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....

\(A=\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}\)

\(=\sqrt{\left(\sqrt{5}\right)^2-4\sqrt{5+2^2}}-\sqrt{\left(\sqrt{5}\right)^2+4\sqrt{5}+2^2}\)

\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}+2\right)^2}\)

\(=\left|\sqrt{5}-2\right|-\left|\sqrt{5}+2\right|\)

\(=\sqrt{5}-2-\left(\sqrt{5}+2\right)\)

\(=-4\)

\(B=\sqrt[3]{9}.\sqrt[3]{-3}+\left(1+\sqrt{2}\right)^2\)

\(=-\sqrt[3]{27}+3+2\sqrt{2}\)

\(=-3+3+2\sqrt{2}\)

\(=2\sqrt{2}\)

a)  ĐK:  \(x\ge5\)

 \(\sqrt{4x-20}+\frac{1}{3}\sqrt{9x-45}-\frac{1}{5}\sqrt{16x-80}=0\)

\(\Leftrightarrow\)\(\sqrt{4\left(x-5\right)}+\frac{1}{3}\sqrt{9\left(x-5\right)}-\frac{1}{5}\sqrt{16\left(x-5\right)}=0\)

\(\Leftrightarrow\)\(2\sqrt{x-5}+\sqrt{x-5}-\frac{4}{5}\sqrt{x-5}=0\)

\(\Leftrightarrow\)\(\frac{11}{5}\sqrt{x-5}=0\)

\(\Leftrightarrow\)\(x-5=0\)

\(\Leftrightarrow\)\(x=5\) (t/m)

Vậy

b)  \(-5x+7\sqrt{x}=-12\)

\(\Leftrightarrow\)\(5x-7\sqrt{x}-12=0\)

\(\Leftrightarrow\)\(\left(\sqrt{x}+1\right)\left(5\sqrt{x}-12\right)=0\)

đến đây tự làm

c) d) e) bạn bình phương lên

f)  \(VT=\sqrt{3\left(x^2+2x+1\right)+9}+\sqrt{5\left(x^4-2x^2+1\right)+25}\)

             \(=\sqrt{3\left(x+1\right)^2+9}+\sqrt{5\left(x^2-1\right)^2}\)

           \(\ge\sqrt{9}+\sqrt{25}=8\)

Dấu "=" xảy ra  \(\Leftrightarrow\)\(\hept{\begin{cases}x+1=0\\x^2-1=0\end{cases}}\)\(\Leftrightarrow\)\(x=-1\)

Vậy...