59 độ

27 độ

54 độ

27 độ 

\(53^08'\simeq53^0\)

\(26^034'\simeq27^0\)

1) Vì x=25 thỏa mãn ĐKXĐ nên Thay x=25 vào biểu thức \(A=\dfrac{\sqrt{x}-2}{x+1}\), ta được:

\(A=\dfrac{\sqrt{25}-2}{25+1}=\dfrac{5-2}{25+1}=\dfrac{3}{26}\)

Vậy: Khi x=25 thì \(A=\dfrac{3}{26}\)

2) Ta có: \(B=\dfrac{\sqrt{x}-3}{\sqrt{x}+1}+\dfrac{2x+8\sqrt{x}-6}{x-\sqrt{x}-2}\)

\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}+\dfrac{2x+8\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x-5\sqrt{x}+6+2x+8\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)

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

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

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

câu 3 chứ

22/ \(\omega A=8\pi\)

\(A^2=x^2+\dfrac{v^2}{\omega^2}\Leftrightarrow A^2=3,2^2+\dfrac{\left(4,8\pi\right)^2}{\omega^2}\)

\(\Leftrightarrow\omega^2A^2=3,2^2\omega^2+23,04\pi^2\Leftrightarrow64\pi^2=3,2^2.\omega^2+23,04\pi^2\Leftrightarrow\omega=2\pi\left(rad/s\right)\)

\(\Rightarrow f=\dfrac{\omega}{2\pi}=\dfrac{2\pi}{2\pi}=1\left(Hz\right)\Rightarrow D.1Hz\)

23/ \(\omega A=20;\omega^2A=80\Rightarrow\left\{{}\begin{matrix}\omega=4\left(rad/s\right)\\A=5cm\end{matrix}\right.\)

\(\Rightarrow v=\omega\sqrt{A^2-x^2}=4.\sqrt{5^2-4^2}=12\left(cm/s\right)\Rightarrow A.12cm/s\)

Câu 1: D

Câu 2: C

Câu 3: C

Câu 4: D

Câu 5: A

 1: D

 2: C

 3: C

 4: D

 5: A

nMg = 2.4/24 = 0.1 (mol) 

Mg + 2HCl => MgCl2 + H2 

0.1.......0.2..........0.1........0.1

mMgCl2 = 0.1*95 = 9.5 (g) 

VH2 = 0.1*22.4 = 2.24 (l) 

CM HCl = 0.2/0.4 = 0.5 (M) 

Câu 17 :

\(a) Mg + 2HCl \to MgCl_2 + H_2\\ b) n_{MgCl_2} = n_{H_2} = n_{Mg} = \dfrac{2,4}{24} = 0,1(mol)\\ m_{MgCl_2} = 0,1.95 = 9,5(gam)\\ V_{H_2} = 0,1.22,4 = 2,24(lít)\\ c) n_{HCl} =2 n_{Mg} = 0,2(mol)\\ C_{M_{HCl}} = \dfrac{0,2}{0,4} = 0,5M\)