\(Al+4HNO_3\rightarrow Al\left(NO_3\right)_3+NO\uparrow+2H_2O\)

\(Al+6HNO_3\rightarrow Al\left(NO_3\right)_3+3NO_2+3H_2O\)

Gọi \(\left\{{}\begin{matrix}n_{NO}=x\\n_{NO_2}=y\end{matrix}\right.\) ( mol )

\(n_{hhk}=x+y=\dfrac{8,96}{22,4}=0,4\left(mol\right)\left(1\right)\)

\(\dfrac{M_{hhk}}{M_{H_2}}=16,75\Rightarrow M_{hhk}=16,75.2=33,5\)

\(\Leftrightarrow\dfrac{30x+46y}{x+y}=33,5\)

\(\Leftrightarrow30x+46y=13,4\left(2\right)\)

\(\left(1\right);\left(2\right)\Rightarrow\left\{{}\begin{matrix}x=0,3125\\y=0,0875\end{matrix}\right.\)

Theo pt: \(n_{Al}=x+\dfrac{y}{3}=0,3125+\dfrac{0,0875}{3}=\dfrac{41}{120}\)

\(\Rightarrow m_{Al}=27.\dfrac{41}{120}=9,225\left(g\right)\)

Sửa đề: 150g `->` 150 ml

\(m_{H_2SO_4}=150.1,2=180\left(g\right)\)

\(n_{BaCl_2}=\dfrac{200.15,6\%}{208}=0,15\left(mol\right)\) ; \(n_{H_2SO_4}=\dfrac{180}{98}=1,83\left(mol\right)\)

\(BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+H_2O\)

0,15          1,83                                  ( mol )

0,15           0,15           0,15                  ( mol )

\(m_{BaSO_4}=0,15.233=34,95\left(g\right)\)

Sau p.ứ H₂SO₄ còn dư

\(m_{H_2SO_4\left(dư\right)}=\left(1,83-0,15\right).98=164,64\left(g\right)\)

\(A=1+5+5^2+5^3+...+5^{50}\)

\(\Rightarrow5A=5+5^2+5^3+5^4+...+5^{51}\)

\(5A-A=4A=\left(5+5^2+5^3+5^4+...+5^{51}\right)-\left(1+5+5^2+5^3+...+5^{50}\right)\)\(\Rightarrow4A=5^{51}-1\)

\(\Rightarrow A=\dfrac{5^{51}-1}{4}\)

\(\left(1-x\right)^2+\left(x-y\right)^2+\left(y-z\right)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}x=1\\x=y\\y=z\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=1\\z=1\end{matrix}\right.\)

`=>x=y=z=1`

Ta có: \(n^2+n+1=n\left(n+1\right)+1\)

Ta có: \(n\left(n+1\right)\) là tích của 2 số tự nhiên liên tiếp nên  có chữ số tận cùng là 0,2,6

`=>` \(n\left(n+1\right)+1\) có chữ số tận cùng là 1,3,7

Mà 1,3,7 không chia hết cho 5

`=>`\(n^2+n+1\) không chia hết cho 5

`b.`

`@` Cho \(x=0\Rightarrow y=m-2\)

\(\Rightarrow A\left(0;m-2\right)\in\left(d\right)\)

`@` Cho \(y=0\Rightarrow x=\dfrac{2-m}{3m-2}\)

\(\Rightarrow B\left(\dfrac{2-m}{3m-2};0\right)\in\left(d\right)\)

Vậy \(\left(d\right)\) cắt trục Ox tại \(B\left(\dfrac{2-m}{3m-2};0\right)\)

        \(\left(d\right)\) cắt trục Oy tại \(A\left(0;m-2\right)\)

\(\Rightarrow OB=\left|\dfrac{2-m}{3m-2}\right|;OA=\left|m-2\right|\)

Ta có:

\(S_{OAB}=\dfrac{1}{2}.OA.OB\)

\(\Leftrightarrow\dfrac{1}{2}=\dfrac{1}{2}.\left|m-2\right|.\left|\dfrac{2-m}{3m-2}\right|\) ; \(\left(m\ne\dfrac{2}{3}\right)\)

\(\Leftrightarrow\left|\left(m-2\right)\left(\dfrac{2-m}{3m-2}\right)\right|=1\)

\(\Leftrightarrow\left|-\dfrac{\left(m-2\right)^2}{3m-2}\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}-\dfrac{\left(m-2\right)^2}{3m-2}=1\\-\dfrac{\left(m-2\right)^2}{3m-2}=-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}m^2-m+2=0\left(vô.lý\right)\\m^2-7m+6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}m=6\\m=1\end{matrix}\right.\) ( tm )

Vậy \(m=\left\{6;1\right\}\) thì \(S_{OAB}=\dfrac{1}{2}\)

`a.`Xét tam giác AEB và tam giác AFC, có:

\(\widehat{A}\) : chung

\(\widehat{AEB}=\widehat{AFC}=90^o\left(gt\right)\)

Vậy tam giác AEB đồng dạng tam giác AFC ( g.g )

`b.` \(\Rightarrow\dfrac{AE}{AF}=\dfrac{AB}{AC}\)

Xét tam giác AEF và tam giác ABC, có:

\(\dfrac{AE}{AF}=\dfrac{AB}{AC}\left(cmt\right)\)

\(\widehat{A}\) : chung

Vậy tam giác AEF đồng dạng tam giác ABC ( c.g.c )

\(\Rightarrow\widehat{AEF}=\widehat{ABC}\) ( 2 góc tương ứng )

`c.` \(\Rightarrow\dfrac{S_{AEF}}{S_{ABC}}=\left(\dfrac{AE}{AB}\right)^2=\left(\dfrac{3}{6}\right)^2=\left(\dfrac{1}{2}\right)^2=\dfrac{1}{4}\)

\(\Leftrightarrow S_{ABC}=4S_{AEF}\)