\(f\left(x\right)=-2\)

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

\(\Leftrightarrow-x^2=-2-2=-4\)

\(\Leftrightarrow x^2=4\)

\(\Leftrightarrow\left[\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

\(-12\left(x-5\right)+7\left(3-x\right)=5\)

\(-12x+60+21-7x=5\)

\(-19x+81=5\)

\(-19x=5-81=-76\)

\(x=-\frac{76}{-19}=4\)

seventh

\(\left(x-2\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

Theo bài ra ta có: \(x^2+2y^2-z^2=-12\)

\(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\Rightarrow\frac{x^2}{4}=\frac{y^2}{9}=\frac{z^2}{25}\Rightarrow\frac{x^2}{4}=\frac{2y^2}{18}=\frac{z^2}{25}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{x^2}{4}=\frac{2y^2}{18}=\frac{z^2}{25}=\frac{x^2+2y^2-z^2}{4+18-25}=-\frac{12}{-3}=4\)

\(\Rightarrow\left\{\begin{matrix}\frac{x}{2}=4\rightarrow x=4\cdot2=8\\\frac{y}{3}=4\rightarrow y=4\cdot3=12\\\frac{z}{5}=4\rightarrow z=4\cdot5=20\end{matrix}\right.\)

1/ 4Al + 3O₂ \(\rightarrow\) 2Al₂O₃

2/ \(n_{Al}=\frac{2,7}{27}=0,1\left(mol\right)\)

Theo phương trình hóa học:

\(n_{O_2}=\frac{3}{4}n_{Al}=\frac{3}{4}\cdot0,1=0,075\left(mol\right)\)

ở đktc:

\(V_{O_2}=0,075\cdot22,4=1,68\left(l\right)\)

\(\frac{6}{5}\)