Hùng Nguyễn

Tắt nhé, nhớ ghi lời giải

\(n_{Fe_2O_3}=0,2\left(mol\right);n_{H_2}=0,4\left(mol\right)\)

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

(mol) 1 3 2 3

(mol) 2/15 0,4 4/15 0,4

Tỉ lệ: 0,2 / 1 > 0,4 / 3

=> Fe2O3 dư

m Fe = n.M = 4/15 . 56 =224/15 (g)

\(n_{H_2SO_4}=0,3\left(mol\right)\)

\(PTHH:Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

(mol) 1 1 1 1

(mol) 4/15 4/15

Tỉ lệ: \(\frac{\frac{4}{15}}{1}< \frac{0,3}{1}\Rightarrow H_2SO_4.du\)

Khối lượng muối FeSO4 là:

m FeSO4 = n.M = 4/15 .152 = 40,53 (g)

\(ĐK:x\ge2,26\)

\(\Leftrightarrow2\sqrt{81-7x^3}=18-x^3\)

\(\Leftrightarrow4.\left(81-7x^3\right)=\left(18-x^3\right)^2\)

\(\Leftrightarrow324-28x^3=324-36x^3+x^5\)

\(\Leftrightarrow8x^3-x^5=0\)

\(\Leftrightarrow x^3\left(8-x^2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^3=0\\8-x^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^3=0\\x^2=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(KTM\right)\\x=2\sqrt{2}\left(TM\right)\end{matrix}\right.\)

Vậy \(S=\left\{2\sqrt{2}\right\}\)

Nhầm tý :)), dòng thứ 4 nhé!!

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

\(=\frac{\sqrt{2}\left(\sqrt{7}-1-\sqrt{7}-1\right)}{2}=\frac{\sqrt{2}.\left(-2\right)}{2}=-\sqrt{2}\)

\(=\frac{\sqrt{16-4\sqrt{7}}}{2}-\frac{\sqrt{16+4\sqrt{7}}}{2}\)

\(=\frac{\sqrt{2\left(8-2\sqrt{7}\right)}}{2}-\frac{\sqrt{2\left(8+2\sqrt{7}\right)}}{2}\)

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

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

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

\(=\frac{\sqrt{2}\left(1-\sqrt{7}-1-\sqrt{7}\right)}{2}=\frac{\sqrt{2}.-2\sqrt{7}}{2}=\frac{-2\sqrt{14}}{2}=-\sqrt{14}\)

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Bài thi cấp tỉnh 4 đ :))