a}\right)^2=2a+2a\)\(\Leftrightarrow\sqrt{a^2-1}< \sqrt{a^2}\)\(\Rightarrow\sqrt{a+1}+\sqrt{a-1}< 2\sqrt{a}\)(2)

Từ (1) và (2)\(\Rightarrow\frac{2}{\sqrt{a+1}+\sqrt{a-1}}>\frac{2}{2\sqrt{a}}=\frac{1}{\sqrt{a}}\)\(\Rightarrow\sqrt{a+1}-\sqrt{a-1}>\frac{1}{\sqrt{a}}\)

\(\Rightarrow\sqrt{a+1}>\frac{1}{\sqrt{a}}+\sqrt{a-1}\)

Có: \(a^3+b^3=c^3\Leftrightarrow\left(\frac{a}{c}\right)^3+\left(\frac{b}{c}\right)^3=1.\)
Đặt : \(\frac{a}{c}=x;\frac{b}{c}=y\). Suy ra \(0< x< 1;0< y< 1\).
Vì vậy: \(x^{2010}< x^3;y^{2010}< y^3.\)
Từ đó: \(x^{2010}+y^{2010}< x^3+y^3< 1\).
Suy ra: \(\left(\frac{a}{c}\right)^{2010}+\left(\frac{b}{c}\right)^{2010}< 1\)hay: \(a^{2010}+b^{2010}< c^{2010}.\)
 

a^2010+b^2001<c^2010

\(=\left(\frac{\sqrt{1+m}}{\sqrt{1+m}-\sqrt{1-m}}+\frac{\sqrt{1-m}\cdot\sqrt{1-m}}{\sqrt{1-m}\cdot\left(\sqrt{1+m}-\sqrt{1-m}\right)}\right)\cdot\frac{\sqrt{1-m^2}-1}{m}\)

\(=\frac{\sqrt{1+m}+\sqrt{1-m}}{\sqrt{1+m}-\sqrt{1-m}}\cdot\frac{\sqrt{1-m^2}-1}{m}\)

\(=\frac{\left(\sqrt{1+m}+\sqrt{1-m}\right)^2}{\left(\sqrt{1+m}-\sqrt{1-m}\right)\left(\sqrt{1+m}+\sqrt{1-m}\right)}\cdot\frac{\sqrt{1-m^2}-1}{m}\)

\(=\frac{1+m-m+1+2\sqrt{1-m^2}}{2m}\cdot\frac{\sqrt{1-m^2}-1}{m}\)

\(=\frac{\sqrt{1-m^2}+1}{m}\cdot\frac{\sqrt{1-m^2}-1}{m}=\frac{1-m^2-1}{m^2}=-1\)