ĐKXĐ: \(a\ge0,a\ne9\)

a)\(P=\frac{2\sqrt{a}}{\sqrt{a}+3}+\frac{\sqrt{a}+1}{\sqrt{a}-3}+\frac{3+7\sqrt{a}}{9-a}.\)

\(=\frac{2\sqrt{a}\left(\sqrt{a}-3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}+\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}-\frac{3+7\sqrt{a}}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)

\(=\frac{2\sqrt{a}\left(\sqrt{a}-3\right)+\left(\sqrt{a}+1\right)\left(\sqrt{a}+3\right)-\left(3+7\sqrt{a}\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)

\(=\frac{2a-6\sqrt{a}+a+3\sqrt{a}+\sqrt{a}+3-3-7\sqrt{a}}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)

\(=\frac{3a-9\sqrt{a}}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}=\frac{3\sqrt{a}\left(\sqrt{a}-3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)

\(=\frac{3\sqrt{a}}{\sqrt{a}+3}\)

b) Để P<1 hay \(\frac{3\sqrt{a}}{\sqrt{a}+3}< 1\Leftrightarrow3\sqrt{a}< \sqrt{a}+3\Leftrightarrow\sqrt{a}< \frac{3}{2}\Leftrightarrow0\le a< \frac{9}{4}\)(thỏa mãn ĐKXĐ)

Vậy với \(0\le a< \frac{9}{4}\)thì P<1.

(p/s đừng ti ck cho câu trả lời này)

Ai giả đc giúp mk với

\(H=\sqrt{2+\sqrt{3}}.\sqrt{14-5\sqrt{3}}+\sqrt{2}.\)

\(2.H=\sqrt{2\left(2+\sqrt{3}\right)}.\sqrt{2\left(14-5\sqrt{3}\right)}+2\sqrt{2}\)

\(2H=\sqrt{3+2\sqrt{3}+1}.\sqrt{25-2.5.\sqrt{3}+3}+2\sqrt{2}\)

\(2H=\sqrt{\left(\sqrt{3}+1\right)^2}.\sqrt{\left(5-\sqrt{3}\right)^2}+2\sqrt{2}\)

\(2H=\left(\sqrt{3}+1\right)\left(5-\sqrt{3}\right)+2\sqrt{2}\)

\(2H=5\sqrt{3}-3+5-\sqrt{3}+2\sqrt{2}\)

\(2H=2+4\sqrt{3}+2\sqrt{2}\)

\(H=1+2\sqrt{3}+\sqrt{2}.\)

(P/S :  đừng k cho câu trả lời này nhé)

\(dkxd\Leftrightarrow\hept{\begin{cases}x\ge0\\\sqrt{x}-2\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}}\)

\(A=\left(\frac{\sqrt{x}}{x-4}-\frac{2}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\right):\frac{1}{\sqrt{x}+2}.\)

\(=\left(\frac{\sqrt{x}}{x-4}-\frac{2\left(\sqrt{x}+2\right)}{x-4}+\frac{\sqrt{x}-2}{x-4}\right):\frac{1}{\sqrt{x}+2}\)

\(=\frac{\sqrt{x}-2\sqrt{x}-4+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}+2}{1}\)

\(=\frac{-6\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=-\frac{6}{\sqrt{x}-2}\)

\(A=\)\(\left(\frac{\sqrt{x}}{x-4}+\frac{2}{2-\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right)\)\(:\frac{1}{\sqrt{x}+2}\)

a,ĐKXĐ:\(\hept{\begin{cases}x\ge0\\2-\sqrt{x}\\x-4\ne0\end{cases}\ne0}\)\(\Rightarrow\)\(\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)

\(A=\)\(\left(\frac{\sqrt{x}}{x-4}+\frac{2}{2-\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right)\)\(:\frac{1}{\sqrt{x}+2}\)

\(A=\)\(\left(\frac{\sqrt{x}}{x-4}-\frac{2}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\right)\)\(.\left(\sqrt{x}+2\right)\)

