\(\dfrac{M_A}{M_{H_2}}=37,5\) \(\Rightarrow M_A=37,5.2=75\) \((g/mol)\)

\(m_O=7,5-2,4-0,5-1,4=3,2\left(g\right)\)

Gọi CTTQ A: \(C_xH_yO_zN_t\)

\(\left\{{}\begin{matrix}n_C=\dfrac{2,4}{12}=0,2\left(mol\right)\\n_H=\dfrac{0,5}{1}=0,5\left(mol\right)\\n_N=\dfrac{1,4}{14}=0,1\left(mol\right)\\n_O=\dfrac{3,2}{16}=0,2\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow x:y:z:t=0,2:0,5:0,2:0,1\)

                       \(=2:5:2:1\)

CTĐG A: \(\left(C_2H_5O_2N\right)n=75\)

                       \(\Rightarrow n=1\)

`=>` CTPT A: \(C_2H_5O_2N\)

CTHH đúng: Mg;H₂;H₂S;MgSO₄;Cu(NO₃)₂;Fe₂(SO₄)₃;P₂O₅

CTHH sai: 

S₂ `->` S

Cl `->` Cl₂

KS₂ `->` K₂S

AlNO₃ `->` Al(NO₃)₃

SO₄ `->` SO₃;SO₂

CaPO₄ `->` Ca₃(PO₄)₂

NaCO₃ `->` Na₂CO₃

CO₃ `->` CO₂;CO

\(\widehat{xOz}=\widehat{xOy}+\widehat{yOz}=30^o+50^o=80^o\)

Ot là tia phân giác

\(\Rightarrow\widehat{xOt}=\widehat{tOz}=\dfrac{80^o}{2}=40^o\)

\(\left(x-1\right)^3=\dfrac{27}{8}\)

\(\Rightarrow x-1=\sqrt[3]{\dfrac{27}{8}}\)

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

\(\Leftrightarrow x=\dfrac{5}{2}\)

Gọi \(x;y;z\) lần lượt là số học sinh khối \(6,7,8\) \(\left(x;y;z>0\right)\)

\(\Rightarrow z-y=15\)

Ta có: 

\(\dfrac{x}{8}=\dfrac{y}{7}=\dfrac{z}{6}=\dfrac{z-y}{6-7}=\dfrac{15}{-1}\) ??

Đề sai ạ!!

a.\(A=\left(2-\dfrac{5}{2}+1\dfrac{1}{3}\right)-\left(5,3-\dfrac{1}{5}\right)-\left(\dfrac{4}{6}-\dfrac{5}{3}\right)\)

\(A=\left(2-\dfrac{5}{2}+\dfrac{4}{3}\right)-\left(\dfrac{53}{10}-\dfrac{1}{5}\right)-\left(\dfrac{4}{6}-\dfrac{5}{3}\right)\)

\(A=\dfrac{5}{6}-\dfrac{51}{10}+1\)

\(A=-\dfrac{49}{15}\)

b.\(B=\left(\dfrac{11}{3}+\dfrac{5}{6}-\dfrac{1}{2}\right)+\left(\dfrac{4}{8}-\dfrac{1}{2}\right).\left(\dfrac{1}{3}-\dfrac{1}{2}\right)\)

\(B=4+0.\left(\dfrac{1}{3}-\dfrac{1}{2}\right)\)

\(B=4+0\)

`B=4`