1.2 với \(x\ge0,x\in Z\)

A=\(\dfrac{2\sqrt{x}+7}{\sqrt{x}+2}=2+\dfrac{3}{\sqrt{x}+2}\in Z< =>\sqrt{x}+2\inƯ\left(3\right)=\left(\pm1;\pm3\right)\)

*\(\sqrt{x}+2=1=>\sqrt{x}=-1\)(vô lí)

*\(\sqrt{x}+2=-1=>\sqrt{x}=-3\)(vô lí
*\(\sqrt{x}+2=3=>x=1\)(TM)

*\(\sqrt{x}+2=-3=\sqrt{x}=-5\)(vô lí)

vậy x=1 thì A\(\in Z\)

\(\left(2x+\dfrac{1}{2}y\right)^3=\left(2x\right)^3+3.\left(2x\right)^2.\dfrac{1}{2}y+3.2x.\left(\dfrac{1}{2}y\right)^2+\left(\dfrac{1}{2}\right)^3=8x^3+6x^2y+\dfrac{3}{2}xy^2+\dfrac{1}{8}y^3\)

Câu 1:

\((1)Mg+H_2SO_4\to MgSO_4+H_2\\ (2)MgSO_4+2NaOH\to Mg(OH)_2\downarrow+Na_2SO_4\\ (3)Mg(OH)_2+2HCl\to MgCl_2+2H_2O\\ (4)MgCl_2+2AgNO_3\to Mg(NO_3)_2+2AgCl\downarrow\)

Câu 2:

\(a,Mg,Al,Cu,Ag\\ b,Mg,Al\\ Mg+2HCl\to MgCl_2+H_2\\ 2Al+6HCl\to 2AlCl_3+3H_2\)

\(c,\) Dùng Al vì Al bị thụ động hóa với \(H_2SO_4\) đặc nguội

b

\(\text{B.}50^0\)