\(n_{HCl}=\dfrac{300.10\%}{36,5}=\dfrac{60}{73}\left(mol\right)\)

Pt: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

30/73 \(\leftarrow\) 60/73 \(\rightarrow\) 30/73 ​ ​\(\rightarrow\) 30/73​

\(m_{Fe}=\dfrac{30}{73}.56=23,01\left(g\right)\)

\(\Sigma_{m_{dd\left(spu\right)}}=23,01+300-\dfrac{30}{73}.2=322,188\left(g\right)\)

\(C\%_{FeCl_2}=\dfrac{\dfrac{30}{73}.127}{322,188}.100=16,2\%\)

a) \(\left(x-1\right)\left(x^2+x+1\right)=x^3-1\)

b) \(8x^3-y^3=\left(2x\right)^3-y^3=\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)

- Trường hợp 1: A nằm giữa O và B.

Ta có: OB=OA+ AB=8 +2 =10 cm.

- Trường hợp 2: B nằm giữa O và A. Ta có: OB+ BA= OA

=> OB=OA-BA => OB= 8 – 2 = 6(cm).

Vậy bài toán có hai đáp số là 10 cm và 6 cm.

Theo gt: \(m_{Cu}+m_{Fe}=26,4\)(1)

mà \(m_{Cu}=1,2m_{Fe}\) (2)

(1)(2) \(\Rightarrow m_{Fe}=12\left(g\right)\)

\(\Rightarrow m_{Cu}=26,4-12=14,4\left(g\right)\)

\(n_{Fe}=\dfrac{12}{56}=\dfrac{3}{14}\left(mol\right)\)

\(n_{Cu}=\dfrac{14,4}{64}=0,225\left(mol\right)\)

Pt: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)

3/28\(\leftarrow\) 9/28 \(\leftarrow\) 3/14

Pt: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

0,225\(\leftarrow\) 0,225 \(\leftarrow\) 0,225

\(\Sigma_{n_{H_2}}=\dfrac{9}{28}+0,225=\dfrac{153}{280}\left(mol\right)\)

\(V_{H_2}=\dfrac{153}{280}.22,4=12,14\left(l\right)\)

b) \(m_{Fe_2O_3}=\dfrac{3}{28}.160=17,14\left(g\right)\)

\(m_{CuO}=0,225.80=18\left(g\right)\)

Cách 2:

Theo gt: p + e + n = 58

mà p = e

\(\Rightarrow2p+n=58\)

\(\Rightarrow n=58-2p\)

mà \(p\le n\le1,5p\)

\(\Rightarrow p\le58-2p\le1,5p\)

\(\Rightarrow\left\{{}\begin{matrix}58-2p\ge p\\58-2p\le1,5p\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}p\le19,3\\p\ge16,6\end{matrix}\right.\)

\(\Rightarrow16,6\le p\le19,3\) ; M_X < 40

loại loại nhận

X là Kali ( K )

Theo gt: p + e + n = 58

mà p = e

\(\Rightarrow2p+n=58\)

\(\Rightarrow n=58-2p\)

mà \(p\le n\le1,5p\)

\(\Rightarrow p\le58-2p\le1,5p\)

\(\Rightarrow-58\le-4,5p\le-58\)

\(\Rightarrow p>13\)

mà M_X < 40

loại loại nhận

Vậy X là Kali ( K )

\(\left(a+b+c\right)^2=3\left(ab+ac+bc\right)\)

\(\Leftrightarrow a^2+b^2+c^2+2ab+2ac+2bc=3ab+3ac+3bc\)

\(\Leftrightarrow a^2+b^2+c^2-ab-ac-bc=0\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\)

\(\left[{}\begin{matrix}\left(a-b\right)^2=0\\\left(a-c\right)^2=0\\\left(b-c\right)^2=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}a=b\\a=c\\b=c\end{matrix}\right.\) \(\Rightarrow a=b=c\)

a) \(x^2+2x+1=\left(x+1\right)^2\)

b) \(x^2+6x+9=\left(x+3\right)^2\)

c) \(x^2-6x+9=\left(x-3\right)^2\)

d) \(x^2-2x+1=\left(x-1\right)^2\)

e) \(4x^2+4xy+y^2=\left(2x+y\right)^2\)

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

\(\Leftrightarrow x^2-2x+1=0\)

\(\Leftrightarrow\left(x-1\right)^2=0\)

\(\Rightarrow x=1\)

b) \(\left(x^2-36\right)-\left(x+6\right)=0\)

\(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+6=0\\x-7=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)

Vậy..

c) \(\left(2x-1\right)^2-\left(4x^2-1\right)=0\)

\(\Leftrightarrow4x^2-4x+1-4x^2+1=0\)

\(\Leftrightarrow-4x+2=0\)

\(\Rightarrow x=\dfrac{1}{2}\)

d) \(x^2\left(x^2-4\right)-\left(x^2-4\right)=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(x^2-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x^2-4=0\\x^2-1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\pm2\\x=\pm1\end{matrix}\right.\)

Vậy..

\(\sqrt{4x-4\sqrt{xy}+y}\)

\(=\sqrt{4x-4\sqrt{x}.\sqrt{y}+y}\)

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