1 Yesterday was cooler than today

2 Salim is the most beautiful in her team

3 His house is the highest in my city

4 No one in her class is as intelligent as Lan

5 BJ is one of the greatest tennis players in the world

1. Yesterday was colder than today.

2. Salim is the most beautiful in her team.

3. His house is the highest house in my city.

4.No one in her class is more intelligent than Lan

8. BJ isthe greatest tennis player in the world.

1) Yesterday was colder than today.

2) Salim is the most beautiful in her team.

3) His house is the highest house in my city.

4) No one in my class is more intelligent than Lan.

5) BJ is one of the greatest tennis players in the world.

\(B=2+2^2+2^3+2^4+...+2^{99}+2^{100}=2\left(1+2^2+2^3+2^4\right)+...+2^{96}\left(1+2^2+2^3+2^4\right)=2.31+2^6.31+...+2^{96}.31=31\left(2+2^6+...+2^{96}\right)⋮31\)

B=2+22+23+24+...+299+2100=2(1+22+23+24)+...+296(1+22+23+24)=2.31+26.31+...+296.31=31(2+26+...+296)⋮31

a b c a

D A B A 

c a a c

14 C

15 A

16 A

17 C

18 A

đây nha bn

14 c

15 a

16 a

17 c

19 B

20 D

21 A

22 A

b d a a 

đây nha bn

19 B

20 D

21 A

22 A

\(\Leftrightarrow sin\left(2x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{3}=\dfrac{\pi}{6}+k2\pi\\2x+\dfrac{\pi}{3}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{12}+k\pi\\x=\dfrac{\pi}{4}+k\pi\end{matrix}\right.\)

Nghiệm dương nhỏ nhất là \(x=\dfrac{\pi}{4}\approx0.79\)

Đáp án C

a: BC=căn 6^2+8^2=10cm

bD là phân giác

=>AD/AB=CD/BC

=>AD/3=CD/5=(AD+CD)/(3+5)=8/8=1

=>AD=3cm; CD=5cm

b: Xét ΔBHA vuông tại H và ΔBAC vuông tại A có

góc B chung

=>ΔBHA đồng dạng với ΔBAC

=>BH/BA=BA/BC

=>BH*BC=BA^2

c: Xét ΔBHA có BI là phân giác

nên IH/IA=BH/BA

=>IH/IA=BA/BC=AD/DC

Em bị deadline dí à :))?? Mấy cái này chỉ cần quy đồng, nhóm hay chuyển vế gì là ra mà, em phải tự làm trước , câu nào không hiểu hay không biết làm thì mới đăng chứ!

Bài 2: 

a) Ta có: \(\dfrac{5}{3}+\dfrac{1}{6}:x=1\)

\(\Leftrightarrow\dfrac{1}{6}:x=\dfrac{2}{3}\)

hay \(x=\dfrac{1}{4}\)

b) Ta có: \(\dfrac{3}{2}-\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{2}\)

\(\Leftrightarrow\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=1\\\dfrac{1}{2}x-\dfrac{1}{3}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=\dfrac{4}{3}\\\dfrac{1}{2}x=-\dfrac{2}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{4}{3}\end{matrix}\right.\)