Gọi số mol Na, K là a, b

=> 23a + 39b = 8,5

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: 2Na + 2H₂O --> 2NaOH +H₂

______a--------------------------->0,5a

2K + 2H₂O --> 2KOH + H₂

b------------------------->0,5b

=> 0,5a + 0,5b = 0,15

=> a = 0,2; b = 0,1

=> \(\left\{{}\begin{matrix}\%Na=\dfrac{0,2.23}{8,5}.100\%=54,12\%\\\%K=\dfrac{0,1.39}{8,5}.100\%=45,88\%\end{matrix}\right.\)

15.

\(\Delta'=m^2+m-2>0\Leftrightarrow\left[{}\begin{matrix}m>1\\m< -2\end{matrix}\right.\)

Đáp án B

16.

\(\dfrac{\pi}{2}< a< \pi\Rightarrow\dfrac{\pi}{4}< \dfrac{a}{2}< \dfrac{\pi}{2}\Rightarrow\dfrac{\sqrt{2}}{2}< sin\dfrac{a}{2}< 1\Rightarrow\dfrac{1}{2}< sin^2\dfrac{a}{2}< 1\)

\(sina=\dfrac{3}{5}\Leftrightarrow sin^2a=\dfrac{9}{25}\Leftrightarrow4sin^2\dfrac{a}{2}.cos^2\dfrac{a}{2}=\dfrac{9}{25}\)

\(\Leftrightarrow sin^2\dfrac{a}{2}\left(1-sin^2\dfrac{a}{2}\right)=\dfrac{9}{100}\Leftrightarrow sin^4\dfrac{a}{2}-sin^2\dfrac{a}{2}+\dfrac{9}{100}=0\)

\(\Rightarrow\left[{}\begin{matrix}sin^2\dfrac{a}{2}=\dfrac{1}{10}< \dfrac{1}{2}\left(loại\right)\\sin^2\dfrac{a}{2}=\dfrac{9}{10}\end{matrix}\right.\)

\(\Rightarrow sin\dfrac{a}{2}=\dfrac{3\sqrt{10}}{10}\)

17.

Áp dụng công thức trung tuyến:

\(AM=\dfrac{\sqrt{2\left(AB^2+AC^2\right)-BC^2}}{2}=\dfrac{\sqrt{201}}{2}\)

18.

\(\Leftrightarrow x^2+2x+4>m^2+2m\) ; \(\forall x\in\left[-2;1\right]\)

\(\Leftrightarrow m^2+2m< \min\limits_{\left[-2;1\right]}\left(x^2+2x+4\right)\)

Xét \(f\left(x\right)=x^2+2x+4\) trên \(\left[-2;1\right]\)

\(-\dfrac{b}{2a}=-1\in\left[-2;1\right]\) ; \(f\left(-2\right)=4\) ; \(f\left(-1\right)=3\) ; \(f\left(1\right)=7\)

\(\Rightarrow\min\limits_{\left[-2;1\right]}\left(x^2+2x+4\right)=f\left(1\right)=3\)

\(\Rightarrow m^2+2m< 3\Leftrightarrow m^2+2m-3< 0\)

\(\Rightarrow-3< m< 1\Rightarrow m=\left\{-2;-1;0\right\}\)

Đáp án C

19. Hình đa giác là bát giác đều như hình vẽ

\(S=8S_{OAB}=8.\dfrac{1}{2}.IB.OA=4.y_B.x_A=4.\dfrac{\sqrt{2}}{2}.1=2\sqrt{2}\)

Cả 4 đáp án đều không chính xác?

20.

\(M\in\Delta\Rightarrow a+b+1=0\Rightarrow b=-a-1\Rightarrow M\left(a;-a-1\right)\)

\(\left\{{}\begin{matrix}\overrightarrow{AM}=\left(a+1;-a-4\right)\\\overrightarrow{BM}=\left(a-1;-a-2\right)\end{matrix}\right.\)

\(AM+BM=\sqrt{\left(a+1\right)^2+\left(-a-4\right)^2}+\sqrt{\left(-a-2\right)^2+\left(a-1\right)^2}\)

\(AM+BM\ge\sqrt{\left(a+1-a-2\right)^2+\left(-a-4+a-1\right)^2}=\sqrt{26}\)

Dấu "=" xảy ra khi:

\(\left(a+1\right)\left(a-1\right)=\left(-a-4\right)\left(-a-2\right)\Leftrightarrow a=-\dfrac{3}{2}\Rightarrow b=\dfrac{1}{2}\)

\(\Rightarrow ab=-\dfrac{3}{4}\)

Chọn B

1. handy (= convenient)

2. She drives cautiously

1. handy

2. She drives cautiously

1. handy

2. She drives cautiously

Chúc bạn học tốt ^^ !!!

mình làm những bài bn chưa lm nhé

9B

10A

bài 2

have repainted

bàii 3

ride - walikking

swimming

watch

2 have repainted

4 have done 

8 have stepped

9 have you ever played

7) 1 riding -walking

2 swimming

3 watching

8)

2 my mother doesn't like listening to music

5 doing more exercises makes you healthier

còn lại ok nha

chịu ???????????????????

Bạn có thể đăng lại rồi chia nhỏ câu hỏi ra được không ạ? nếu có thể thì đăng full toàn bộ văn bản lên nữa để làm được câu 4