mn giúp mik vs ah mik rất cần
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
There are many ways to keep our body fit. There are a lot of activities to do. You can ride bike, go jogging, skipping, or some other activities. You need to do exercises that can help you gain sweat. Moreover, you can also have a healthy diet. A healthy diet is really important. If you give in your body a large amount of fat, that can make your body bigger. Besides, don't eat to much sweets and fast foods. That doesn't let you to losing weight, moreover you can gain weight. So in my opinion, we need to exercise recently and have a good diet each day. That can make your body fit!
1.
⇒ He doesn't work as hard as me
⇒ He is lazier than me.
2.
⇒ Your bedroom is more uncomforable than my bedroom.
⇒ Your bedroom is not as comfortable as my bedroom.
3.
⇒ Mary speaks English better than John.
⇒ Mary doesn't speak English as worse as John.
4. No boy in my class are as intelligent as Jack.
5. They suggest that he should play football.
1. -> He doesn't work as hard as me.
-> He works lazier than me.
2. -> Your bedroom isn't as comfortable as mine.
-> Your bedroom is uncomfortable than mine.
3. -> Mary speaks English better than John.
-> Mary doesn't speak English as bad as John.
4. -> No boy is as intelligent as Jack in my class.
5. -> They suggest playing football.
Bài 5:
\(A=2A-A=2^2+2^3+...+2^{107}-2-2^2-...-2^{2016}=2^{107}-2\)
\(2\left(A+2\right)=2^{2x}\\ \Rightarrow2\left(2^{107}-2+2\right)=2^{2x}\\ \Rightarrow2^{108}=2^{2x}\\ \Rightarrow2x=108\\ \Rightarrow x=54\)
Bài 3:
Gọi số học sinh lớp 7A, 7B lần lượt là a,b
Ta có: \(\left\{{}\begin{matrix}\dfrac{x}{8}=\dfrac{y}{9}\\y-x=5\end{matrix}\right.\)
Áp dụng TCDTSBN ta có:
\(\dfrac{x}{8}=\dfrac{y}{9}=\dfrac{y-x}{9-1}=\dfrac{5}{1}=5\)
\(\dfrac{x}{8}=5\Rightarrow x=40\\ \dfrac{y}{9}=5\Rightarrow y=45\)
Vậy số học sinh lớp 7A, 7B lần lượt là 40, 45 học sinh
\(M=\sqrt{\dfrac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\dfrac{4}{\left(2+\sqrt{5}\right)^2}}=\dfrac{2}{\left|2-\sqrt{5}\right|}-\dfrac{2}{\left|2+\sqrt{5}\right|}\)
\(=\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{\sqrt{5}+2}=\dfrac{2\left(\sqrt{5}+2\right)-2\left(\sqrt{5}-2\right)}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\)
\(=\dfrac{8}{1}=8\)
Lm ơn giúp mik đii mà mik bt ơn bn đó nhiều lắm . Mik đang rất cần
Ta có: MN là đường trung bình
nên MN//CD
mà CD\(\perp\)AH
nên AH\(\perp\)MN
Ta có:MN là đường trung bình của ΔACD
⇒MN//CD
mà AH⊥CD(đường cao AH)
⇒AH⊥MN
25xy mũ 2+55xy mũ 2+75 xy mũ 2=155 xy mũ 2
thông cảm nha telex của mình bị lỗi rồi T-T