a) = 5/14

b) = 25/36

c) = 37/45

a) \(\text{2}\sqrt{\text{18}}-9\sqrt{50}+3\sqrt{8}\)

= \(\text{6}\sqrt{\text{2}}-45\sqrt{2}+6\sqrt{2}\)

= \(-33\sqrt{2}\)

b) = \(7-2.\sqrt{7}.\sqrt{3}+3+7.2\sqrt{21}\)

= \(10-2\sqrt{21}+14\sqrt{21}\)

= \(10+12\sqrt{21}\)

a) 28 + 55 = 83

45 + 49 = 94

37 + 53 = 90

b) 61 - 18 = 43

53 - 26 = 27

92 - 84 = 8

\(\begin{array}{*{20}{c}}{a)\,\,}\\{}\\{}\end{array}\begin{array}{*{20}{c}}{ + \begin{array}{*{20}{c}}{28}\\{55}\end{array}}\\\hline{\,\,\,\,83}\end{array}\)                    \(\begin{array}{*{20}{c}}{ + \begin{array}{*{20}{c}}{45}\\{49}\end{array}}\\\hline{\,\,\,\,94}\end{array}\)                            \(\begin{array}{*{20}{c}}{ + \begin{array}{*{20}{c}}{37}\\{53}\end{array}}\\\hline{\,\,\,\,90}\end{array}\)

\(\begin{array}{*{20}{c}}{b)\,\,}\\{}\\{}\end{array}\begin{array}{*{20}{c}}{ - \begin{array}{*{20}{c}}{61}\\{18}\end{array}}\\\hline{\,\,\,43}\end{array}\)                    \(\begin{array}{*{20}{c}}{ - \begin{array}{*{20}{c}}{53}\\{26}\end{array}}\\\hline{\,\,\,27}\end{array}\)                           \(\begin{array}{*{20}{c}}{ - \begin{array}{*{20}{c}}{92}\\{84}\end{array}}\\\hline{\,\,\,08}\end{array}\)

a) 0,25 mol =\(\dfrac{A}{6,022\cdot10^{23}}^{ }\)

=>A= 0,25 * \(^{6,022\cdot10^{23}}\)

=>A=1,5055*10^23 .vậy số phân tử có trong 0,25 mol HCl là 1,5055*10^23(đoạn này ko có cx đc)

b)n_CO2=3,36 : 22,4=0,15(mol) ->0,15mol =\(\dfrac{A}{6,022\cdot10^{23}}^{ }\)

=>A=0,15*\(^{6,022\cdot10^{23}}\)

=>A=9,033*10²²

c)m=n*M ->n=m/M -> 4,6=n*M ->n=4,6/M=4,6/23=0,2(mol)

->0,2 mol =\(\dfrac{A}{6,022\cdot10^{23}}^{ }\)

->A=0,2*\(^{6,022\cdot10^{23}}\)

=1,2044*10²³

11:

a: Số cần tìm là 3;2;1;0

b: 0

10:

\(2\dfrac{1}{2}=2.5>2.25\)

\(6>5\)

nên căn 6>căn 5

-căn 5<0

0<căn 5

=>-căn 5<căn 5

a) \(\dfrac{9}{11}-\dfrac{3}{11}=\dfrac{9-3}{11}=\dfrac{6}{11}\)

b) \(\dfrac{10}{4}-\dfrac{5}{4}=\dfrac{10-5}{4}=\dfrac{5}{4}\)

c) \(\dfrac{22}{15}-\dfrac{8}{15}=\dfrac{22-8}{15}=\dfrac{14}{15}\)

a) \(\dfrac{3}{5}:\dfrac{3}{4}=\dfrac{3}{5}\cdot\dfrac{4}{3}=\dfrac{4}{5}\)

b) \(\dfrac{2}{5}:\dfrac{3}{10}=\dfrac{2}{5}\cdot\dfrac{10}{3}=\dfrac{4}{3}\)

c) \(\dfrac{1}{8}:\dfrac{1}{6}=\dfrac{1}{8}\cdot6=\dfrac{6}{8}=\dfrac{3}{4}\)

a: \(=\dfrac{3}{5}\cdot\dfrac{4}{3}=\dfrac{12}{15}=\dfrac{4}{5}\)

b: \(=\dfrac{2}{5}\cdot\dfrac{10}{3}=\dfrac{20}{15}=\dfrac{4}{3}\)

c: \(=\dfrac{1}{8}\cdot6=\dfrac{6}{8}=\dfrac{3}{4}\)

a, \(\dfrac{3}{5}:\dfrac{3}{4}=\dfrac{3}{5}x\dfrac{4}{3}=\dfrac{12}{15}\)

b, \(\dfrac{2}{5}:\dfrac{3}{10}=\dfrac{2}{5}x\dfrac{10}{3}=\dfrac{4}{3}\)

c, \(\dfrac{1}{8}:\dfrac{1}{6}=\dfrac{1}{8}x\dfrac{6}{1}=\dfrac{3}{4}\)

a) $\frac{1}{6} + \frac{2}{3} = \frac{1}{6} + \frac{4}{6} = \frac{{1 + 4}}{6} = \frac{5}{6}$

b) $\frac{3}{5} + \frac{9}{{20}} = \frac{{12}}{{20}} + \frac{9}{{20}} = \frac{{12 + 9}}{{20}} = \frac{{21}}{{20}}$

c) $\frac{3}{8} + \frac{3}{{24}} = \frac{9}{{24}} + \frac{3}{{24}} = \frac{{9 + 3}}{{24}} = \frac{{12}}{{24}} = \frac{1}{2}$