Gọi số hs là \(x(x\in \mathbb{N^*},hs)\)

\(15=3.5\\ 28=2^2.7\\ 63=3^2.7\\ \Rightarrow BCNN\left(15,28,63\right)=2^2.3^2.5.7=1260\\ \Rightarrow x\in BC\left(15,28,63\right)=B\left(1260\right)=\left\{0;1260;2520;3780;...\right\}\)

Mà \(2500\le x\le3000\Rightarrow x=2520\)

Vậy trường có 2520 hs

\(=\left(x^3+4x^2-6x^2-24x+9x+36\right):\left(x+4\right)\\ =\left[x^2\left(x+4\right)-6x\left(x+4\right)+9\left(x+4\right)\right]:\left(x+4\right)\\ =x^2-6x+9\)

\(A=\dfrac{-4x^2+x^2-2x+1-x^2-2x-1}{\left(1-x\right)\left(1+x\right)}=\dfrac{-4x\left(x+1\right)}{\left(1-x\right)\left(1+x\right)}=\dfrac{4x}{x-1}\\ C=\dfrac{-x^2-4x-4+x^2-4x+4-4x^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{-4x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{4x}{2-x}\\ E=\dfrac{x^2-9-x^2+4x-4-x^2+9}{\left(x-2\right)\left(x+3\right)}=\dfrac{-\left(x-2\right)^2}{\left(x-2\right)\left(x+3\right)}=\dfrac{2-x}{x+3}\)

\(\Leftrightarrow\dfrac{x^2-1}{x}\ge0\Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{x}\ge0\)

Trường hợp 1:

\(\left\{{}\begin{matrix}\left(x-1\right)\left(x+1\right)\ge0\\x>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge1\\x\le-1\end{matrix}\right.\\x>0\end{matrix}\right.\Leftrightarrow x\ge1\)

Trường hợp 2:

\(\left\{{}\begin{matrix}\left(x-1\right)\left(x+1\right)\le0\\x< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-1\le x\le1\\x< 0\end{matrix}\right.\Leftrightarrow-1\le x< 0\)

Vậy hệ có nghiệm \(S=[1;+\infty)\cup [-1;0)\)

\(e,=\dfrac{1}{x-1}-\dfrac{2x}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{x^2-2x+1}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{\left(x-1\right)^2}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{x-1}{x^2+1}\\ f,=\dfrac{3x-1}{2\left(3x+1\right)}+\dfrac{3x+1}{2\left(3x-1\right)}-\dfrac{6x}{\left(3x-1\right)\left(3x+1\right)}\\ =\dfrac{9x^2-6x+1+9x^2+6x+1-12x}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{2\left(3x-1\right)^2}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{3x-1}{3x+1}\)

\(g,=\dfrac{x}{x\left(x-2\right)}-\dfrac{x^2+4x}{x\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x\left(x+2\right)}\\ =\dfrac{x^2+2x-x^2-4x-2x+4}{x\left(x-2\right)\left(x+2\right)}=\dfrac{-4x+4}{x\left(x-2\right)\left(x+2\right)}\\ h,=\dfrac{2x^2+1-x^2+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{1}{x^2-x+1}\)

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

Trường hợp 1: \(x< -2\)

\(PT\Leftrightarrow-x-\dfrac{1}{2}+x+2=3\\ \Leftrightarrow\dfrac{3}{2}=3\left(vn\right)\)

Với \(-2\le x< -\dfrac{1}{2}\)

\(PT\Leftrightarrow-x-\dfrac{1}{2}-x-2=3\\ \Leftrightarrow-2x=\dfrac{11}{2}\Leftrightarrow x=-\dfrac{11}{4}\left(ktm\right)\Leftrightarrow x\in\varnothing\)

Với \(x\ge-\dfrac{1}{2}\)

\(PT\Leftrightarrow x+\dfrac{1}{2}-x-2=3\Leftrightarrow-\dfrac{3}{2}=3\left(vn\right)\)

Vậy PT vô nghiệm

Áp dụng BĐT cauchy, ta có:

\(\sqrt{\left(2y+2z-x\right)\cdot3x}\le\dfrac{2z+2y-x+3x}{2}=\dfrac{2\left(x+y+z\right)}{2}=x+y+z\\ \Leftrightarrow\sqrt{2y+2z-x}\le\dfrac{x+y+z}{\sqrt{3x}}\\ \Leftrightarrow\sqrt{\dfrac{x}{2y+2z-x}}\ge\dfrac{\sqrt{x}}{\dfrac{x+y+z}{\sqrt{3x}}}=\dfrac{x\sqrt{3}}{x+y+z}\)

\(\Leftrightarrow S=\sum\sqrt{\dfrac{x}{2y+2z-x}}\ge\sqrt{3}\left(\dfrac{x}{x+y+z}+\dfrac{y}{x+y+z}+\dfrac{z}{x+y+z}\right)\\ \Leftrightarrow S\ge\sqrt{3}\cdot\dfrac{x+y+z}{x+y+z}=\sqrt{3}\)

Dấu \("="\Leftrightarrow x=y=z\) hay tam giác đều

BT nhận biết:

Trích mẫu thử từng chất trong từng trường hợp

\(a,\) Dùng quỳ tím:

- Hóa đỏ: \(HCl,H_2SO_4\)

- Hóa xanh: \(NaOH\)

- Ko đổi màu: \(Na_2SO_4\)

Cho \(BaCl_2\) vào nhóm quỳ hóa đỏ, tạo KT trắng là \(H_2SO_4\), ko hiện tượng là \(HCl\)

\(H_2SO_4+BaCl_2\to BaSO_4\downarrow+2HCl\)

\(b,\) Dùng dd \(BaCl_2\), tạo KT trắng: \(Na_2CO_3\)

Cho \(AgNO_3\) vào các mẫu thử còn lại, tạo KT trắng: \(NaCl\)

Còn lại là \(NaNO_3\)

\(Na_2CO_3+BaCl_2\to BaCO_3\downarrow+2NaCl\\ AgNO_3+NaCl\to AgCl\downarrow+NaNO_3\)

\(c,\) Dùng quỳ tím, quỳ hóa xanh: \(NaOH\)

Cho các mẫu thử còn lại vào dd \(BaCl_2\)

- Tạo KT trắng: \(Na_2SO_4\)

- Ko ht: \(NaCl\)

\(BaCl_2+Na_2SO_4\to 2NaCl+BaSO_4\downarrow\)

Bài 3:

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ a,Zn+2HCl\to ZnCl_2+H_2\\ \Rightarrow n_{Zn}=0,15(mol);n_{HCl}=0,3(mol)\\ b,m_{Zn}=0,15.65=9,75(g)\\ c,C_{M_{HCl}}=\dfrac{0,3}{0,1}=3M\)