\(\dfrac{A}{B}=\dfrac{2\sqrt{x}-1}{\sqrt{x}+2}:\dfrac{\sqrt{x}-3}{\sqrt{x}+2}=\dfrac{2\sqrt{x}-1}{\sqrt{x}+3}\)

\(\Rightarrow\)\(\dfrac{2\sqrt{x}-1}{\sqrt{x}-3}< \dfrac{1}{2}\)

\(\dfrac{2\sqrt{3}-\sqrt{21}}{2-\sqrt{7}}+\dfrac{6}{3+\sqrt{3}}\)

\(=\dfrac{\sqrt{3}\left(2-\sqrt{7}\right)}{2-\sqrt{7}}+\dfrac{6\left(3-\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\)

\(=\sqrt{3}+\dfrac{6\left(3-\sqrt{3}\right)}{3^2-\left(\sqrt{3}\right)^2}=\sqrt{3}+\dfrac{6\left(3-\sqrt{3}\right)}{6}=\sqrt{3}+3-\sqrt{3}=3\)

Vậy...

Gọi muối đó là:\(M_2CO_3\)

\(M_2CO_3+2HCl\rightarrow2MCl+H_2O+CO_2\) (1)

Từ (1)\(\Rightarrow n_{M_2CO_3}=n_{CO_2}\)

\(\Leftrightarrow2.\dfrac{2,12}{2M+60}=\dfrac{0,38}{12+2.16}\)

\(\Leftrightarrow M=216\)(ko có M thỏa mãn)

Vậy sai đề :v.

\(n_M=\dfrac{11,2}{M}\left(mol\right),n_{HCl}=0,4\left(mol\right)\)

 \(M+2HCl\rightarrow MCl_2+H_2\) (1)

Từ (1)\(\Rightarrow2n_M=n_{HCl}\Leftrightarrow2.\dfrac{11,2}{M}=0,4\)

\(\Leftrightarrow M=56\)

Vậy M là Fe

Ý B

\(g_h=\dfrac{1}{2}g_{MĐ}\)

\(\Leftrightarrow\dfrac{GM}{\left(R+h\right)^2}=\dfrac{1}{2}\dfrac{GM}{R^2}\)

\(\Leftrightarrow h\simeq2651\left(km\right)\)

B8)

Tiết diện của dây: \(S=\dfrac{\rho.l}{R}=\dfrac{0,4.10^{-6}.3}{2}=6.10^{-7}\left(m^2\right)=0,6\left(mm^2\right)\)

Bán kính thiết diện dây:\(S=\pi R^2\Rightarrow R=\sqrt{\dfrac{S}{\pi}}\simeq0,4\left(mm\right)\)

Ý A

B9)\(R_1=R_2\Leftrightarrow\rho_1.\dfrac{l}{S_1}=\rho_2.\dfrac{l}{S_2}\)

\(\Leftrightarrow\rho_1.\dfrac{l}{\dfrac{\pi d_1^2}{4}}=\rho_2.\dfrac{l}{\dfrac{\pi d^2_2}{4}}\)

\(\Leftrightarrow d_2^2=\dfrac{\rho_2.d_1^2}{\rho_1}\simeq6.10^{-7}\left(m\right)\Rightarrow d_2\simeq7,861.10^{-4}\left(m\right)=0,786\left(mm\right)\)

Ý D

\(3x^2-6x+3=0\Leftrightarrow x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)