3:

a: BC^2=AB^2+AC^2

=>ΔABC vuông tại A

b: Xét ΔBCD có

BA vừa là đường cao, vừa là trung tuyến

=>ΔBCD cân tại B

c: OA=1/3*BA=1cm
OC=căn 4^2+1^2=căn 17(cm)

\(d,\left(1-\dfrac{4}{7}\right)\left(\dfrac{-3}{8}-\dfrac{-5}{7}\right)=\dfrac{3}{7}.\dfrac{19}{56}=\dfrac{57}{392}\\ e,\left(-\dfrac{5}{6}\right)^2=\dfrac{25}{36}\\ c,\dfrac{4}{3}-\dfrac{1}{3}:5+\dfrac{3}{4}=\dfrac{4}{3}-\dfrac{1}{15}+\dfrac{3}{4}=\dfrac{19}{15}+\dfrac{3}{4}=\dfrac{121}{60}\)

a: \(=\dfrac{3}{11}+\dfrac{18}{66}=\dfrac{18}{66}+\dfrac{18}{66}=\dfrac{36}{66}=\dfrac{6}{11}\)

b: \(=\dfrac{2}{9}-\dfrac{5}{12}+\dfrac{3}{4}\)

\(=\dfrac{8}{36}-\dfrac{15}{36}+\dfrac{27}{36}=\dfrac{20}{36}=\dfrac{5}{9}\)

c: \(=\dfrac{3\cdot7-5}{35}\cdot\dfrac{-40-30}{16}\)

\(=\dfrac{16}{35}\cdot\dfrac{-70}{16}=-2\)

d: \(=\dfrac{1}{5}\cdot\dfrac{2-7}{4}-\dfrac{3}{4}\cdot\dfrac{4-11}{12}\)

\(=\dfrac{1}{5}\cdot\dfrac{-5}{4}-\dfrac{3}{4}\cdot\dfrac{-7}{12}\)

\(=\dfrac{-1}{4}+\dfrac{21}{48}=\dfrac{-12+21}{48}=\dfrac{9}{48}=\dfrac{3}{16}\)

a: \(\Leftrightarrow x^2=16\)

hay \(x\in\left\{4;-4\right\}\)

b: \(\Leftrightarrow\dfrac{20}{6-5x}=\dfrac{5}{14}\)

=>6-5x=56

=>5x=-50

hay x=-10

c: =>7/17=x/119=-35/y

=>x=49; y=-85

d: =>x/27=1/9+1/9=2/9

hay x=6

\(a,\dfrac{x}{-2}=\dfrac{-8}{x}\)

\(\Leftrightarrow x^2=16\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

\(b,\dfrac{5}{14}=\dfrac{20}{6-5x}\)

\(\Leftrightarrow30-25x=280\)

\(\Leftrightarrow25x=30-280=-250\)

\(\Leftrightarrow x=-10\)

\(c,\dfrac{7}{17}=\dfrac{x}{119}=\dfrac{-35}{y}\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{119.7}{17}=49\\y=\dfrac{-35.17}{7}=-85\end{matrix}\right.\)

\(d,\dfrac{x}{27}+\dfrac{-3}{27}=\dfrac{1}{9}\)

\(\dfrac{x-3}{27}=\dfrac{1}{9}\)

\(9x-27=27\)

\(9x=54\Rightarrow x=6\)

mình không biết hình như là về chủ đề an toàn giao thông thì phải

ko nha bn

\(a,\dfrac{-1}{3}-x=\dfrac{1}{5}-\dfrac{1}{-2}\\ \Rightarrow\dfrac{-1}{3}-x=\dfrac{7}{10}\\ \Rightarrow x=-\dfrac{31}{30}\\ b,\dfrac{3}{7}:x=\dfrac{1}{4}-\dfrac{3}{2}\\ \Rightarrow\dfrac{3}{7}:x=-\dfrac{5}{4}\\ \Rightarrow x=-\dfrac{12}{35}\\ c,\dfrac{4}{9}.x+\dfrac{1}{3}=\dfrac{5}{6}\\ \Rightarrow\dfrac{4}{9}.x=\dfrac{1}{2}\\ \Rightarrow x=\dfrac{9}{8}\\ d,\dfrac{2}{5}+\dfrac{3}{5}.x=\dfrac{14}{15}\\ \Rightarrow\dfrac{3}{5}.x=\dfrac{8}{15}\\ \Rightarrow x=\dfrac{8}{9}\)

a: \(=\left(\dfrac{2}{5}-\dfrac{2}{5}\right)+\left(\dfrac{5}{4}+\dfrac{5}{4}\right)-\dfrac{5}{2}=\dfrac{10}{4}-\dfrac{5}{2}=0\)

b: \(=\left(-\dfrac{2}{11}-\dfrac{9}{11}\right)+\left(\dfrac{6}{7}+\dfrac{1}{7}\right)+\left(\dfrac{1}{2}+\dfrac{1}{3}\right)=\dfrac{5}{6}\)

c: \(=\dfrac{7}{19}\left(\dfrac{5}{13}+\dfrac{8}{13}-3\right)=\dfrac{7}{19}\cdot\left(-2\right)=-\dfrac{14}{19}\)

d: \(=\dfrac{12}{11}+\dfrac{9}{22}\left(\dfrac{7}{15}+\dfrac{8}{15}\right)=\dfrac{12}{11}+\dfrac{9}{22}=\dfrac{24+9}{22}=\dfrac{33}{22}=\dfrac{3}{2}\)

My sister likes collecting stamps, so she has a big stamp collection.

-Sơn Tinh:Là 1 người chính trực,chug thủy,hùng mạnh,...(có toàn bộ yếu tố tốt).Cho dù có khó khăn bao nhiu cx vượt qua thể hiện ý chí kiên định và quyết tâm vô cùng.Bảo vệ nhân dân ta khỏi cái thiên tai lũ lụt,hoạn nạn,bão táp,...ST là biểu hiện cho sự kiên định ý chí ko khuất phục của nhân dân ta,lun đánh bại cái xấu và bảo vệ điều tốt.ST là niềm tự hào cả dân tộc VN

D 2-1-2-1-1-1-1-2-1-1

E

1 move

2 Are you watching

3 helps

4 am using

5 Does - do

6 Do - have

7 aren't holding

8 belongs

D. 2-1-2-1-1-1-1-2-1-1

E.

1. move

2.Are you watching

3. helps

4. am using

5. Does - do

6. Do - have

7. aren't holding

8. belongs