MN ơi giải giúp e vs
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11. Have you ever been
12. haven't done
13. have you seen - has already done
14. have just decided
15. has been
16. hasn't had
17. hasn't played
18. haven't had
19. haven't seen
20. have just realized
Đây là thì HTHT nhé, cấu trúc rất dễ nhớ thôi nè :3
S+ have/has + Vp2 (nói về một việc đã bắt đầu trong quá khứ và vẫn tiếp diễn đến bh)
Dấu hiệu nhận biết: for + một khoảng thời gian, since + một thời gian cụ thể trong quá khứ
vd: She hasn't played badminton for 2 years.
He hasn't gone to school since 3 weeks ago.
2. She didn't buy the house because of.
3, She only accepted the job because it brought high salary .
4, We couldn't sleep because of the hot weather .
5, Because the World War II happened, women took over business for their absent husbands .
6, We didn't go fishing because of the rough sea .
7, She was very angry because of his bad behavior .
8, He couldn't sleep because he worried .
9, Because he drove too fast, he caused a serious accident .
Bài 1:
\(n_{CuO}=\dfrac{56}{80}=0,7\left(mol\right)\)
PTHH: CuO + 2HCl → CuCl2 + H2O
Mol: 0,7 1,4
\(m_{ddHCl}=\dfrac{1,4.36,5.100}{14,6}=350\left(g\right)\)
Bài 2:
\(n_{Na_2SO_3}=\dfrac{12,6}{126}=0,1\left(mol\right)\)
PTHH: Na2SO3 + 2HCl → 2NaCl + SO2 + H2O
Mol: 0,1 0,1
\(V_{SO_2}=0,1.22,4=2,24\left(l\right)\)
1) \(\sqrt{2x-5}=7\)
\(\left(\sqrt{2x-5}\right)^2=7^2\)
\(2x-5=49\)
\(2x=54\)
\(x=27\)
2) \(3+\sqrt{x-2}=4\)
\(\sqrt{x-2}=1\)
\(\left(\sqrt{x-2}\right)^2=1^2\)
\(x-2=1\)
\(x=3\)
1) \(\sqrt{2x-5}=7\left(đk:x\ge\dfrac{5}{2}\right)\)
\(\Leftrightarrow2x-5=49\Leftrightarrow2x=54\Leftrightarrow x=27\left(tm\right)\)
2) \(3+\sqrt{x-2}=4\left(đk:x\ge2\right)\)
\(\Leftrightarrow\sqrt{x-2}=1\Leftrightarrow x-2=1\Leftrightarrow x=3\)
3) \(\Leftrightarrow\sqrt{\left(x-1\right)^2}=1\Leftrightarrow\left|x-1\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)
4) \(\Leftrightarrow\sqrt{\left(x-2\right)^2}=1\Leftrightarrow\left|x-2\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
5) \(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=\sqrt{\left(x+4\right)^2}\)
\(\Leftrightarrow\left|2x-1\right|=\left|x+4\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=x+4\\2x-1=-x-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
6) \(ĐK:x\ge-2\)
\(\Leftrightarrow5\sqrt{x+2}-3\sqrt{x+2}-\sqrt{x+2}=\sqrt{x+7}\)
\(\Leftrightarrow\sqrt{x+2}=\sqrt{x+7}\)
\(\Leftrightarrow x+2=x+7\Leftrightarrow2=7\left(VLý\right)\)
Vậy \(S=\varnothing\)
7) \(ĐK:x\ge-1\)
\(\Leftrightarrow5\sqrt{2x+1}+3\sqrt{x+1}=4\sqrt{x+1}+4\sqrt{2x+1}\)
\(\Leftrightarrow\sqrt{2x+1}=\sqrt{x+1}\)
\(\Leftrightarrow2x+1=x+1\Leftrightarrow x=0\left(tm\right)\)
Câu 36 C
Câu 37 C
Câu 38 C