Ta có: \(n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: 

CaCO₃ + 2HCl ---> CaCl₂ + CO₂ + H₂O (1)

CaO + 2HCl ---> CaCl₂ + H₂O

a. Theo PT₍₁₎: \(n_{CaCO_3}=n_{CO_2}=0,15\left(mol\right)\)

=> \(m_{CaCO_3}=0,15.100=15\left(g\right)\)

=> \(\%_{m_{CaCO_3}}=\dfrac{15}{17,8}.100\%=84,27\%\%\)

\(\%_{m_{CaO}}=100\%-84,27\%=15,73\%\)

b. Ta có: \(m_{CaO}=17,8-15=2,8\left(g\right)\)

=> \(n_{CaO}=\dfrac{2,8}{56}=0,05\left(mol\right)\)

Theo PT₍₁₎: \(n_{HCl}=2.n_{CaCO_3}=2.0,15=0,3\left(mol\right)\)

Theo PT₍₂₎: \(n_{HCl}=2.n_{CaO}=2.0,05=0,1\left(mol\right)\)

=> \(n_{HCl_{PỨ}}=0,3+0,1=0,4\left(mol\right)\)

Đổi 200ml = 0,2 lít

=> \(C_{M_{HCl}}=\dfrac{0,4}{0,2}=2M\)

B

đúng ko v bn :)

\(B=\left(1-\frac{2}{5}\right)\left(1-\frac{2}{7}\right)\left(1-\frac{2}{9}\right)....\left(1-\frac{2}{99}\right)\)

\(B=\frac{3}{5}\cdot\frac{5}{7}\cdot\frac{7}{9}\cdot...\cdot\frac{97}{99}\)

\(B=\frac{3\cdot5\cdot7\cdot...\cdot97}{5\cdot7\cdot9\cdot...\cdot99}=\frac{3}{99}=\frac{1}{33}\)

Vậy B = \(\frac{1}{33}\)

\(\left[1-\frac{2}{5}\right]\left[1-\frac{2}{7}\right]\left[1-\frac{2}{9}\right]...\left[1-\frac{2}{99}\right]\)

\(=\frac{3}{5}\cdot\frac{5}{7}\cdot\frac{7}{9}\cdot...\cdot\frac{97}{99}\)

\(=\frac{3\cdot5\cdot7\cdot...\cdot97}{5\cdot7\cdot9\cdot...\cdot99}=\frac{3}{99}=\frac{1}{33}\)

a47b chia hết cho 5 => b=0 hoặc 5

b = 0 thì a470 chia hết cho 9 => a+4+7+0 chia hết cho 9

=> a+11 chia hết cho 9 => a=7 ( vì 0 < a < = 9 )

b = 5 thì a475 chia hết cho 9 => a+4+7+5 chia hết cho 9

=> a+16 chia hết cho 9 => a=2 ( vì 0 < a < = 9 )

Vậy ............

Tk mk nha

Trước 21:15 nhà!!!!!

Là cái muôi (dụng cụ nhà bếp) nhoa!

A

BÀi 4:

2: \(\frac{5}{4-\sqrt{14}}+\frac{1}{3-\sqrt7}-\frac{6}{3+\sqrt7}-\frac{\sqrt7-5}{2}\)
\(=\frac{5\left(4+\sqrt{14}\right)}{16-14}+\frac{3+\sqrt7}{\left(3-\sqrt7\right)\left(3+\sqrt7\right)}-\frac{6\left(3-\sqrt7\right)}{\left(3+\sqrt7\right)\left(3-\sqrt7\right)}-\frac{\sqrt7-5}{2}\)

\(=\frac{20+5\sqrt{14}}{2}+\frac{3+\sqrt7}{2}-\frac{18-6\sqrt7}{2}-\frac{\sqrt7-5}{2}=\frac{20+5\sqrt{14}+3+\sqrt7-18+6\sqrt7-\sqrt7+5}{2}=\frac{10+5\sqrt{14}+6\sqrt7}{2}\)

3: \(\frac{1}{1+\sqrt2}+\frac{1}{\sqrt2+\sqrt3}+\cdots+\frac{1}{\sqrt{2020}+\sqrt{2021}}\)

\(=\frac{-1+\sqrt2}{\left(1+\sqrt2\right)\left(-1+\sqrt2\right)}+\frac{-\sqrt2+\sqrt3}{\left(\sqrt3-\sqrt2\right)\left(\sqrt3+\sqrt2\right)}+...+\frac{-\sqrt{2020}+\sqrt{2021}}{\left(\sqrt{2020}+\sqrt{2021}\right)\left(\sqrt{2021}-\sqrt{2020}\right)}\)

