\(pthh:3Fe+2O_2\overset{t^o}{--->}Fe_3O_4\)

Ta có: \(n_{Fe_3O_4}=\dfrac{92,8}{232}=0,4\left(mol\right)\)

Theo pt: \(\left\{{}\begin{matrix}n_{Fe}=3n_{Fe_3O_4}=1,2\left(mol\right)\\n_{O_2}=2n_{Fe_3O_4}=0,8\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=1,2.56=67,2\left(g\right)\\V_{O_2}=0,8.22,4=17,92\left(lít\right)\end{matrix}\right.\)

\(A=\left(\dfrac{5\sqrt{x}-6}{x-9}-\dfrac{2}{\sqrt{x}+3}\right):\left(1+\dfrac{6}{x-9}\right)\left(x\ge0;x\ne9\right)\)

\(=\left[\dfrac{5\sqrt{x}-6}{x-9}-\dfrac{2\left(\sqrt{x}-3\right)}{x-9}\right].\dfrac{x-9}{x-3}\)

\(=\dfrac{5\sqrt{x}-6-2\sqrt{x}+6}{x-9}.\dfrac{x-9}{x-3}\)

\(=\dfrac{3\sqrt{x}}{x-9}.\dfrac{x-9}{x-3}\)

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

\(x:7+15=-17\)

\(\Leftrightarrow\dfrac{1}{7}x=-32\)

\(\Leftrightarrow x=-224\)

\(\dfrac{\dfrac{\sqrt{x}+4}{\sqrt{x}-1}}{\dfrac{1}{\sqrt{x}-1}}>\dfrac{x}{4}+5\left(x\ge0;x\ne1\right)\)

\(\Leftrightarrow\dfrac{\sqrt{x}+4}{\sqrt{x}-1}.\left(\sqrt{x}-1\right)>\dfrac{x+20}{4}\)

\(\Leftrightarrow\sqrt{x}+4>\dfrac{x+20}{4}\)

\(\Leftrightarrow\dfrac{4\sqrt{x}-x-4}{4}>0\)

\(\Leftrightarrow\dfrac{-\left(\sqrt{x}-2\right)^2}{4}>0\) (vô lí)

Vậy không có nghiệm x nào thỏa mãn \(\dfrac{A}{B}>\dfrac{x}{4}+5\)

\(4\sqrt{x}-x-4>0\)

\(\Leftrightarrow-\left(x-4\sqrt{x}+4\right)>0\)

\(\Leftrightarrow-\left(\sqrt{x}-2\right)^2>0\) (vô lí)

Vậy bất phương trình vô nghiệm

Có 2³ = 8

Mà 15 chia hết cho 3 

=> 2¹⁵ chia hết cho 8

Mà 424 cũng chia hết có 8

=> 2¹⁵ + 424 chia hết cho 8

\(A=\dfrac{2-x}{1-2x}+\dfrac{1+2x}{3x}\)

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

\(A=\dfrac{6x-3x^2+1-4x^2}{3x\left(1-2x\right)}=\dfrac{-7x^2+6x+1}{3x\left(1-2x\right)}\)

\(A=\dfrac{\left(x-1\right)\left(x+\dfrac{1}{7}\right)}{3x\left(1-2x\right)}\)

Có: \(\left\{{}\begin{matrix}\left\{{}\begin{matrix}x-1< 0\\x+\dfrac{1}{7}>0\end{matrix}\right.\\\left\{{}\begin{matrix}3x>0\\1-2x< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(x+\dfrac{1}{7}\right)< 0\\3x\left(1-2x\right)< 0\end{matrix}\right.\)

\(\Rightarrow\dfrac{\left(x-1\right)\left(x+\dfrac{1}{7}\right)}{3x\left(1-2x\right)}>0\) hay \(\dfrac{\left(x-1\right)\left(x+\dfrac{1}{7}\right)}{3x\left(1-2x\right)}\ge1\)

Vậy GTNN của A là 1

a. Xét \(2\Delta\) vuông: \(\Delta AEC\) và \(\Delta ADB\) có:

\(\left\{{}\begin{matrix}BD=CE\left(gt\right)\\\widehat{A}.chung\end{matrix}\right.\)

\(\Rightarrow\Delta AEC=\Delta ADB\) (cạnh huyền - góc nhọn)

\(\Rightarrow\) AC = AB

\(\Rightarrow\Delta ABC\) cân

b. Vì \(\Delta ABC\) cân, suy ra: \(\widehat{ABC}=\widehat{ACB}\)

Chứng minh tương tự câu a, suy ra: \(\Delta BEC=\Delta CDB\) 

\(\Rightarrow\widehat{ECB}=\widehat{DBC}\)

\(\Rightarrow\Delta HBC\) cân

Sai đề e nhé