bài này dễ ẹt ak 

nhưng giúp mình bài này đi 

chotam giac abc . co canh bc=12cm, duong cao ah=8cm

a> tinh s tam giac abc

b> tren canh bc lay diem e sao cho be=3/4bc. tinh s tam giac abe va s tam giac ace ( bằng nhiều cách )

c> lay diem chinh giua cua canh ac va m . tinh s tam giac ame

1.

\(\frac{\sqrt{b^2+2a^2}}{ab}+\frac{\sqrt{c^2+2b^2}}{bc}+\frac{\sqrt{a^2+2c^2}}{ca}\ge\sqrt{3}\left(1\right)\)

Ta có ab+bc+ca=abc nên \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\)

\(\left(1\right)\Leftrightarrow\sqrt{\frac{1}{a^2}+\frac{2}{b^2}}+\sqrt{\frac{1}{b^2}+\frac{2}{c^2}}+\sqrt{\frac{1}{c^2}+\frac{2}{a^2}}\ge\sqrt{3}\)

Trong mặt phẳng với hệ tọa độ Oxy, với các Vecto

\(\overrightarrow{u}=\left(\frac{1}{a};\frac{\sqrt{2}}{b}\right);\left|\overrightarrow{u}\right|=\sqrt{\frac{1}{a^2}+\frac{2}{b^2}}\)

\(\overrightarrow{v}=\left(\frac{1}{b};\frac{\sqrt{2}}{c}\right)\Rightarrow\left|\overrightarrow{v}\right|=\sqrt{\frac{1}{b^2}+\frac{2}{c^2}}\)

\(\overrightarrow{w}=\left(\frac{1}{c};\frac{\sqrt{2}}{a}\right)\Rightarrow\left|\overrightarrow{w}\right|=\sqrt{\frac{1}{c^2}+\frac{2}{a^2}}\)

Ta có \(\overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c};2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\right)=\left(1;\sqrt{2}\right)\)

=> \(\left|\overrightarrow{u}\right|+\left|\overrightarrow{v}\right|+\left|\overrightarrow{w}\right|=\sqrt{1+2}=\sqrt{3}\)

Mặt khác \(\left|\overrightarrow{u}\right|+\left|\overrightarrow{v}\right|+\left|\overrightarrow{w}\right|\ge\left|\overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}\right|\)

\(\Rightarrow\frac{\sqrt{b^2+2a^2}}{ab}+\frac{\sqrt{c^2+2b^2}}{bc}+\frac{\sqrt{a^2+2c^2}}{ac}\ge\sqrt{3}\)

Dấu "=" xảy ra <=> a=b=c

a)

$3^x-y^3=1$

$\Leftrightarrow 3^x=y^3+1$

$\Leftrightarrow 3^x=(y+1)(y^2-y+1)$

$\gcd(y+1,y^2-y+1)=\gcd(y+1,3)$

Vì $3^x$ chỉ có ước nguyên tố là $3$ nên

$y+1=3^m,\quad y^2-y+1=3^n\qquad (m,n\in\mathbb N,\ m+n=x)$

Ta có $y^2-y+1-(y+1)(y-2)=3$ nên $\gcd(y+1,y^2-y+1)\mid 3$

Suy ra $\gcd(y+1,y^2-y+1)=1$ hoặc $3$.

Nếu $\gcd=1$ thì $y+1=1$

$\Rightarrow y=0$

$\Rightarrow 3^x=1$

$\Rightarrow x=0$.

Nếu $\gcd=3$ thì $3\mid y+1$

$\Rightarrow y\equiv2\pmod3$

$\Rightarrow y^2-y+1\equiv4-2+1\equiv3\equiv0\pmod3$

Lại có $y^2-y+1=(y+1)^2-3y$ nên $9\nmid (y^2-y+1)$

Suy ra $y^2-y+1=3$

$\Rightarrow y^2-y-2=0$

$\Rightarrow y=2$.

Khi đó $3^x=2^3+1=9$ $\Rightarrow x=2$.

Vậy $\boxed{(x,y)=(0,0)\ \text{hoặc}\ (2,2).}$

b/

$a+b+c=0$

$\Rightarrow c=-(a+b)$

$ab+2c^2=ab+2(a+b)^2$

$=2a^2+5ab+2b^2$

$=(2a+b)(a+2b)$

Tương tự $bc+2a^2=(2a+b)(a-b)$

$ca+2b^2=(a+2b)(b-a)$

Suy ra $(ab+2c^2)(bc+2a^2)(ca+2b^2)$

$=(2a+b)^2(a+2b)^2(a-b)(b-a)$

$=-(2a+b)^2(a+2b)^2(a-b)^2$ $\le 0$

Do đó $N=1-(ab+2c^2)(bc+2a^2)(ca+2b^2)$$=1+(2a+b)^2(a+2b)^2(a-b)^2$

$\ge 1$$>0$

=> N là số dương.