\(4x^2-8x+4=2\left(1-x\right)\left(1+x\right)\)

\(\Leftrightarrow4x^2-8x+4=2-2x^2\)

\(\Leftrightarrow6x^2-8x+2=0\)

\(\Leftrightarrow3x^2-4x+1=0\)

\(\Leftrightarrow3x^2-3x-x+1=0\)

\(\Leftrightarrow3x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\3x-1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)

Hình như 2y + x = 5 chứ ? ' '

\(2x\left(x-\dfrac{1}{7}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)

\(\dfrac{x^2-2x+2}{x-3}\)

\(=\dfrac{x^2-2x+1-4+5}{x-3}\)

\(=\dfrac{\left(x-1\right)^2-4+5}{x-3}\)

\(=\dfrac{\left(x-1-2\right)\left(x-1+2\right)}{x-3}+\dfrac{5}{x-3}\)

\(=x+1+\dfrac{5}{x-3}\)

À nhầm

số mol của HCl khi phản ứng với MgO là 0,2mol

=> V_HCl = (0,2+0,2)/2 = 0,4(l)

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

0,1 <--- 0,2 <----------------- 0,1

\(MgO+2HCl\rightarrow MgCl_2+H_2O\)

0,05 ---> 0,1

\(m_{Mg}=0,1.24=2,4\left(g\right)\)

\(\Rightarrow m_{MgO}=4,4-2,4=2\left(g\right)\)

n_MgO = 0,05(mol)

\(V_{HCl}=\dfrac{0,1+0,2}{2}=0,15\left(l\right)\)

2) Ta có : 10^5000 + 125=100...00+125=100...00125

Có tổngcác chữ số là 1+1+2+5=9 chia hết cho 9

Do 10^500 chia hết cho 125 và 125 chia hết cho 125

=> 10^5000+125 chia hết cho 5

a) \(\left(x+1\right)\left(x-2\right)< 0\)

\(\Rightarrow\left\{{}\begin{matrix}x+1>0\\x-2< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>-1\\x< 2\end{matrix}\right.\)\(\Rightarrow2>x>-1\)

\(\Rightarrow\left\{{}\begin{matrix}x+1< 0\\x-2>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x< -1\\x>2\end{matrix}\right.\)=> VN

b) \(\left(x-2\right)\left(x+\dfrac{2}{3}\right)>0\)

\(\Rightarrow\left\{{}\begin{matrix}x-2>0\\x+\dfrac{2}{3}>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>2\\x>-\dfrac{2}{3}\end{matrix}\right.\)\(\Rightarrow x>2\)

\(\Rightarrow\left\{{}\begin{matrix}x-2< 0\\x+\dfrac{2}{3}< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x< 2\\x< -\dfrac{2}{3}\end{matrix}\right.\)\(\Rightarrow x< -\dfrac{2}{3}\)

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

\(X+2HCl\rightarrow XCl_2+H_2\)

M_X ................................ 1mol

16,8...............................0,3mol

=> M_X = 56

=> X là Fe