Ta có: \(2003^{2003}+1=2003^{2002+1}+1và2003^{2004}+1=2003^{2003+1}+1\)

\(\Rightarrow A>B\)

\(\dfrac{15}{30}:\dfrac{17}{51}=\dfrac{15}{30}x\dfrac{51}{17}=\dfrac{3}{2}\)

\(x-\dfrac{2}{3}=\dfrac{4}{5}\\ x=\dfrac{4}{5}+\dfrac{2}{3}\\ x=\dfrac{22}{15}\)

1. How is your first week at secondary school?

2. The students are  doing some  creative drawings in the art  club.

\(\dfrac{5}{3}x\dfrac{4}{7}x\dfrac{7}{8}x\dfrac{3}{5}\\ =\dfrac{5}{3}x\dfrac{3}{5}x\dfrac{4}{7}x\dfrac{7}{8}\\ =1x\dfrac{1}{2}\\ =\dfrac{1}{2}\)

Số học sinh nữ khối 4 có là:

  \(115-59=56\\\) (hs)

Mỗi lớp có số hs nữ là:

   \(56:4=14\) ( hs )

               Đ/S...

Chiều dài của ao đó là:

  \(70:2-15=20\) (m)

                  Đ/S:...

Đề hỏi j bn?

1. D

2. A

3. C

4. D

5. B

\(A=\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+..+\dfrac{1}{44.49}\right)\left(\dfrac{1-3-5-7-..-49}{89}\right)\\ A=\dfrac{1}{5}\left(\dfrac{5}{4.9}+\dfrac{5}{9.14}+..+\dfrac{5}{44.49}\right)\left(\dfrac{1-3-5-7-...-49}{89}\right)\\ A=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{49}\right)\left(\dfrac{1-3-5-7-...-49}{89}\right)\)

\(A=\dfrac{9}{196}\left(\dfrac{1-3-5-7-...-49}{89}\right)\)

Ta đặt: \(P=1-3-5-7-...-49\\ =1-\left(3+5+7+..+49\right)\\ =1-624\\ =-623\\ \Rightarrow\dfrac{9}{196}.-\dfrac{623}{89}=-\dfrac{9}{28}.\)