\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

0,1<-----------------------------------0,15

\(m_{Al}=0,1.27=2,7\left(g\right)\)

Vậy chọn B.

ĐK: \(\left\{{}\begin{matrix}y\ne1\\y\ne3\end{matrix}\right.\)

Biểu thức trở thành: \(\dfrac{\left(y+5\right)\left(y-3\right)}{\left(y-1\right)\left(y-3\right)}-\dfrac{\left(y+1\right)\left(y-1\right)}{\left(y-1\right)\left(y-3\right)}=-\dfrac{8}{\left(y-1\right)\left(y-3\right)}\)

\(\Leftrightarrow\dfrac{y^2-3y+5y-15}{\left(y-1\right)\left(y-3\right)}-\dfrac{y^2-1}{\left(y-1\right)\left(y-3\right)}+\dfrac{8}{\left(y-1\right)\left(y-3\right)}=0\)

\(\Leftrightarrow y^2+2y-15-y^2+1+8=0\\ \Leftrightarrow2y=6\\ \Leftrightarrow y=\dfrac{6}{2}=3\left(loại\right)\)

Vậy không có giá trị y để hai biểu thức trên bằng nhau.

a 

ĐK: \(x^2-2x+1>0\)

PT \(\Leftrightarrow\sqrt{\left(x-1\right)^2}+x-6x+9=0\)

\(\Leftrightarrow\left|x-1\right|-5x+9=0\\ \Leftrightarrow\left|x-1\right|=-9+5x\\ \Leftrightarrow\left[{}\begin{matrix}x-1=-9+5x\\1-x=-9+5x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=\dfrac{10}{6}\left(nhận\right)\end{matrix}\right.\)

b

ĐK: \(\left\{{}\begin{matrix}2x^2-3>0\\4x-3>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>\dfrac{\sqrt{6}}{2}\\x< -\dfrac{\sqrt{6}}{2}\end{matrix}\right.\\x>\dfrac{3}{4}\end{matrix}\right.\Leftrightarrow x>\dfrac{\sqrt{6}}{2}\)

PT \(\Leftrightarrow2x^2-3=4x-3\)

\(\Leftrightarrow2x^2-4x=0\\ \Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=2\left(nhận\right)\end{matrix}\right.\)

c

ĐK: \(\left\{{}\begin{matrix}1-x^2\ge0\\x-1\ge0\end{matrix}\right.\Leftrightarrow x=1\)

PT \(\Leftrightarrow1-x^2=x-1\)

\(\Leftrightarrow x^2+x-2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-2\left(loại\right)\end{matrix}\right.\)

Giả sử \(2\sqrt{2}+\sqrt{3}=x\left(x\in Q\right)\)

\(\Leftrightarrow\left(2\sqrt{2}+\sqrt{3}\right)^2=x^2\\ \Leftrightarrow11+4\sqrt{6}=x^2\\ \Leftrightarrow\sqrt{6}=\dfrac{x^2-11}{4}\)

Vì \(\sqrt{6}\) là số vô tỉ nên \(\dfrac{x^2-11}{4}\) là số vô tỉ \(\Rightarrow\) \(x^2\) là số vô tỉ, \(\Rightarrow x\) là số vô tỉ (vô lý)

Vậy \(2\sqrt{2}+\sqrt{3}\) là số vô tỉ

Giả sử \(\sqrt{3}-\sqrt{2}=x\left(x\in Q\right)\)  

\(\Leftrightarrow\left(\sqrt{3}-\sqrt{2}\right)^2=x^2\\ \Rightarrow5-2\sqrt{6}=x^2\\ \Rightarrow\sqrt{6}=\dfrac{5-x^2}{2}\)

Vì \(\sqrt{6}\) là số vô tỉ nên \(\dfrac{5-x^2}{2}\Rightarrow\) \(x^2\)là số vô tỉ, \(\Rightarrow x\) là số vô tỉ (vô lý)

Vậy \(\sqrt{3}-\sqrt{2}\) là số vô tỉ

a

ĐK: \(x\ge1\left(\sqrt{x-1}\ge0\right)\)

\(PT\Leftrightarrow\sqrt{x^2-x-2x+2}=\sqrt{x-1}\\ \Leftrightarrow\sqrt{x\left(x-1\right)-2\left(x-1\right)}=\sqrt{x-1}\\ \Leftrightarrow\sqrt{\left(x-2\right)\left(x-1\right)}=\sqrt{x-1}\\ \Leftrightarrow\left(\sqrt{x-1}\right)\left(\sqrt{x-2}-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=0\\\sqrt{x-2}=1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=3\left(nhận\right)\end{matrix}\right.\)

b

ĐK: \(\left\{{}\begin{matrix}x^2-4x+4>0\\4x^2-4x+9>0\end{matrix}\right.\)

PT \(\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(2x-3\right)^2}\)

\(\Leftrightarrow\left|x-2\right|=\left|2x-3\right|\\ \Leftrightarrow\left[{}\begin{matrix}x-2=2x-3\\x-2=3-2x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=\dfrac{5}{3}\left(nhận\right)\end{matrix}\right.\)

