\(\dfrac{1}{2}\sqrt{48a}-2\sqrt{45}-\dfrac{a}{3}\sqrt{\dfrac{27}{a}}+5\sqrt{1\dfrac{4}{5}}\)

\(=2\sqrt{3a}-6\sqrt{5}-\sqrt{3a}+3\sqrt{5}\)

=\(=\sqrt{3a}-3\sqrt{5}\)

Gọi M(x\(_o\),y\(_o\)) là tọa độ giao điểm của (d2)và (d3)

\(\Leftrightarrow\left\{{}\begin{matrix}y_o=2x_o+4\\y_o=-3x_o-1\end{matrix}\right.\)

\(\Rightarrow2x_o+4=-3x_o-1\)

\(\Leftrightarrow x_o=-1\Rightarrow y_o=2\)

Vậy M(-1,2) là tọa độ giao điểm của (d2) và (d3)

Để 3 đường thẳng đồng quy tại 1 điểm khi M(-1,2)

\(\Rightarrow2=-1.\left(m+2\right)-3m\)

\(\Leftrightarrow2=-m-2-3m\)

\(\Leftrightarrow-4m=4\)

\(\Leftrightarrow m=-1\)

Vậy m=-1 thì 3 đường thẳng đồng quy

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

\(\Leftrightarrow2x^2-2x+1=4x^2-4x+1\)

\(\Leftrightarrow2x^2-2x=0\)

\(\Leftrightarrow2x\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)

Vậy S=\(\left\{0,1\right\}\)

a)\(đkx\ge1,x\ne-1\)

\(\sqrt{\dfrac{x-1}{x+1}}=2\)

\(\Leftrightarrow\dfrac{x-1}{x+1}=4\)

\(\Leftrightarrow x-1=4x-4\)

\(\Leftrightarrow x=1\)(nhận)

Vậy S=\(\left\{1\right\}\)

c)đk\(25x^2-10x+1=\) \(\left(5x-1\right)^2\ge0\Leftrightarrow x\ge\dfrac{1}{5}\)

\(\sqrt{25x^2-10x+1}+2x=1\)

\(\Leftrightarrow\sqrt{\left(5x-1\right)^2}+2x=1\)

\(\Leftrightarrow5x-1+2x=1\)

\(\Leftrightarrow x=\dfrac{2}{7}\)(nhận)

Vậy S=\(\left\{\dfrac{2}{7}\right\}\)

raining đổi thành rains

3)b)\(Ca\left(OH\right)_2+2SO_2\rightarrow Ca\left(HSO_3\right)_2\)

2)a)\(NaOH+CO_2\rightarrow NaHCO_3\)

1)\(BaO\)\(_3\)\(\rightarrow^{t^o}BaO+CO_2\)

 \(BaO+H_2O\rightarrow Ba\left(OH\right)_2\)

\(Ba\left(OH\right)_2+CO_2\rightarrow BaCO_3+H_2O\)

\(BaO+HCl\rightarrow BaCl_2+H_2O\)

\(Ba\left(OH\right)_2+2HNO_3\rightarrow Ba\left(NO_3\right)_2+2H_2O\)

\(BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\)

2)a)\(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)

3)b)\(Ca\left(OH\right)_2+SO_2\rightarrow CaSO_3+H_2O\)

4)a)\(PTHH:CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)

           TL :  1      :    2          :         1         :   1   (mol)

           BR   1   ->     2         ->       1      ->    1   (mol)                           

b)n\(_{CO_2}=\dfrac{V}{22,4}=\dfrac{22,4}{22,4}=1\left(mol\right)\)

n\(_{NaOH}=\dfrac{m}{M}=\dfrac{100}{40}=2,5\left(mol\right)\)

So sánh \(\dfrac{n_{NaOH}}{2}>\dfrac{n_{CO_2}}{1}\left(\dfrac{2,5}{2}>\dfrac{1}{1}\right)\)=> n\(_{NaOH}dư\),tính theo n\(_{CO_2}\)

m\(_{Na_2CO_3}=n.M=1.106=106\left(g\right)\)

Ta có t có n\(_{NaOHdư}=n_{NaOH_{đề}}-n_{NaOH_{Phảnứng}}=2,5-2=0,5\left(mol\right)\)

=>\(m_{NaOHdư}=n.M=0,5.40=20\left(g\right)\)

a) Khi \(x=9\)=>\(\sqrt{x}=3\)

=>A=\(\dfrac{2.3+1}{3}=\dfrac{7}{3}\)

Vậy A=\(\dfrac{7}{3}\)khi x=9

b)\(B=\dfrac{x-3\sqrt{x}+4}{x-2\sqrt{x}}-\dfrac{1}{\sqrt{x}-2}\)

\(=\dfrac{x-3\sqrt{x}+4-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x-4\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)

c)\(P=\dfrac{B}{A}=\dfrac{\sqrt{x}-2}{\sqrt{x}}.\dfrac{\sqrt{x}}{2\sqrt{x}+1}=\dfrac{\sqrt{x}-2}{2\sqrt{x}+1}\)

T có\(\left|P\right|>P\)=>\(\left[{}\begin{matrix}P>P\left(loại\right)\\-P>P\end{matrix}\right.\)

T có \(-P>P\)

\(\Rightarrow-\dfrac{\sqrt{x}-2}{2\sqrt{x}+1}>\dfrac{\sqrt{x}-2}{2\sqrt{x}+1}\)

\(\Leftrightarrow\dfrac{\sqrt{x}-2}{2\sqrt{x}+1}+\dfrac{\sqrt{x}-2}{2\sqrt{x}+1}< 0\)

\(\Leftrightarrow\dfrac{2\sqrt{x}-4}{2\sqrt{x}+1}< 0\)

Do \(x\ge0\Rightarrow2\sqrt{x}+1>0\) nên\(2\sqrt{x}-4< 0\)

\(\Rightarrow\sqrt{x}< 2\)

\(\Leftrightarrow x< 4\) \(\forall x,x\ge0\)

\(\Rightarrow0\le x< 4\)

Vậy\(\left|P\right|>P\) khi \(0\le x< 4\)

a) \(\dfrac{12}{1+\sqrt{5}}+\dfrac{15}{\sqrt{5}}-\dfrac{\sqrt{20}-5}{2-\sqrt{5}}\)

=\(\dfrac{12\left(1-\sqrt{5}\right)}{-4}+\dfrac{15\sqrt{5}}{5}-\dfrac{\left(\sqrt{20}-5\right)\left(2+\sqrt{5}\right)}{-1}\)

=\(-3+3\sqrt{5}-\sqrt{5}+3\sqrt{5}+4\sqrt{5}+10-10-5\sqrt{5}\)

=\(5\sqrt{5}-3\)

b)\(\dfrac{2\sqrt{x}}{\sqrt{x}-1}-\dfrac{3x}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}}\)

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

=\(\dfrac{-x+\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)