g12hg5221hjoplj./k';ll;

it so hard

A B C H

\(cosC=cos30^0=\frac{AC}{BC}=\frac{\sqrt{3}}{2}\)

\(\Rightarrow\)\(BC=\frac{AC.2}{\sqrt{3}}=\frac{16}{\sqrt{3}}\)

\(tanC=tan30^0=\frac{AB}{AC}=\frac{1}{\sqrt{3}}\)

\(\Rightarrow\)\(AB=\frac{AC}{\sqrt{3}}=\frac{8}{\sqrt{3}}\)

\(sinC=sin30^0=\frac{AH}{AC}=\frac{1}{2}\)

\(\Rightarrow\)\(AH=\frac{AC}{2}=4\)

k mk đi 

ai k mk 

mk k lại

thanks

Ok, vậy đừng dùng olm nữa

ai giúp mik vs : cảm ơn mn nhé >3

ai giúp mik đi huhu

\(Q=\frac{x+y-2\sqrt{xy}}{\sqrt{x}-\sqrt{y}}\div\frac{1}{\sqrt{x}-\sqrt{y}}\)

\(=\frac{x-2\sqrt{xy}+y}{\sqrt{x}-\sqrt{y}}\div\frac{1}{\sqrt{x}-\sqrt{y}}\)

\(=\frac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\sqrt{x}-\sqrt{y}}\times\left(\sqrt{x}-\sqrt{y}\right)\)

\(=x-y\)

\(Q=\frac{x+y-2\sqrt{xy}}{\sqrt{x}-\sqrt{y}}\div\frac{1}{\sqrt{x}-\sqrt{y}}\)

\(=\frac{x-2\sqrt{xy}+y}{\sqrt{x}-\sqrt{y}}\div\frac{1}{\sqrt{x}-\sqrt{y}}\)

\(=\frac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\sqrt{x}-\sqrt{y}}\times\left(\sqrt{x}-\sqrt{y}\right)\)

\(=x-y\)

\(\sqrt{x+2}+x=4\)

\(\text{Đ}K\text{X}\text{Đ}\)\(:\)  \(x\ge-2\)

\(\sqrt{x+2}=4-x\)

\(\Leftrightarrow\hept{\begin{cases}4-x\ge0\\x+2=\left(4-x\right)^2\end{cases}}\)

\(\Leftrightarrow x=2\) 

................

ĐK:  \(-2\le x\le4\)

\(\sqrt{x+2}+x=4\)

\(\Leftrightarrow\)\(\sqrt{x+2}=4-x\)

\(\Leftrightarrow\)\(x+2=16-8x+x^2\)

\(\Leftrightarrow\)\(x^2-9x+14=0\)

\(\Leftrightarrow\)\(\left(x-2\right)\left(x-7\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2\left(t/m\right)\\x=7\left(L\right)\end{cases}}\)

Vậy....