GTTĐ < 0 nếu x < 0 ấy bạn: )

\(\left|x^2-3x\right|=5x\\ \Leftrightarrow\left[{}\begin{matrix}x^2-3x=5x\\x^2-3x=-5x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2-8x=0\\x^2+2x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x\left(x-8\right)=0\\x\left(x+2\right)=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=8\\x=-2\end{matrix}\right.\)

\(\left|x^2+5x\right|=6x\\ \Leftrightarrow\left[{}\begin{matrix}x^2+5x=6x\\x^2+5x=-6x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2-x=0\\x^2+11x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x\left(x-1\right)=0\\x\left(x+11\right)=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-11\end{matrix}\right.\)

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

theo tỉ lệ PT thôi: )

\(n_{NaOH}=\dfrac{40.20\%}{100\%}:40=0,2\left(mol\right)\)

\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

0,2------>0,1-------->0,1

\(m_{ddH_2SO_4}=\dfrac{0,1.98.100\%}{25\%}=39,2\left(g\right)\)

\(m_{Na_2SO_4}=0,1.142=14,2\left(g\right)\)

bình tĩnh: )

Với `x>0`, \(x\ne1\) có:

\(P=\left(\dfrac{x-1}{\sqrt{x}}\right):\left(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\\ =\dfrac{x-1}{\sqrt{x}}:\dfrac{x-1+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\\ =\dfrac{x-1}{\sqrt{x}}:\dfrac{x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\\ =\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)1\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)

Các PT:

\(2Na+2HCl\rightarrow2NaCl+H_2\)

\(Ca+2HCl\rightarrow CaCl_2+H_2\)

\(2Na+2H_2O\rightarrow2NaOH+H_2\)

\(Ca+2H_2O\rightarrow Ca\left(OH\right)_2+H_2\)

Từ tỉ lệ PT có: \(m_1:m_2=n_{H_2}:n_{H_2O}=\dfrac{6,72}{22,4}:\dfrac{21,6}{18}=\dfrac{1}{4}\)

Bài 9

a

\(\left(1\right)S+O_2\underrightarrow{t^o}SO_2\\ \left(2\right)SO_2+\dfrac{1}{2}O_2\rightarrow\left(xt,t^o\right)SO_3\\ \left(3\right)SO_3+H_2O\rightarrow H_2SO_4\\ \left(4\right)H_2SO_4+Fe\rightarrow FeSO_4+H_2\\ \left(5\right)FeSO_4+BaCl_2\rightarrow BaSO_4+FeCl_2\)

b

\(\left(1\right)Cu+Cl_2\rightarrow CuCl_2\\ \left(2\right)CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2+2NaCl\\ \left(3\right)Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\\ \left(4\right)CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

Bài 8

Hòa tan hỗn hợp bằng dung dịch HCl loãng dư, lọc chất rắn không tan ta thu được kim loại đồng.

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

Nhiệt phân dung dịch muối ta thu được kim loại sắt.

\(FeCl_2\underrightarrow{đpdd}Fe+Cl_2\)

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

\(\left(x+3\right)^2-\left(2x-5\right)\left(x+3\right)\\ =\left(x+3\right)\left(x+3-2x+5\right)\\ =\left(x+3\right)\left(8-x\right)\)

\(\left(3x+2\right)^2+\left(3x-2\right)^2-2\left(9x^2-4\right)\\ =9x^2+12x+4+9x^2-12x+4-18x^2+8\\ =4+4+8\\ =16\)