2. i will meet you at six o'clock

3. i will call him now

4. i will ask her

5. i will not buy it

6. i will get my umbrella

2, I will meet you at six o 'clock

3 , I will call him now

4 , I will ask her

5, I won't buy it

6 , I will get my umbrella

đề sai à

đúng mà bn

1 What will they do tomorrow afternoon?

2 Who will go camping next Sunday?

3 Where will we meat?

4 What time will you we back?

5 How will she go to the zoo?

mk sẽ lm nhưng bn cs k cho k ko

Bài này đề sửa thành: \(H=a+4b+1\) mk ms lm được ạ

Ta có: \(a=111...1\) (2020 chữ số 1)

\(a=111...1\cdot100...0+111...1\)

\(a=b.\left(9b+1\right)+b\)

Thay vào:

\(H=a+4b+1=b\left(9b+1\right)+b+4b+1=9b^2+6b+1=\left(3b+1\right)^2\)

=> đpcm

TH1 : 

\(\hept{\begin{cases}-x+2>0\\x-10< 0\end{cases}}\)    

\(\hept{\begin{cases}-x>-2\\x< 10\end{cases}}\)     

\(\hept{\begin{cases}x< 2\\x< 10\end{cases}\Rightarrow x< 2}\)       

TH2 : 

\(\hept{\begin{cases}-x+2< 0\\x-10>0\end{cases}}\)               

\(\hept{\begin{cases}-x< -2\\x>10\end{cases}}\)                              

\(\hept{\begin{cases}x>2\\x>10\end{cases}\Rightarrow x>10}\)                              

Ta có: \(\frac{6}{15.18}+\frac{6}{18.21}+\frac{6}{21.24}+...+\frac{6}{87.90}\)

\(=2\left(\frac{3}{15.18}+\frac{3}{18.21}+\frac{3}{21.24}+...+\frac{3}{87.90}\right)\)

\(=2\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+\frac{1}{21}-\frac{1}{24}+...+\frac{1}{87}-\frac{1}{90}\right)\)

\(=2\left(\frac{1}{15}-\frac{1}{90}\right)\)

\(=2\cdot\frac{1}{18}=\frac{1}{9}\)

\(\frac{6}{15.18}+\frac{6}{18.21}+\frac{6}{21.24}+.......+\frac{6}{87.90}\)

\(=2.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+\frac{1}{21}-\frac{1}{24}+.......+\frac{1}{87}-\frac{1}{90}\right)\)

\(=2.\left(\frac{1}{15}-\frac{1}{90}\right)\)

\(=2.\frac{1}{18}\)

\(=\frac{1}{9}\)

\(8x\left(x-2017\right)-2x+4034=0\)\(\Leftrightarrow8x\left(x-2017\right)-2\left(x-2017\right)=0\)

    \(\Leftrightarrow2\left(x-2017\right)\cdot\left(4x-1\right)=0\)\(\Leftrightarrow\hept{\begin{cases}x-2017=0\\4x-1=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=2017\\x=\frac{1}{4}\end{cases}}\)

          Vậy \(x=2017\)hoặc \(x=\frac{1}{4}\)

8x( x - 2017 ) - 2x + 4034 = 0

<=> 8x( x - 2017 ) - 2( x - 2017 ) = 0

<=> ( 8x - 2 )( x - 2017 ) = 0

<=> \(\orbr{\begin{cases}8x-2=0\\x-2017=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=2017\end{cases}}\)