PTHH:

\(FeO+CO\underrightarrow{t^o}Fe+CO_2\\ Fe_2O_3+3CO\underrightarrow{t^o}2Fe+3CO_2\\ Fe_3O_4+4CO\underrightarrow{t^o}3Fe+4CO_2\\ CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)

Từ PTHH có: \(n_{CO_2}=n_{CaCO_3}=n_{kt}=\dfrac{8}{100}=0,08\left(mol\right)\)

\(\Rightarrow n_{CO.pứ}=n_{CO_2}=0,08\left(mol\right)\)

Theo định luật bảo toàn khối lượng:

\(x=m_{Fe}=m_{hh.A}+m_{CO}-m_{CO_2}=5,64+0,08.28-44.0,08=4,36\left(g\right)\)

\(\dfrac{2x}{3y}=-\dfrac{1}{3}\\ \Rightarrow3y=2x:-\dfrac{1}{3}=\dfrac{2x.3}{-1}=-6x\\ \Rightarrow y=-\dfrac{6x}{3}=-2x\)

Thế \(y=-2x\) vào \(2x+3y^2=\dfrac{161}{4}\) được:

\(2x+3.\left(-2x\right)^2=\dfrac{161}{4}\\ \Leftrightarrow2x+12x^2-\dfrac{161}{4}=0\\ \Leftrightarrow48x^2+8x-161=0\\ \Leftrightarrow\left(48x^2+92x\right)+\left(-84x-161\right)=0\\ \Leftrightarrow4x\left(12x+23\right)-7\left(12x+23\right)=0\\ \Leftrightarrow\left(4x-7\right)\left(12x+23\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{7}{4}\Rightarrow y=-\dfrac{2.7}{4}=-\dfrac{7}{2}\\x=-\dfrac{23}{12}\Rightarrow y=-2.-\dfrac{23}{12}=\dfrac{23}{6}\end{matrix}\right.\)

Vậy phương trình có nghiệm \(\left\{x;y\right\}=\left\{\dfrac{7}{4};-\dfrac{7}{2}\right\}\) hoặc \(\left\{x;y\right\}=\left\{-\dfrac{23}{12};\dfrac{23}{6}\right\}\)

6 How_do__you go to school every morning, tuan?

7 we__are__going to open a new library .

8 the often_have__ english lesson on monday and thursday.

9_Does__your brother like pop music?

10 Lan _is_watching tv at the moment.

11 he__is_doing his homework at present.

12 i__am__ having dinner at jane's house.

1

a

\(=x\left(x^2+xy\right)-z\left(x^2+xy\right)\\ =\left(x-z\right)\left(x^2+xy\right)\)

b

\(=12\left(xy-xz\right)+3x\left(xy-xz\right)\\ =\left(12+3x\right)\left(xy-xz\right)\\ =x\left(12+3x\right)\left(y-z\right)\)

c

\(=\dfrac{1}{2}\left(x^4+2x^2y^2+y^4\right)-\dfrac{1}{2}.4x^2y^2\\ =\dfrac{1}{2}\left(x^4+2x^2y^2+y^4-4x^2y^2\right)\\ =\dfrac{1}{2}\left(x^4-2x^2y^2+y^4\right)\\ =\dfrac{1}{2}\left(x^2-y^2\right)^2\\ =\dfrac{1}{2}\left(x^2-y^2\right)\left(x^2-y^2\right)\\ =\dfrac{1}{2}\left(x-y\right)\left(x+y\right)\left(x-y\right)\left(x+y\right)\)

d

\(=2\left(x-y\right)^2-2.\left(5xy\right)^2\\ =2\left[\left(x-y\right)^2-\left(5xy\right)^2\right]\\ =2\left(x-y-5xy\right)\left(x-y+5xy\right)\)

2

\(5x^2z-10xyz+5y^2z\\ =5x^2z-5xyz-5xyz+5y^2z\\ =\left(5x^2z-5xyz\right)-\left(5xyz-5y^2z\right)\\ =5xz\left(x-y\right)-5yz\left(x-y\right)\\ =\left(5xz-5yz\right)\left(x-y\right)\\ =5z\left(x-y\right)\left(x-y\right)\\=5z\left(x-y\right)^2\)

Thế \(x=124;y=24;z=2\) vào biểu thức được:

\(5.2\left(124-24\right)^2=10.\left(100\right)^2=10.10000=100000\)

\(2\sqrt{8\sqrt{3}}+2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\\ =2.2\sqrt{2\sqrt{3}}+2\sqrt{5\sqrt{3}}-3.2\sqrt{5\sqrt{3}}\\ =4\sqrt{2\sqrt{3}}+2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}\\ =4\sqrt{2\sqrt{3}}+\left(2-6\right)\sqrt{5\sqrt{3}}\\ =4\sqrt{2\sqrt{3}}-4\sqrt{5\sqrt{3}}\)