Chưa cho lí do vì sao: 2x - 5 = 0; 3y - 7 = 0

Có: \(\left(2x-5\right)^{2022}\ge0;\left(3y-7\right)^{2024}\ge0\)

\(\Rightarrow\left(2x-5\right)^{2022}+\left(3y-7\right)^{2024}\ge0\) (Mà đề cho bằng 0)

\(\Rightarrow\left\{{}\begin{matrix}\left(2x-5\right)^{2022}=0\\\left(3y-7\right)^{2024}=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=\dfrac{7}{3}\end{matrix}\right.\)

\(Q=\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}+\dfrac{2\sqrt{x}}{x-4}\left(x\ge0;x\ne4\right)\)

\(=\dfrac{2\left(2-\sqrt{x}\right)}{4-x}+\dfrac{2+\sqrt{x}}{4-x}-\dfrac{2\sqrt{x}}{4-x}\)

\(=\dfrac{4-2\sqrt{x}+2+\sqrt{x}-2\sqrt{x}}{4-x}\)

\(=\dfrac{6-3\sqrt{x}}{4-x}\)

\(=\dfrac{3\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)

\(=\dfrac{3}{\sqrt{x}+2}\)

D

CO₂ chứ chị

Ý là e ko thấy BaO nên làm thiếu :))

C.

\(H_2SO_4+BaCO_3--->BaSO_4\downarrow+CO_2\uparrow+H_2O\)

E ko thấy BaO, :))

\(BaO+HCl--->BaCl_2\downarrow+H_2O\)