\(\dfrac{36}{64}=\dfrac{9}{16}\) ; \(-\dfrac{12}{81}=-\dfrac{4}{27}\)

Quy đồng:

\(\dfrac{9}{16}=\dfrac{243}{432}\) ; \(-\dfrac{4}{27}=-\dfrac{64}{432}\)

Chọn phương án B.

\(\dfrac{5}{12}-\dfrac{3}{-16}+\dfrac{3}{4}\)

\(=\dfrac{5}{12}+\dfrac{3}{16}+\dfrac{3}{4}\)

\(=\dfrac{20}{48}+\dfrac{9}{48}+\dfrac{36}{48}\)

\(=\dfrac{65}{48}\)

15% của 72 kg là:

\(72\times\dfrac{15}{100}=10,8\) (kg)

Bài 1:

e) \(-5\dfrac{1}{7}+3\dfrac{2}{5}=-\dfrac{36}{7}+\dfrac{17}{5}=-\dfrac{180}{35}+\dfrac{119}{35}=-\dfrac{61}{35}\)

f) \(-2\dfrac{1}{3}-1\dfrac{2}{7}=-\dfrac{7}{3}-\dfrac{9}{7}=-\dfrac{49}{21}-\dfrac{27}{21}=-\dfrac{76}{21}\)

Bài 2:

a) \(\dfrac{3}{4}x=1\Rightarrow x=1:\dfrac{3}{4}\Rightarrow x=\dfrac{4}{3}\)

b) \(\dfrac{4}{7}x=\dfrac{9}{8}-0,125\)

\(\dfrac{4}{7}x=\dfrac{9}{8}-\dfrac{1}{8}\)

\(\dfrac{4}{7}x=\dfrac{8}{8}=1\)

\(x=1:\dfrac{4}{7}\)

\(x=\dfrac{7}{4}\)

a) \(NH_4F+NaNO_2\xrightarrow[]{t^\circ}NaF+N_2\uparrow+2H_2O\)

Ion RG: \(NH_4^++NO_2^-\xrightarrow[]{t^\circ}N_2\uparrow+2H_2O\)

b) \(9FeCl_2+3KMnO_4+6H_2O\rightarrow5FeCl_3+4Fe\left(OH\right)_3\downarrow+3KCl+3MnO_2\downarrow\)

Ion RG: \(9Fe^{2+}+3MnO_4^-+6H_2O\rightarrow5Fe^{3+}+4Fe\left(OH\right)_3\downarrow+3MnO_2\downarrow\)

c) \(MnO_2+4HCl\text{ (đặc)}\rightarrow MnCl_2+Cl_2\uparrow+2H_2O\)

Ion RG: \(MnO_2+4H^++2Cl^-\rightarrow Mn^{2+}+Cl_2\uparrow+2H_2O\)

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

Ion RG: \(2Cl^-+2H_2O\xrightarrow[cmn]{đpdd}2OH^-+H_2\uparrow+Cl_2\uparrow\)

e) \(CH_3COONa+NaOH\xrightarrow[CaO]{t^\circ}Na_2CO_3+CH_4\uparrow\)

Ion RG: \(CH_3COO^-+OH^-\rightarrow CO_3^{2-}+CH_4\uparrow\)