\(A=\)\(\left(\frac{\sqrt{x}-2\left(\sqrt{x}+2\right)+\sqrt{x}-2}{x-4}\right)\)\(.\left(\sqrt{x}+2\right)\)

\(A=\)\(\left(\frac{\sqrt{x}-2\sqrt{x}-4+\sqrt{x}-2}{x-4}\right)\)\(.\left(\sqrt{x}+2\right)\)

\(A=\)\(\left(\frac{-6}{x-4}\right)\)\(.\left(\sqrt{x}+2\right)\)

\(A=\)\(\frac{-6}{\sqrt{x}-2}\)

b,\(x=9-4\sqrt{5}\)\(\Rightarrow\)\(A=\)\(\frac{-6}{\sqrt{9-4\sqrt{5}}-2}\)\(=\frac{-6}{\sqrt{5-2.2\sqrt{5}+4}-2}\)

\(A=\)\(\frac{-6}{\sqrt{\left(\sqrt{5}-2\right)^2}-2}\)\(=\frac{-6}{\sqrt{5}-2-2}\)\(=\frac{-6}{\sqrt{5}-4}\)

c,\(A>-1\)\(\Rightarrow\)\(\frac{-6}{\sqrt{x}-2}\)\(>-1\)\(\Rightarrow\)\(\frac{-6}{\sqrt{x}-2}+1>0\)

\(\Leftrightarrow\)\(\frac{-6+\sqrt{x}-2}{\sqrt{x}-2}>0\)

\(\Leftrightarrow\)\(\frac{\sqrt{x}-8}{\sqrt{x}-2}>0\)

HCl     +    NaOH ------> NaCl         +      H2O

200g         100g

3,65%         2%

                0,05 mol       0,05mol

m HCl = 7,3g  => n HCl = 0,2 (mol) 

m NaOH = 2g => n NaOH = 0,05 (mol)

=> HCl dư 

=> n NaCl = n NaOH = 0,05 (mol)

=> m NaCl = 0,05 * (23+35,5) = 2,925 g

m dung dịch = (m dd HCl) + (m dd NaOH) = 200 + 100 =300

C% = (m chất tan / m dung dịch) * 100 = (2,925 / 300)*100= 0,975%

m_HCl = \(\frac{md.C\%}{100\%}=\frac{200.3,65\%}{100\%}=7,3\)(g)

n_HCl = \(\frac{m}{M}=\frac{7,3}{36,5}=0,2\)(mol)

m_NaOH = \(\frac{md.C\%}{100\%}=\frac{100.2\%}{100\%}=2\)(g)

n_NaOH = \(\frac{m}{M}=\frac{2}{40}=0,05\)(mol)

Khi cho HCl tác dụng với NaOH, ta có PTHH:

                     HCl + NaOH \(\rightarrow\)NaCl + H₂O 

                       0,2    :   0,05       

Xét tỉ lệ \(\frac{0,2}{1}>\frac{0,05}{1}\)=> HCl dư, dựa vào n_NaOH để tính

     HCl + NaOH \(\rightarrow\)   NaCl + H₂O

     0,05   \(\leftarrow\)0,05  \(\rightarrow\)0,05 : 0,05     (mol)

m_NaCl = n.M = 0,05 . 58,5 = 2,925 (g)

m_dd = m_dHCl + m_dNaOH = 200 + 100 = 300 (g)

C%_HCl = \(\frac{mt}{md}.100\%\)= \(\frac{2,925}{300}.100\%\)= 0,975%

Xem lời giải tại :

Câu hỏi của Lê Quỳnh Như - Hóa học lớp 9 | Học trực tuyến

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Xem lời gải rõ ở đây :

Cho glucozơ lên men thành ancol etylic. Dẫn khí CO2 tạo thành qua dung dịch nước vôi trong dư, thu được 50 gam kết tủa. ?

https://moon.vn/hoi-dap/cho-glucozo-len-men-thanh-ancol-etylic-dan-khi-co2-tao-thanh-qua-dung-dich-nuoc-voi--419265

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