56083   | 123

  688     | 455

    733   | 

    118

=> \(56083:123=455\left(dư.118\right)\)

các số  lẻ có 3 chữ số là 101 - 999

số lẻ chia 5 dư 2 là 107, 117, 127,......997 ( có chữ số tận cùng là 7)

các số tự nhiên lẻ có 3 chữ số mà mỗi số chia cho 5 dư 2 là

(997 - 107) :10 + 1 = 90 số

\(4,=\dfrac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{5-2\sqrt{6}-9}=\dfrac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{-4-2\sqrt{6}}\\ =\dfrac{3\left(3-\sqrt{2}-\sqrt{3}\right)}{2+\sqrt{6}}=\dfrac{\left(9-3\sqrt{2}-3\sqrt{3}\right)\left(\sqrt{6}-2\right)}{2}\\ =\dfrac{9\sqrt{6}-18-6\sqrt{3}+6\sqrt{2}-9\sqrt{2}+6\sqrt{3}}{2}\\ =\dfrac{9\sqrt{6}-3\sqrt{2}-18}{2}\)

\(7,=\dfrac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-2-\sqrt{3}\\ =\sqrt{3}+2+\sqrt{2}+1-2-\sqrt{3}=1+\sqrt{2}\)

\(10,\dfrac{1}{\sqrt{a}+\sqrt{a+2}}=\dfrac{\sqrt{a}-\sqrt{a+2}}{a-a-2}=\dfrac{\sqrt{a-2}-\sqrt{a}}{2}\)

Do đó \(\dfrac{1}{\sqrt{1}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{5}}+...+\dfrac{1}{\sqrt{47}+\sqrt{49}}\)

\(=\dfrac{\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+...+\sqrt{49}-\sqrt{47}}{2}=\dfrac{-1+\sqrt{49}}{2}=\dfrac{7-1}{2}=3\)

10, \(\dfrac{1}{\sqrt{1}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{5}}+...+\dfrac{1}{\sqrt{17}+\sqrt{19}}=\dfrac{\sqrt{1}-\sqrt{3}}{\left(\sqrt{1}+\sqrt{3}\right)\left(\sqrt{1}-\sqrt{3}\right)}+\dfrac{\sqrt{3}-\sqrt{5}}{\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{3}-\sqrt{5}\right)}+...+\dfrac{\sqrt{17}-\sqrt{19}}{\left(\sqrt{17}+\sqrt{19}\right)\left(\sqrt{17}-\sqrt{19}\right)}=\dfrac{1-\sqrt{3}+\sqrt{3}-\sqrt{5}+...+\sqrt{17}-\sqrt{19}}{-2}=-\dfrac{1-\sqrt{19}}{2}\)

didn't use to drive

used to be

aren't used to drinking

used to read

Bài 3 : 

\(n_{Na2O}=\dfrac{9,3}{62}=0,15\left(mol\right)\)

 \(Na_2O+H_2O\rightarrow2NaOH|\)

     1           1                2       

   0,15                        0,3

a) \(n_{NaOH}=\dfrac{0,15.2}{1}=0,3\left(mol\right)\)

⇒ \(m_{NaOH}=0,3.40=12\left(g\right)\)

\(C_{ddNaOH}=\dfrac{12.100}{300}=4\)⁰/₀

b) \(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O|\)

       1             2                   1            1

      0,15       0,3

\(n_{CO2}=\dfrac{0,3.1}{2}=0,15\left(mol\right)\)

\(V_{CO2\left(dktc\right)}=0,15.22,4=3,36\left(l\right)\)

 Chúc bạn học tốt             

Bài 1:

\(a.\left(1\right)C+O_2\underrightarrow{to}CO_2\\ \left(2\right)CO_2+Na_2O\rightarrow Na_2CO_3\\ \left(3\right)Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+CO_2+H_2O\\ Na_2SO_4+Ba\left(OH\right)_2\rightarrow NaSO_4+2NaOH\\ b.\left(1\right)4P+5O_2\underrightarrow{to}2P_2O_5\\ \left(2\right)P_2O_5+3H_2O\rightarrow2H_3PO_4\\ \left(3\right)H_3PO_4+3NaOH\rightarrow Na_3PO_4+3H_2O\\ \left(4\right)Na_3PO_4+3AgNO_3\rightarrow Ag_3PO_4+3NaNO_3\)

\(c.\left(1\right)CaCO_3\underrightarrow{to}CaO+CO_2\\ \left(2\right)CaO+H_2O\rightarrow Ca\left(OH\right)_2\\ \left(3\right)Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\\ \left(4\right)CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\\ \left(5\right)CaCl_2+2AgNO_3\rightarrow Ca\left(NO_3\right)_2+2AgCl\)

Bài 4 : 

\(n_{CuO}=\dfrac{3,2}{80}=0,04\left(mol\right)\)

Pt : \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O|\)

          1            1                1            1

        0,04       0,04           0,04

a) \(n_{H2SO4}=\dfrac{0,04.1}{1}=0,04\left(mol\right)\)

⇒ \(m_{H2SO4}=0,04.98=3,92\left(g\right)\)

\(m_{ddH2SO4}=\dfrac{3,92.100}{4,9}=80\left(g\right)\)

b) \(n_{CuSO4}=\dfrac{0,04.1}{1}=0,04\left(mol\right)\)

⇒ \(m_{CuSO4}=0,04.160=6,4\left(g\right)\)

\(m_{ddspu}=3,2+80=83,2\left(g\right)\)

\(C_{CuSO4}=\dfrac{6,4.100}{83,2}=7,7\)⁰/₀

 Chúc bạn học tốt

Linh said that she was working in that restaurant.

Your younger brother goes to school by bike, doesn't he?

Some fresh flowers have been cut from the garden by my mother.

It's very wonderful to spend the weekend in the countryside.

I haven't seen my grandfather for five months.