b

B

B

A

D

Câu 11:

Gọi tọa độ chân đường cao kẻ từ A xuống BC là H(x;y)

=>\(AH\perp\)BC

A(-1;2); B(0;3); C(5;-2)

\(\overrightarrow{BC}=\left(5;-5\right);\overrightarrow{BH}=\left(x;y-3\right)\)

\(\overrightarrow{AH}=\left(x+1;y-2\right)\)

B,H,C thẳng hàng nên ta có: \(\dfrac{x}{5}=\dfrac{y-3}{-5}\)

=>x=-y+3

=>x+y=3(1)

AH\(\perp\)BC

=>\(\overrightarrow{AH}\cdot\overrightarrow{BC}=0\)

=>\(5\left(x+1\right)+\left(-5\right)\cdot\left(y-2\right)=0\)

=>\(\left(x+1\right)-\left(y-2\right)=0\)

=>x+1-y+2=0

=>x-y=-3(2)

Từ (1) và (2) ta có hệ phương trình

\(\left\{{}\begin{matrix}x+y=3\\x-y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=0\\x+y=3\end{matrix}\right.\)

=>x=0 và y=3

=>Chọn A

Câu 12: 

B(-1;3); C(3;1); A(x;y)

\(\overrightarrow{AB}=\left(-1-x;3-y\right)\); \(\overrightarrow{AC}=\left(3-x;1-y\right)\)

\(AB=\sqrt{\left(-1-x\right)^2+\left(3-y\right)^2}=\sqrt{\left(y-3\right)^2+\left(x+1\right)^2}\)

\(AC=\sqrt{\left(3-x\right)^2+\left(1-y\right)^2}=\sqrt{\left(x-3\right)^2+\left(y-1\right)^2}\)

ΔABC vuông cân tại A

=>AB\(\perp\)AC và AB=AC

=>\(\left\{{}\begin{matrix}\overrightarrow{AB}\cdot\overrightarrow{AC}=0\\AB=AC\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(-1-x\right)\left(3-x\right)+\left(3-y\right)\left(1-y\right)=0\\\left(y-3\right)^2+\left(x+1\right)^2=\left(x-3\right)^2+\left(y-1\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(x+1\right)\left(x-3\right)+\left(y-3\right)\left(y-1\right)=0\\y^2-6y+9+x^2+2x+1=x^2-6x+9+y^2-2y+1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(x+1\right)\left(x-3\right)+\left(y-3\right)\left(y-1\right)=0\\-6y+2x=-6x-2y\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(x+1\right)\left(x-3\right)+\left(y-3\right)\left(y-1\right)=0\\8x=4y\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2x\\x^2-2x-3+y^2-4y+3=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2x\\x^2-2x+y^2-4y=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2x\\x^2-2x+4x^2-4\cdot2x=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2x\\5x^2-10x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x\left(x-2\right)=0\\y=2x\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\left(x-2\right)=0\\y=2x\end{matrix}\right.\)

\(x\left(x-2\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

Khi x=0 thì \(y=2\cdot0=0\)

Khi x=2 thì \(y=2\cdot2=4\)

=>Chọn B

Câu 13: A(0;4); B(3;4); C(3;0)

\(AB=\sqrt{\left(3-0\right)^2+\left(4-4\right)^2}=3\)

\(AC=\sqrt{\left(3-0\right)^2+\left(0-4\right)^2}=\sqrt{3^2+4^2}=5\)

\(BC=\sqrt{\left(3-3\right)^2+\left(0-4\right)^2}=4\)

Vì \(AB^2+BC^2=AC^2\)

nên ΔABC vuông tại B

=>\(R=\dfrac{AC}{2}=2,5\)

=>Chọn A

11 a
12 b
13 a

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 

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

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