\(\frac{\left(2+\sqrt{3}\right)^n-\left(2-\sqrt{3}\right)^n}{2\sqrt{3}}=\frac{A+B\sqrt{3}-A+B\sqrt{3}}{2\sqrt{3}}=B\)( A,B thuộc Z )

trong tam giac vuong ABH  ta co \(AH=\sin B\cdot AB\) \(\Rightarrow AH=8\sqrt{3}\)

\(BH=\cos B\cdot AB=8\)

trong tam giac AHC co \(HC^2+AH^2=AC^2\Rightarrow HC^2=14^2-\left(8\sqrt{3}\right)^2=4\Rightarrow HC=2\)

        \(\Rightarrow BC=BH+HC=8+2=10\)

\(\Rightarrow SABC=\frac{1}{2}BC\cdot AH=\frac{1}{2}\cdot10\cdot8\sqrt{3}=40\sqrt{3}\)

\(E^3=182+\sqrt{33125}+182-\sqrt{33125}+3\sqrt[3]{182^2-33125}\left(E\right)\)

   =\(364-3E\)

\(\Rightarrow E^3+3E-364=0\) 

\(\Leftrightarrow E^3-7E^2+7E^2-49E+52E-364=0\)

\(\Leftrightarrow\left(E-7\right)\left(E^2+7E+52\right)=0\)

\(\Rightarrow E=7\)

ta có \(E^3=\left(\sqrt[3]{182+\sqrt{33125}}+\sqrt[3]{182-\sqrt{33125}}\right)^3\)

\(E^3=\left(182+\sqrt{33125}\right)+\left(182-\sqrt{33125}\right)+3\cdot E\cdot\sqrt[3]{33124-33125}\)

\(E^3=364-3E\)

giải phương trình \(E^3+3E-364=0\)

suy ra E= 7

trong tam giac ABH co\(AH=AB\cdot\sin B\) \(\Rightarrow AH=60\cdot\sin30=30\)

trong tam giac AHC co \(\sin C=\frac{AH}{AC}\Rightarrow AC=\frac{30}{\sin130}\approx39\)(vi \(gocC=180-20-30=130\)

TRONG TAM GIAC APC CO\(PC=AC\cdot\sin A=39\cdot\sin20\approx13,34\)

                                               \(AP=\cos A\cdot39\approx36,65\)

\(\Rightarrow AP+BP=AB\Rightarrow BP=60-36.65=23.35\)

(gam). => m =14,8 – 6,4 = 8,4 (gam).

      \(Fe+2HCI->FeCl_2+H_2\)                  

             => V = 0,15.22,4 = 3,36 (lít)

chọn D

sai đề 

4 PT(1) -3 PT(2)

Nhấn chéo hay vế cũng đ.c ^_^

\(\Leftrightarrow\)\(\left(x+y\right)^2\ge4xy\)

\(\Leftrightarrow\)\(x^2-2xy+y^2\ge0\)

\(\Leftrightarrow\)\(\left(x-y\right)^2\ge0\) ( luôn đúng với mọi x, y >0)

Vậy \(\frac{x+y}{xy}\ge\frac{4}{x+y}\)

\(\frac{x+8}{\sqrt{x-1}}=\frac{x-1+9}{\sqrt{x-1}}=\sqrt{x-1}+\frac{9}{\sqrt{x-1}}\)

theo cô-si có

\(\sqrt{x-1}+\frac{9}{\sqrt{x-1}}\ge2\sqrt{\sqrt{x-1}\times\frac{9}{\sqrt{x-1}}}=2\sqrt{9}=6\)

\(\frac{x+8}{\sqrt{x-1}}\ge6\Leftrightarrow\sqrt{x-1}\ge0\Leftrightarrow x\ge1\)

vậy\(\frac{x+8}{\sqrt{x-1}}\ge6\)với mọi \(x\ge1\)