9 I am interested in watching cartoons 

10 for cheap and clean, and effective source of energy

11 sgould switch off electrical appliances when they are not in use

12 in rural areas is lower than in cities

9 I am interested in watching cartoons 

10 for cheap and clean, and effective source of energy

11 you switch off electrical appliances when they are not in use

12 in rural areas is lower than in cities

17:

\(\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}\)

\(\Rightarrow\left|\overrightarrow{AC}\right|=a\) vì tam giác ABC đều

18.

\(\overrightarrow{BC}+\overrightarrow{CA}=\overrightarrow{BA}\)

\(\Rightarrow\left|\overrightarrow{ BA}\right|=2a\) vì tam giác ABC đều

19 . 

\(\overrightarrow{CA}+\overrightarrow{AB}=\overrightarrow{CB}\)

\(\Rightarrow\left|\overrightarrow{CB}\right|=3a\) vì tam giác ABC đều

20.

\(\overrightarrow{BA}+\overrightarrow{AC}=\overrightarrow{BC}\)

\(\Rightarrow\left|\overrightarrow{BC}\right|=4a\) vì tam giác ABC đều

Mình chữa nốt 14, 15, 16, 21

14, \(\overrightarrow{BD}+\overrightarrow{CA}=\overrightarrow{BA}+\overrightarrow{CD}\)

\(\Leftrightarrow\) \(\overrightarrow{BD}-\overrightarrow{BA}=\overrightarrow{CD}-\overrightarrow{CA}\)

\(\Leftrightarrow\) \(\overrightarrow{AD}=\overrightarrow{AD}\) (luôn đúng)

15, \(\overrightarrow{CA}+\overrightarrow{DB}=\overrightarrow{CB}+\overrightarrow{DA}\)

\(\Leftrightarrow\) \(\overrightarrow{CA}-\overrightarrow{CB}=\overrightarrow{DA}-\overrightarrow{DB}\)

\(\Leftrightarrow\) \(\overrightarrow{BA}=\overrightarrow{BA}\) (luôn đúng)

16, \(\overrightarrow{DC}+\overrightarrow{AB}=\overrightarrow{DB}+\overrightarrow{AC}\)

\(\Leftrightarrow\) \(\overrightarrow{DC}-\overrightarrow{DB}=\overrightarrow{AC}-\overrightarrow{AB}\)

\(\Leftrightarrow\) \(\overrightarrow{BC}=\overrightarrow{BC}\) (luôn đúng)

21,

Ta có: \(\left|\overrightarrow{F}\right|=\left|\overrightarrow{F1}+\overrightarrow{F2}\right|\)

\(\Leftrightarrow\) \(F^2=F1^2+F2^2+2\overrightarrow{F1}\cdot\overrightarrow{F2}\)

\(\Leftrightarrow\) \(F^2=F1^2+F2^2+2F1\cdot F2\cdot cos\left(\overrightarrow{F1},\overrightarrow{F2}\right)\)

\(\Leftrightarrow\) \(F=\sqrt{500^2+500^2+2\cdot500\cdot500\cdot cos60^o}\)

\(\Leftrightarrow\) \(F\approx866\left(N\right)\)

Chúc bn học tốt!

nhìu quá cho bớt lại đc hơm , 20 21 thôi 

?? bài dou

Số ở giữa là 10

vậy ô đó sẽ có các số như sau

         9

8      10      12

        11 

kb đi rồi chỉ cho

16.

\(A=\dfrac{16^2.16^3}{4^8}=\dfrac{4^4.4^6}{4^8}=\dfrac{4^{10}}{4^8}=4^2=16\)

\(B=\dfrac{8^2.8^3}{2^{11}}=\dfrac{2^6.2^9}{2^{11}}=\dfrac{2^{15}}{2^{11}}=2^4=16\)

17. 

\(6,673\)

18.

\(\Rightarrow x=\dfrac{-3}{5}.15=-9\)

19.

\(x=12:\dfrac{3}{4}=16\)

20.Áp dụng t/c dtsbn ta có;

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{x+y}{8+12}=\dfrac{40}{20}=2\)

\(\dfrac{x}{8}=2\Rightarrow x=16\\ \dfrac{y}{12}=2\Rightarrow y=24\)

21.

Áp dụng t/c dtsbn ta có;

\(\dfrac{x}{4}=\dfrac{y}{9}=\dfrac{x-y}{4-9}=\dfrac{15}{-5}=-3\)

\(\dfrac{x}{4}=-3\Rightarrow x=-12\\ \dfrac{y}{9}=-3\Rightarrow y=-27\)

22.

gọi số học sinh nam, nữ lần lượt là a,b

Ta có: \(\left\{{}\begin{matrix}\dfrac{a}{6}=\dfrac{b}{7}\\a+b=39\end{matrix}\right.\)

Áp dụng t/c dtsbn ta có;

\(\dfrac{a}{6}=\dfrac{b}{7}=\dfrac{a+b}{6+7}=\dfrac{39}{13}=3\)

\(\dfrac{a}{6}=3\Rightarrow a=18\\ \dfrac{b}{7}=3\Rightarrow b=21\)

Vậy ...

23.\(\sqrt{16}=4\)

24.\(\sqrt{x}=5\Rightarrow x=25\)

25.B

26.A