ai giải giùm em bài này với em cảm ơn ạ.
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
TT
2
NT
0
TT
2
DT
Đỗ Thanh Hải
CTVVIP
25 tháng 5 2021
Đề dài thế này sao giải thích nhanh cho e đc
Part 1
1 C
2 B
3 D
4 C
5 B
6 A
Part 2
1 T
2 F
3 F
4 F
V
1 That old house has just been bought
2 If he doesn't take these pills, he won't be better
3 I suggest taking a train
4 Spending the weekend in the countryside is very wonderful
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 6 2021
Sử dụng công thức: \(cos\alpha=sin\left(90^0-\alpha\right)\)
AH
Akai Haruma
Giáo viên
1 tháng 5 2023
Lời giải:
a.
\((2-3x^2)^5=\sum\limits_{k=0}^52^k(-3x^2)^{5-k}=\sum\limits_{k=0}^52^k(-3)^{5-k}x^{10-2k}\)
b.
$10-2k=6$
$\Leftrightarrow k=2$
Hệ số gắn với $x^6$ là: \(2^2(-3)^{5-2}=-108\)
KY
3 tháng 8 2021
He asked me to know what the time was.
She asked me when we would meet again.
She asked him If he was crazy.
He asked me to know when they had lived.
He asked her If she would be at the party.
She asked me If I could meet her at the station.
The teacher asked me to know who knew the answer
She asked him to know why he didn't help her.
He asked me If I had seen that car.
The mother asked the twins If they had tidied up their room.
Tom asked how much that computer was
LT
3 tháng 8 2021
11. what the time was.
12. when they would meet again.
13. if he was crazy.
14. where they had lived.
15. if she would be at the party.
16. if I could meet her at the station.
17. who knew the answer.
18. why I didn't help her.
19. if I had seen that car.
20. if they had tidied up their room.
Bài 3:
asked how much that computer was.
b: Để hai đường thẳng song song thì m-4=1
hay m=5
\(b,\Leftrightarrow\left\{{}\begin{matrix}m-4=1\\m-1\ne3\end{matrix}\right.\Leftrightarrow m=5\\ c,\Leftrightarrow A\left(3;0\right)\in\left(d_2\right)\Leftrightarrow3m-12+m-1=0\Leftrightarrow m=\dfrac{13}{4}\\ d,\text{PT giao Ox và Oy: }\left\{{}\begin{matrix}y=0\Leftrightarrow x=\dfrac{1-m}{m-4}\Leftrightarrow OA=\left|\dfrac{m-1}{m-4}\right|\\x=0\Leftrightarrow y=m-1\Leftrightarrow OB=\left|m-1\right|\end{matrix}\right.\\ \text{Kẻ }OH\perp\left(d\right)\Leftrightarrow\dfrac{1}{OH^2}=\dfrac{1}{OA^2}+\dfrac{1}{OB^2}=\dfrac{\left(m-4\right)^2}{\left(m-1\right)^2}+\dfrac{1}{\left(m-1\right)^2}\\ \text{Đặt }OH^2=t\Leftrightarrow\dfrac{1}{t}=\dfrac{m^2-8m+17}{m^2-2m+1}\\ \Leftrightarrow m^2t-8mt+17t=m^2-2m+1\\ \Leftrightarrow m^2\left(t-1\right)-2m\left(4t-1\right)+17t-1=0\\ \Leftrightarrow\Delta'=\left(4t-1\right)^2-\left(t-1\right)\left(17t-1\right)\ge0\\ \Leftrightarrow-t^2+10t\ge0\Leftrightarrow0\le t\le10\\ \Leftrightarrow OH_{max}=\sqrt{10}\Leftrightarrow\dfrac{m^2-2m+1}{m^2-8m+17}=10\Leftrightarrow...\)