- Mình dùng cách lớp 8 để làm câu b được không :)?

- Tham khảo câu b:

https://olm.vn/hoi-dap/tim-kiem?q=+++++++++++A=2006%5E2005+1/2006%5E2006+1B=2006%5E2006+1/2006%5E2007+1so+s%C3%A1nh+A+v%C3%A0+B&id=520258

:< dễ lém á , mng giúp mìn ii 

a)A = B

b)A>B

hỏi Huỳnh Thị Huyền Trang ấy

Ta có: \(C=\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+...+\dfrac{2006}{2007}}{\dfrac{2006}{1}+\dfrac{2005}{2}+\dfrac{2004}{3}+...+\dfrac{1}{2006}}\)

\(=\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+...+\dfrac{2006}{2007}}{1+\left(1+\dfrac{2005}{2}\right)+\left(1+\dfrac{2004}{3}\right)+...+\left(1+\dfrac{1}{2006}\right)}\)

\(=\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+...+\dfrac{2006}{2007}}{\dfrac{2007}{2007}+\dfrac{2007}{2}+\dfrac{2007}{3}+...+\dfrac{2007}{2006}}\)

\(=\dfrac{2006}{2007}\)

Ta có:

\(2006A=\dfrac{2006^{2007}+2016}{2006^{2007}+1}=1+\dfrac{2005}{2006^{2007}+1}\)

\(2006B=\dfrac{2006^{2006}+2006}{2006^{2006}+1}=1+\dfrac{2005}{2006^{2006}+1}\)

Do \(\dfrac{2005}{2006^{2006}+1}>\dfrac{2005}{2006^{2007}+1}\Rightarrow1+\dfrac{2005}{2006^{2006}+1}>1+\dfrac{2005}{2006^{2007}+1}\)

\(\Rightarrow2006A< 2006B\Rightarrow A< B\)

Mình sẽ giải cách ngắn hơn cách bạn đạt nha:

Nếu:

\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)

\(A=\dfrac{2006^{2006}+1}{2006^{2007}+1}< 1\)

\(A< \dfrac{2006^{2006}+1+2005}{2006^{2007}+1+2005}\Rightarrow A< \dfrac{2006^{2006}+2006}{2006^{2007}+2006}\Rightarrow A< \dfrac{2006\left(2006^{2005}+1\right)}{2006\left(2006^{2006}+1\right)}\Rightarrow A< \dfrac{2006^{2005}+1}{2006^{2006}+1}=B\)\(A< B\)

A=2006^2005+1/2006^2006+1

B=2006^2006+1/2006^2007+1

Có : 2006A = 2006^2006+2006/2006^2006+1

 = 1 + 2005/2006^2006+1 2006B

= 2006^2007+2006/2006^2007+1

= 1 + 2005/2006^2007+1

Vì : 2006^2006 < 2006^2007

=> 2006^2006+1 < 2006^2007+1

=> 2005/2006^2006+1 > 2005/2006^2007+1

=> 2016A > 2016B

=> A>B

Ta có:

\(A=\frac{2006^{2005}+1}{2006^{2006}+1}\)

\(\Rightarrow2006A=\frac{2006^{2006}+2006}{2006^{2006}+1}=\frac{\left(2006^{2006}+1\right)+2005}{2006^{2006}+1}=1+\frac{2005}{2006^{2006}+1}\)

Ta lại có:

\(B=\frac{2006^{2006}+1}{2006^{2007}+1}\)

\(\Rightarrow2006B=\frac{2006^{2007}+2006}{2006^{2007}+1}=\frac{\left(2006^{2007}+1\right)+2005}{2006^{2007}+1}=1+\frac{2005}{2006^{2007+1}}\)

Ta thấy:

\(\frac{2005}{2006^{2006}+1}>\frac{2005}{2006^{2007}+1}\Rightarrow2006A>2006B\Rightarrow A>B\)

                                                                          Vậy A>B.

Ai k mình, mình k lại.

cho mình xin lỗi sai ở 2 dòng cuối

đầu bài nó như thế chứ không có sai đâu cậu ạ! mk cũng đang hỏi câu này nè 

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