\(=-1+\sqrt2-\sqrt2+\sqrt3-\cdots-\sqrt{2020}+\sqrt{2021}=\sqrt{2021}-1\)

4: \(\frac{1}{1-\sqrt2}-\frac{1}{\sqrt2-\sqrt3}+\frac{1}{\sqrt3-\sqrt4}-\cdots+\frac{1}{\sqrt{143}-\sqrt{144}}\)

\(=\frac{1\left(1+\sqrt2\right)}{\left(1-\sqrt2\right)\left(1+\sqrt2\right)}-\frac{\left(\sqrt2+\sqrt3\right)}{\left(\sqrt2-\sqrt3\right)\left(\sqrt2+\sqrt3\right)}+\frac{\left(\sqrt3+\sqrt4\right)}{\left(\sqrt3-\sqrt4\right)\left(\sqrt3+\sqrt4\right)}+\cdots+\frac{\sqrt{143}+\sqrt{144}}{\left(\sqrt{143}+\sqrt{144}\right)\left(\sqrt{143}-\sqrt{144}\right)}\)

\(=-1-\sqrt2+\sqrt2+\sqrt3-\sqrt3-\sqrt4+\cdots-\sqrt{143}-\sqrt{144}\)

=-1-12

=-13

5: \(\sqrt2\cdot\left(\sqrt3+1\right)\cdot\sqrt{2-\sqrt3}\)

\(=\left(\sqrt3+1\right)\cdot\sqrt{4-2\sqrt3}\)

\(=\left(\sqrt3+1\right)\cdot\sqrt{\left(\sqrt3-1\right)^2}=\left(\sqrt3+1\right)\left(\sqrt3-1\right)=3-1=2\)

6: \(\left(4+\sqrt{15}\right)\cdot\left(\sqrt{10}-\sqrt6\right)\cdot\sqrt{4-\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt5-\sqrt3\right)\cdot\sqrt{8-2\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt5-\sqrt3\right)\left(\sqrt5-\sqrt3\right)=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)=2\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)=2\cdot\left(16-15\right)=2\)

7: \(\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}\)

\(=\frac{1}{\sqrt2}\left(\sqrt{4+2\sqrt3}-\sqrt{4-2\sqrt3}\right)\)

\(=\frac{1}{\sqrt2}\left(\sqrt3+1-\sqrt3+1\right)=\frac{2}{\sqrt2}=\sqrt2\)

Bài 5:

1: \(x\cdot\sqrt{27}+\sqrt8=x\cdot\sqrt{18}+\sqrt{12}\)

=>\(x\cdot3\sqrt3-x\cdot3\sqrt2=2\sqrt3-2\sqrt2\)

=>\(3x\left(\sqrt3-\sqrt2\right)=2\left(\sqrt3-\sqrt2\right)\)

=>3x=2

=>\(x=\frac23\)

2: \(2x\cdot\sqrt3+\sqrt{48}=x\cdot\sqrt{75}+\sqrt{243}\)

=>\(x\cdot2\sqrt3+4\sqrt3=x\cdot5\sqrt3+9\sqrt3\)

=>\(x\cdot\left(-3\sqrt3\right)=9\sqrt3-4\sqrt3=5\sqrt3\)

=>\(x=-\frac53\)

4: ĐKXĐ: x>=2

\(\sqrt{49x-98}-\sqrt{9x-18}-\sqrt{16x-32}=\sqrt{4x-4}\)

=>\(7\cdot\sqrt{x-2}-3\cdot\sqrt{x-2}-4\cdot\sqrt{x-2}=2\sqrt{x-1}\)

=>\(2\sqrt{x-1}=0\)

=>x-1=0

=>x=1(loại)

5: ĐKXĐ: \(x^2-2\ge0\)

=>\(x^2\ge2\)

=>\(\left[\begin{array}{l}x\ge\sqrt2\\ x\le-\sqrt2\end{array}\right.\)

\(1-\sqrt{x^2-2}=0\)

=>\(\sqrt{x^2-2}=1\)

=>\(x^2-2=1\)

=>\(x^2=3\)

=>\(\left[\begin{array}{l}x=\sqrt3\left(nhận\right)\\ x=-\sqrt3\left(nhận\right)\end{array}\right.\)