\(\dfrac{6}{10}=\dfrac{6:2}{10:2}=\dfrac{3}{5}\)

\(\dfrac{70}{90}=\dfrac{70:10}{90:10}=\dfrac{7}{9}\)

\(\dfrac{96}{72}=\dfrac{96:24}{72:24}=\dfrac{4}{3}\)

\(\dfrac{45}{35}=\dfrac{45:5}{35:5}=\dfrac{9}{7}\)

He feels hungry so he wants a loaf of bread and a bacon

chen chúc

chao đảo

trao đổi

Sửa đoạn: \(\dfrac{8}{5}-\dfrac{11}{18}\) là: \(\dfrac{8}{15}-\dfrac{11}{18}\) nhé

\(\dfrac{8}{15}-x:\dfrac{3}{4}=\dfrac{11}{18}\)

\(x:\dfrac{3}{4}=\dfrac{8}{5}-\dfrac{11}{18}\)

\(x:\dfrac{3}{4}=-\dfrac{7}{90}\)

\(x=-\dfrac{7}{90}\cdot\dfrac{3}{4}\)

\(x=-\dfrac{7}{120}\)

(1) \(Cl_2+H_2⇌HCl+HClO\)

(2) \(MnO_2+4HCl\left(\text{đặc}\right)\xrightarrow[]{}MnCl_2+Cl_2\uparrow+H_2O\)

(3) \(Cl_2+2NaOH\xrightarrow[]{}NaCl+NaClO+H_2O\)

(4) \(2NaCl+2H_2O\xrightarrow[cmn]{đpdd}2NaOH+Cl_2\uparrow+H_2\)

1. \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

\(2Mg+O_2\xrightarrow[]{t^\circ}2MgO\)

 0,2 → 0,1

\(\Rightarrow V_{O_2}=0,1\cdot22,4=2,24\left(l\right)\)

2. \(n_{P_2O_5}=\dfrac{28,2}{142}=\dfrac{141}{710}\left(mol\right)\)

\(4P+5O_2\xrightarrow[]{t^\circ}2P_2O_5\)

\(\dfrac{141}{355}\)       ←       \(\dfrac{141}{710}\)

\(\Rightarrow m_P=\dfrac{141}{355}\cdot31\approx12,312\left(g\right)\)

Check lại đi bạn.

\(\sqrt{2}x^2+6x=0\)

\(\Leftrightarrow x\left(\sqrt{2}x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\sqrt{2}x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\sqrt{2}\end{matrix}\right.\)

Vậy \(S=\left\{0;-3\sqrt{2}\right\}\)