1.C

2.D

3.C

4.C

5.C

6.C

7.D

8.D

9.C

10.D

11.A

12.A

1.

\(Na_2O+H_2O\rightarrow2NaOH\)

\(2NaOH+FeCl_2\rightarrow Fe\left(OH\right)_2\downarrow+2NaCl\)

\(Fe\left(OH\right)_2\rightarrow FeO+H_2O\)

\(FeO+2HCl\rightarrow FeCl_2+H_2O\)

áo màu đen hấp thụ nhiệt tốt hơn áo màu trắng
--> áo màu đen nhanh khô hơn trong mùa hè hoặc dưới nắng

9C

10B

11D

12A

13B

14C

15B

16B

Đề bài là: Tính cos2x 

Cảm ơn mn nhiều ạ!

`sin3x sinx+sin(x-π/3) cos (x-π/6)=0`

`<=> 1/2 (cos2x - cos4x) + 1/2(-sin π/6 + sin (2x-π/2)=0`

`<=> cos2x-cos4x-1/2+ sin(2x-π/2)=0`

`<=>cos2x-cos4x-1/2+ sin2x .cos π/2 - cos2x. sinπ/2=0`

`<=> cos2x - cos4x - cos2x = 1/2`

`<=> cos4x = cos(2π)/3`

`<=>` \(\left[{}\begin{matrix}4x=\dfrac{2\text{π}}{3}+k2\text{π}\\4x=\dfrac{-2\text{π}}{3}+k2\text{π}\end{matrix}\right.\)

`<=>` \(\left[{}\begin{matrix}x=\dfrac{\text{π}}{6}+k\dfrac{\text{π}}{2}\\x=-\dfrac{\text{π}}{6}+k\dfrac{\text{π}}{2}\end{matrix}\right.\)

\(sin3x.sinx+sin\left(x-\dfrac{\pi}{3}\right)cos\left(x-\dfrac{\pi}{6}\right)=0\)

\(\Leftrightarrow\dfrac{1}{2}cos2x-\dfrac{1}{2}cos4x+\dfrac{1}{2}sin\left(2x-\dfrac{\pi}{2}\right)+\dfrac{1}{2}sin\left(-\dfrac{\pi}{6}\right)=0\)

\(\Leftrightarrow\dfrac{1}{2}cos2x-\dfrac{1}{2}cos4x-\dfrac{1}{2}cos2x-\dfrac{1}{4}=0\)

\(\Leftrightarrow cos4x+\dfrac{1}{2}=0\)

\(\Leftrightarrow2cos^22x-1+\dfrac{1}{2}=0\)

\(\Leftrightarrow cos^22x=\dfrac{1}{4}\)

\(\Rightarrow cos2x=\pm\dfrac{1}{2}\)

Okie, xinh nên giúp :3 Đùa thui

a/ 5 nguồn mắc nối tiếp \(\left\{{}\begin{matrix}\xi_b=5.\xi=5.4=20\left(V\right)\\r_b=5r=5.0,2=1\left(\Omega\right)\end{matrix}\right.\)

b/ \(R_D=\dfrac{U^2_{dm}}{P_{dm}}=\dfrac{36}{6}=6\left(\Omega\right);I_{dm}=\dfrac{P_{dm}}{U_{dm}}=\dfrac{6}{6}=1\left(A\right)\)

Đèn sáng bình thường \(\Rightarrow I_2=I_D=I_{dm}=1\left(A\right)\)

\(\left(R_1ntR_B\right)//\left(R_2ntR_D\right)\Rightarrow R_{td}=\dfrac{\left(R_1+R_B\right)\left(R_2+R_D\right)}{R_1+R_B+R_2+R_D}=\dfrac{\left(2+4\right)\left(6+6\right)}{2+4+6+6}=4\left(\Omega\right)\)

c/ \(I=\dfrac{\xi_b}{r_b+R_{td}}=\dfrac{20}{1+4}=4\left(A\right)\)

\(I=I_1+I_2\Rightarrow I_1=I-I_2=4-1=3\left(A\right)\Rightarrow P_1=I_1^2.R_1=3^2.2=18\left(W\right)\)

\(m_{Cu}=\dfrac{A_{Cu}.I_B.t}{F.n}=\dfrac{64.3.\left(32.60+10\right)}{96500.2}=...\left(g\right)\)