Vì x tỉ lệ thuận y nên \(\dfrac{x_1}{y_1}=\dfrac{x_2}{y_2}\Rightarrow x_1=\dfrac{y_1\cdot x_2}{y_2}=\dfrac{-4\cdot8}{16}=-2\)

\(a,n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ Fe+2HCl\to FeCl_2+H_2\\ Fe_2O_3+6HCl\to 2FeCl_3+3H_2O\\ \Rightarrow n_{Fe}=0,1(mol)\Rightarrow m_{Fe}=0,1.56=5,6(g)\\ \Rightarrow \%m_{Fe}=\dfrac{5,6}{21,6}.100\%=25,926\%\\ \Rightarrow \%m_{Fe_2O_3}=100\%-25,926\%=74,074\%\\ b,n_{Fe_2O_3}=\dfrac{21,6-5,6}{160}=0,1(mol)\\\Rightarrow \Sigma n_{HCl}=0,1.2+0,1.6=0,8(mol)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,8}{2}=0,4(l)=400(ml)\)

\(a,\left\{{}\begin{matrix}AB=AC\\BM=MC\\AM\text{ chung}\end{matrix}\right.\Rightarrow\Delta AMB=\Delta AMC\left(c.c.c\right)\\ b,\left\{{}\begin{matrix}BE=BM\\\widehat{EBD}=\widehat{MBD}\left(BD\text{ là p/g}\right)\\BD\text{ chung}\end{matrix}\right.\Rightarrow\Delta BED=\Delta BMD\left(c.g.c\right)\\ c,AB=AC\Rightarrow\Delta ABC\text{ cân tại A}\\ \Rightarrow AM\text{ là trung tuyến cũng là đường cao }\\ \Rightarrow AM\perp BC\\ \Rightarrow\widehat{DMB}=90^0\\ \Rightarrow\widehat{BED}=\widehat{BMD}=90^0\left(\Delta BED=\Delta BMD\right)\\ \Rightarrow DE\bot AB\)

Câu 1:

\(b,\sqrt[3]{27}+\sqrt[6]{64}-\sqrt[3]{125}=3+4-5=2\)

Câu 2:

\(a,2>0\Rightarrow\) Hàm đồng biến

\(b,\text{Vì }A\left(0;4\right)\in y=2x+b\Leftrightarrow2\cdot0+b=4\Leftrightarrow b=4\)

\(c,A=\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\\ A=\left(3\sqrt{11}-3\sqrt{2}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\\ A=\left(2\sqrt{11}-3\sqrt{2}\right)\sqrt{11}+3\sqrt{22}=22-3\sqrt{22}+3\sqrt{22}=22\)

\(a,\text{Gọi hstl là }k\\ \Rightarrow k=\dfrac{y}{x}=\dfrac{45}{-15}=-3\\ b,y=kx=-3x\)

\(A=x^2+2x+1+x^2-2x+1-2x^2+2=4\)

\(ĐK:x\ge0\\ \dfrac{8}{\sqrt{x}+2}\in Z\Leftrightarrow\sqrt{x}+2\inƯ\left(8\right)=\left\{-8;-4;-2;-1;1;2;4;8\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{0;2;6\right\}\left(\sqrt{x}\ge0\right)\\ \Leftrightarrow x\in\left\{0;4;36\right\}\)

Đáp án C

\(\dfrac{8^{110}\cdot\left(-9\right)^{71}}{6^{143}\cdot4^{94}}=\dfrac{\left(2^3\right)^{110}\cdot\left[\left(-3\right)^2\right]^{71}}{\left(2\cdot3\right)^{143}\cdot\left(2^2\right)^{94}}=\dfrac{2^{330}\cdot\left(-3\right)^{142}}{2^{143}\cdot3^{143}\cdot2^{188}}=\dfrac{-1}{2\cdot3}=-\dfrac{1}{6}\)

\(f\left(9\right)=2\cdot9+5=18+5=23\\ f\left(-\dfrac{5}{2}\right)=2\left(-\dfrac{5}{2}\right)+5=-5+5=0\)