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26 tháng 4 2022

ta có 1+2+...+n=n*(n+1)/2

áp dụng vào A, ta được:

\(A=5+\dfrac{5}{\dfrac{2.3}{2}}+\dfrac{5}{\dfrac{3.4}{2}}+\dfrac{5}{\dfrac{4.5}{2}}+...+\dfrac{5}{\dfrac{100.101}{2}}\)

\(=5+\dfrac{10}{2.3}+\dfrac{10}{3.4}+..+\dfrac{10}{100.101}\)

\(=5+10\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{100.101}\right)\)

\(=5+10\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{100}-\dfrac{1}{101}\right)\)

\(=5+10\left(\dfrac{1}{2}-\dfrac{1}{101}\right)=\dfrac{1000}{101}\)

19 tháng 12 2021

\(21_{10}=10101_2\)

27 tháng 2 2022

\(\dfrac{24}{5}\)

27 tháng 2 2022

\(\dfrac{24}{5}\)

25 tháng 10 2021

Đề đâu bạn :v ?

25 tháng 10 2021

à thấy ròi :D

 

25 tháng 3 2022

a)\(=>x=\left(-\dfrac{1}{2}+\dfrac{1}{4}\right):2=-\dfrac{1}{8}\)

 

25 tháng 3 2022

a) \(2x-\dfrac{1}{4}=\dfrac{-1}{2}\)

    \(2x=\dfrac{-1}{2}+\dfrac{1}{4}\)

   \(2x=-\dfrac{1}{4}\)

    \(x=-\dfrac{1}{4}\div2\)

    \(x=-\dfrac{1}{8}\)

b) \(\dfrac{13}{3}-\dfrac{1}{3}\cdot\left(x+1\right)=\dfrac{4}{3}\)

              \(\dfrac{1}{3}\cdot\left(x+1\right)=\dfrac{13}{3}-\dfrac{4}{3}\)         

              \(\dfrac{1}{3}\cdot\left(x+1\right)=3\)

                     \(x+1=3\div\dfrac{1}{3}\)

                    \(x+1=9\)

                   \(x=9-1\)

                   \(x=8\)

              

                     

                     

                     

                     

  

23 tháng 11 2021

-1 ; -3; -7; -21; 1; 3 ; 7 ; 21

23 tháng 11 2021

Ư(-21)={-21, -7, -3, -1, 1, 3, 7, 21}  

19 tháng 8 2021

Bạn sai cta nhé, sắp not xắp

Tham khảo

1. A special kind of tea is sold here 

2. All the cars and trucks have been searched 

3. He was put in prison by the goverment last years

4. We will be met by her parents at the station tomorrow 

5. A meeting is being held by Mr. Brown in the hall 

Bài 2 

1 Tea can't be made with cold water

2 Some of my books have been taken away.

3 Some pictures were taken away by the boys.

4 This room may be used for the classroom.

5 This machine mustn't be used after 5:30 p.m 

6 Mr Cole used to be visited at weekends by John.

7 All the homework ought to be done by her

19 tháng 8 2021

M cảm ơn b nhé

16 tháng 2 2022

1.C

2.D

3.A

4.B

5.A

16 tháng 2 2022

1 C

2 D

3 D

4 A

5 D

27 tháng 10 2023

7:

a: ĐKXĐ: x>=0; x<>1

\(D=\dfrac{1}{2\sqrt{x}-2}-\dfrac{1}{2\sqrt{x}+2}+\dfrac{\sqrt{x}}{1-x}\)

\(=\dfrac{1}{2\left(\sqrt{x}-1\right)}-\dfrac{1}{2\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{-2\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{-1}{\sqrt{x}+1}\)

b: Khi x=4/9 thì \(D=\dfrac{-1}{\dfrac{2}{3}+1}=-1:\dfrac{5}{3}=-\dfrac{3}{5}\)

c: |D|=1/3

=>D=-1/3 hoặc D=1/3

=>\(\left[{}\begin{matrix}\dfrac{-1}{\sqrt{x}+1}=\dfrac{-1}{3}\\\dfrac{-1}{\sqrt{x}+1}=\dfrac{1}{3}\left(loại\right)\end{matrix}\right.\)

=>\(\sqrt{x}+1=3\)

=>\(\sqrt{x}=2\)

=>x=4

6:

a: \(C=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{9-x}\right):\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\)

\(=\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}:\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\dfrac{3\sqrt{x}-x+x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\cdot\dfrac{-\sqrt{x}\left(3-\sqrt{x}\right)}{2\sqrt{x}+4}\)

\(=\dfrac{3\left(\sqrt{x}+3\right)}{3+\sqrt{x}}\cdot\dfrac{-\sqrt{x}}{2\sqrt{x}+4}=\dfrac{-3\sqrt{x}}{2\sqrt{x}+4}\)

b: C<-1

=>C+1<0

=>\(\dfrac{-3\sqrt{x}+2\sqrt{x}+4}{2\sqrt{x}+4}< 0\)

=>\(-\sqrt{x}+4< 0\)

=>\(-\sqrt{x}< -4\)

=>\(\sqrt{x}>4\)

=>x>16

27 tháng 10 2023

\(C=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{9-x}\right):\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\\ =\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\dfrac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{1}{\sqrt{x}}\right)\\ =\left(\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}+\dfrac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\dfrac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\\ =\dfrac{3\sqrt{x}-x+x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\dfrac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\cdot\dfrac{-\sqrt{x}\left(3-\sqrt{x}\right)}{2\sqrt{x}+4}\\ =\dfrac{3\left(\sqrt{x}+3\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\cdot\dfrac{-\sqrt{x}\left(3-\sqrt{x}\right)}{2\sqrt{x}+4}\\ =\dfrac{-3\sqrt{x}}{2\sqrt{x}+4}\)

Để `C < -1` Ta có :

 \(\dfrac{-3}{2\sqrt{x}+4}< -1\\ \Leftrightarrow\dfrac{-3}{2\sqrt{x}+4}+1< 0\\ \Leftrightarrow\dfrac{-3}{2\sqrt{x}+4}+\dfrac{2\sqrt{x}+4}{2\sqrt{x}+4}< 0\\ \Leftrightarrow-3+2\sqrt{x}+4< 0\\ \Leftrightarrow2\sqrt{x}+1< 0\\ \Leftrightarrow2\sqrt{x}< -1\\ \Leftrightarrow\sqrt{x}< -\dfrac{1}{2}\\ \Leftrightarrow x< \dfrac{1}{4}\)

 

24 tháng 6 2022

bài hình mà không gửi hình, làm bài = niềm tin hả bạn