khó quá mn giúp với ạ
Khó quá nên mới phải SOS mn ạ , mn giúp e zới
mn giúp em vs ạ. bài này em khó hiểu quá ạ
Kẻ AH⊥BC
ta có: \(VP=AB^2+BC^2-2.AB.BC.cosB=AB^2+BC^2-2.AB.BC.\dfrac{BH}{AB}=AB^2+BC^2-2.BH.BC=AB^2-BH^2+BC^2-2.BH.BC+BH^2=AH^2+\left(BC-BH\right)^2=AH^2+CH^2=AC^2=VT\)
bài khó quá mn ai giúp em vs được ko ạ
She does not go to school at 8 o'clock
He does not have dinner at 9 o'clock
She does not go to bed at 7 o'clock
He does not go home at 4 o'clock
He does not go to bed at 9 o'clock
Mn giúp với khó quá
Mn giúp em bài 11c và bài 4f với ạ mai em nộp rồi Riêng bài 4f thì em có tìm được 1 dạng giải nhưng khó hiểu quá, ai có cách nào dễ hiểu hơn thì giúp em với
11c.
Từ đề bài ta có:
\(\left\{{}\begin{matrix}\dfrac{16a-b^2}{4a}=\dfrac{9}{2}\\16a+4b+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2b^2=-4a\\b=-4a-1\end{matrix}\right.\)
\(\Rightarrow2b^2-b=1\Leftrightarrow2b^2-b-1=0\Rightarrow\left[{}\begin{matrix}b=1\Rightarrow a=-\dfrac{1}{2}\\b=-\dfrac{1}{2}\Rightarrow a=-\dfrac{1}{8}\end{matrix}\right.\)
Có 2 parabol thỏa mãn: \(\left[{}\begin{matrix}y=-\dfrac{1}{2}x^2+x+4\\y=-\dfrac{1}{8}x^2-\dfrac{1}{2}x+4\end{matrix}\right.\)
4f.
Từ đề bài ta có:
\(\left\{{}\begin{matrix}1+b+c=0\\\dfrac{4c-b^2}{4}=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}c=-b-1\\c=\dfrac{b^2}{4}-1\end{matrix}\right.\)
\(\Rightarrow\dfrac{b^2}{4}+b=0\)
\(\Rightarrow\left[{}\begin{matrix}b=0\Rightarrow c=-1\\b=-4\Rightarrow c=3\end{matrix}\right.\)
Có 2 parabol thỏa mãn: \(\left[{}\begin{matrix}y=x^2-1\\y=x^2-4x+3\end{matrix}\right.\)
Mn giúp mk với bài này khó quá
4. achievements
5. competitive
6. climbing
7. player
8. famous
9. information
10. professional
11. competition
12. Congratulation
13. footballer
14. useful
15. loudly
Khó quá , giúp với ạ
1 not going on holiday with you
2 such easy questions that all the students got them right
3 we had had time, we would have visited the museum
4 to Daisy for breaking her vase
5 studying ENglish 5 years ago
6 are thought to be the most popular dance in Brazil
7 sooner had she received the exam result than she phoned her mom
8 It was the absence of leadership that caused most of the problems on the committee
9 you have any complants about the product, return it to the shop
10 The more fondness for the game increased , the more proficiency he has
11 been a dramatical rise in house prices this year
12 believed to have escaped in a stolen car
13 gone out with him for 2 years
14 he was going to meet his sister in front of the station
Khó quá nên mới phải SOS mn ạ , c.ơn mn tr ạ
Giúp em với mn ơi, câu này khó quá
\(1+cota+cot^2a+cot^3a\)
\(=1+\dfrac{cosa}{sina}+\dfrac{cos^2a}{sin^2a}+\dfrac{cos^3a}{sin^3a}\)
\(=\left(1+\dfrac{cosa}{sina}\right)\left(1+\dfrac{cos^2a}{sin^2a}\right)\)
\(=\dfrac{sina+cosa}{sina}.\dfrac{sin^2a+cos^2a}{sin^2a}\)
\(=\dfrac{cosa+sina}{sin^3a}\)