b

B

B

A

D

a, cường độ dđ mạch 

\(I=\dfrac{U}{R_{td}}=\dfrac{12}{10+5}=0,8\left(A\right)\)

\(\Rightarrow U_1=I.R_1=8\left(V\right)\)

\(\Rightarrow U_2=I.R_2=5.0,8=4\left(V\right)\)

b, \(\Rightarrow U_1=\dfrac{4}{2}=2\left(V\right)\)

\(I=I_2=\dfrac{4}{5}=0,8\left(A\right)\)

\(I_1=\dfrac{2}{10}=0,2\left(A\right)\)

\(I_3=I_2-I_1=0,6\left(A\right)\)

\(\Rightarrow R_3=\dfrac{U_1}{I_3}=\dfrac{2}{0,6}=\dfrac{10}{3}\left(\Omega\right)\)

câu 1

a, \(R_{tđ}=\dfrac{R_1R_2}{R_1+R_2}=\dfrac{10.20}{10+20}=\dfrac{20}{3}\left(\Omega\right)\)

b, \(U_{AB}=R_2I_2=20.0,6=12\left(V\right)\)

\(I_1=\dfrac{U_{AB}}{R_1}=\dfrac{12}{10}=1,2\left(A\right)\)

Có: `-C_2021 ^0 +C_2021 ^1 -C_2021 ^2 +....+C_2021 ^2019-C_2021 ^2020 -C_2021 ^2021 =-1-1=-2`

Mà `C_2021 ^0 +C_2021 ^1 +C_2021 ^2 +....+C_2021 ^2019 +C_2021 ^2020 +C_2021 ^2021 =2^2021`

   `=>2(C_2021 ^1 + C_2021 ^3 +C_2021 ^5 +...+C_2021 ^2017 + C_2021 ^2019 )=-2+2^2021`

 `=>C_2021 ^1 + C_2021 ^3 +...+C_2021 ^2017 + C_2021 ^2019 =-1+2^2020`

ko bt bở vì em mới hok đến lớp 5 thui

dung du bamboo

= 6x-8x²-9+12x

=8x²+6x+12x-9

=8x²+18x-9

Mình làm vầy không biết đúng không nha 

1-1/2+1/2-1/3+1/3-1/4+...+1/x-1/x+1=2020/2021

1-1/x+1=2020/2021

1/x+1=1-2020/2021

1/x+1=1/2021

x+1=2021

x=2020 Vậy x = 2020

Mình làm vắn tắt nha nếu chưa hiểu hỏi tớ tớ chỉ các làm cho 

nhớ like cho me nhé . chúc bn học tốt 

\(26,\\ a,\sin45^0=\cos45^0< \sin50^025'< \sin57^048'=\cos32^012'< \sin72^0=\cos18^0< \sin75^0\\ b,\tan37^026'< \tan47^0< \tan58^0=\cot32^0< \tan63^0< \tan66^019'=\cot23^041'\\ 27,\\ A=\dfrac{\left(\sin^226^0+\sin^264^0\right)+2\left(\cos^215^0+\cos^275^0\right)}{\left(\sin^255^0+\cos^255^0\right)+\left(\sin^242^0+\cos^242^0\right)}-\dfrac{\tan81^0}{2\tan81^0}\\ A=\dfrac{\left(\sin^226^0+\cos^226^0\right)+2\left(\sin^215^0+\cos^215^0\right)}{1+1}-\dfrac{1}{2}\\ A=\dfrac{1+2}{2}-\dfrac{1}{2}=2-\dfrac{1}{2}=\dfrac{3}{2}\)

\(28,\\ \sin^2\alpha=1-\cos^2\alpha=1-\dfrac{1}{2}=\dfrac{1}{2}\\ \Leftrightarrow\sin\alpha=\dfrac{\sqrt{2}}{2}\)

Câu 4: D

Câu 5 : D

Câu 6 : A