\(#TyHM\)

a: ĐKXĐ: a>0; a<>1

\(A=\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\cdot\dfrac{a}{\sqrt{a}+1}=\dfrac{a-1}{\sqrt{a}}\cdot\dfrac{a}{\sqrt{a}+1}\)

\(=\sqrt{a}\left(\sqrt{a}-1\right)=a-\sqrt{a}\)

b: A=a-căn a+1/4-1/4

=(căn a-1/2)^2-1/4>=-1/4

Dấu = xảy ra khi a=1/4

2: Để (d)//y=(m²+1)x-4 thì \(\left\{{}\begin{matrix}m^2=1\\m-5\ne-4\end{matrix}\right.\Leftrightarrow m=1\)

Câu 18:

\(n_{NO}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ \text{Đ}\text{ặt}:n_{Al}=a\left(mol\right);n_{Fe}=b\left(mol\right)\left(a,b>0\right)\\ Al+4HNO_3\rightarrow Al\left(NO_3\right)_3+NO+2H_2O\\ Fe+4HNO_3\rightarrow Fe\left(NO_3\right)_3+NO+2H_2O\\ \Rightarrow\left\{{}\begin{matrix}a+b=0,15\\27a+56b=5,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,05\end{matrix}\right.\\ \Rightarrow\%m_{Al}=\dfrac{0,1.27}{5,5}.100\approx49,091\%;\%m_{Fe}\approx50,909\%\\ b,n_{HNO_3}=4.n_{NO}=0,6\left(mol\right)\\ V_{\text{dd}HNO_3}=\dfrac{0,6}{0,3}=2\left(M\right)\)

c) Dung dịch Y là dung dịch nào?

a)\(2Ca+O_2\underrightarrow{t^o}2CaO\)

   \(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)

   \(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3\downarrow+H_2O\)

   \(CaCO_3\underrightarrow{t^o}CaO+CO_2\)

   \(CaO+2HCl\rightarrow CaCl_2+H_2O\)