\(FeCl\Rightarrow\left\{{}\begin{matrix}FeCl_2\\FeCl_3\end{matrix}\right.\)

\(P_2O\Rightarrow P_2O_5\)

\(SiO\Rightarrow SiO_2\)

\(Nn_2O_7\Rightarrow Mn_2O_7\)

(a chỉ viết cthh sai thôi nhé)

\(4.2^x-3=125\)

\(\Leftrightarrow2^x=32\)

\(\Leftrightarrow2^x=2^5\)

\(\Leftrightarrow x=5\)

A thiếu một trường hợp, e thêm giống bn sau nhé

\(1.\left(3x-7\right)\left(x^2+4\right)=0\)

\(\Leftrightarrow3x-7=0\)

\(\Leftrightarrow x=\dfrac{7}{3}\)

\(2.\left[\left(-2x+6\right)^2-1\right]\left(x+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(-2x+6\right)^2-1=0\\x+8=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x+6=1\\x=-8\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-8\end{matrix}\right.\)

Ko e nhé

\(B=\dfrac{2}{\sqrt{2}-1}-\dfrac{2}{\sqrt{2}+1}\)

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

\(=\dfrac{2\sqrt{2}+2-2\sqrt{2}+2}{1}=4\)

Vẫn là toán với lại e hk lớp 9 nên vẫn dùng delta đc nhé, nhưng nếu e thích cách khác cũng đc

1. \(\sqrt{2x+1}=\sqrt{5}\left(x>-\dfrac{1}{2}\right)\)

\(\Leftrightarrow\left|2x+1\right|=5\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\left(loại\right)\end{matrix}\right.\)

Vậy ...

2. \(\sqrt{4\left(x^2-2x-1\right)}=8\)

\(\Leftrightarrow\sqrt{x^2-2x-1}=4\)

\(\Leftrightarrow x^2-2x-1=16\)

\(\Leftrightarrow x^2-2x-17=0\)

\(\Delta'=1-1.\left(-17\right)=18>0\)

Do \(\Delta'>0\) nên phương trình có 2 nghiệm phân biệt:

\(x_1=\dfrac{1+\sqrt{18}}{1}=1+3\sqrt{2}\)

\(x_2=\dfrac{1-\sqrt{18}}{1}=1-3\sqrt{2}\)

\(pthh:\left\{{}\begin{matrix}2Na+2H_2O--->2NaOH+H_2\uparrow\left(1\right)\\2NaOH+H_2SO_4--->Na_2SO_4+2H_2O\left(2\right)\\Ca+H_2SO_4--->CaSO_4\downarrow+H_2\uparrow\left(3\right)\end{matrix}\right.\)

Theo pt₍₁₎: \(n_{H_2}=\dfrac{1}{2}.n_{Na}=\dfrac{1}{2}.\dfrac{2,3}{23}=0,05\left(mol\right)\)

Theo pt₍₃₎: \(n_{H_2}=n_{Ca}=\dfrac{2}{40}=0,05\left(mol\right)\)

\(\Rightarrow V_{H_2}=\left(0,05+0,05\right).22,4=2,24\left(lít\right)\)

b. \(pthh:2KClO_3\xrightarrow[MnO_2]{t^o}2KCl+3O_2\uparrow\)

Theo pt: \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{8}{15}\left(mol\right)\)

\(\Rightarrow m_{KClO_3}=\dfrac{8}{15}.122,5=65,3\left(g\right)\)