Đề bài là gì bạn

\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\\ =\sqrt{2}.\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{2}.\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\\ =\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{4+2\sqrt{3}}}+\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{4-2\sqrt{3}}}\\ =\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{\left(\sqrt{3}+1\right)^2}}+\dfrac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{\left(\sqrt{3}-1\right)^2}}\\ =\dfrac{\left|\sqrt{3}-1\right|}{\left|\sqrt{3}+1\right|}+\dfrac{\left|\sqrt{3}+1\right|}{\left|\sqrt{3}-1\right|}\\ =\dfrac{\left(\sqrt{3}-1\right)^2+\left(\sqrt{3}+1\right)^2}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\\ =\dfrac{4-2\sqrt{3}+4+2\sqrt{3}}{\sqrt{3^2}-1}\\ =\dfrac{8}{2}\\ =4\)

A = \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\) + \(\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)  = \(\dfrac{\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}}{\sqrt{\left(2-\sqrt{3}\right).\left(2+\sqrt{3}\right)}}\)

A = \(\dfrac{2-\sqrt{3}+2+\sqrt{3}}{\sqrt{4-3}}\) = \(\dfrac{4}{1}\) = 4

Điều kiện: \(-\dfrac{65}{8}\le x\le2\)

\(1+8x+8^2=\sqrt{2-x}\\ \Rightarrow2-x=64x^2+1040x+4225\\ \Leftrightarrow64x^2+1041x+4223=0\\ \Leftrightarrow\left[{}\begin{matrix}x\simeq-7,735\\x\simeq-8,531\end{matrix}\right.\)

between 

 Ha Noi will have temperature between 23^oC and 27^oC

a) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\) (*)

Phương trình hóa học

Mg + 2HCl ---> MgCl₂ + H₂ (**)

MgO + 2HCl ---> MgCl₂ + H₂O (***)

b) Từ (*) và (**) ta có \(n_{Mg}=0,15\Leftrightarrow m_{Mg}=0,15.24=3,6\left(g\right)\)

\(\Rightarrow m_{MgO}=10-3,6=6,4\left(g\right)\)

 \(\%Mg=\dfrac{3,6}{10}.100\%=36\%\)

\(\%MgO=\dfrac{6,4}{10}.100\%=64\%\)

c) Xét phản ứng (**) ta có \(m_{MgO}=6,4\left(g\right)\Leftrightarrow n_{MgO}=n_{MgCl_2}=\dfrac{1}{2}n_{HCl}=0,16\left(mol\right)\) (1) 

\(\Leftrightarrow n_{HCl}=0,32\left(mol\right)\)

Tương tự có số mol HCl trong phản ứng (*) là 0,3 mol

\(C_M=\dfrac{0,32+0,3}{0,2}=3,1\left(M\right)\)

d) Từ (1) ; (*) ; (**) ta có : \(n_{MgCl_2}=0,15+0,16=0,31\left(mol\right)\)

\(m_{MgCl_2}=0,31.95=29,45\left(g\right)\)

e) \(C_M=\dfrac{0,31}{0,2}=1,55\left(M\right)\)