Đặt đề toán theo tóm tắt sau và giải bài toán đó: 7 thùng: 2135 quyển vở 5 thùng: ... quyển vở?

0

\(\sqrt[]{2}\) \(\sim1,414\)

sao lại là mẫu giáo

\(A=\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{2001\cdot2005}\)

\(A=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-...+\dfrac{1}{2001}-\dfrac{1}{2005}\)

\(A=1-\dfrac{1}{2005}=\dfrac{2004}{2005}\)

\(B=\dfrac{3}{10\cdot12}+\dfrac{3}{12\cdot14}+...+\dfrac{3}{998\cdot1000}\)

\(\dfrac{2}{3}B=\dfrac{2}{10\cdot12}+...+\dfrac{2}{998\cdot1000}\)

\(\dfrac{2}{3}B=\dfrac{1}{10}-\dfrac{1}{12}+\dfrac{1}{12}-...+\dfrac{1}{998}-\dfrac{1}{1000}\)

\(\dfrac{2}{3}B=\dfrac{1}{10}-\dfrac{1}{1000}=\dfrac{99}{1000}\)

\(B=\dfrac{99}{1000}:\dfrac{2}{3}=\dfrac{297}{2000}\)

\(A=\dfrac{4}{1.5}+\dfrac{4}{5.9}+...+\dfrac{4}{2001.2005}\)

\(\Rightarrow A=4\left(\dfrac{1}{1.5}+\dfrac{1}{5.9}+...+\dfrac{1}{2001.2005}\right)\)

\(\Rightarrow A=4.\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{2001}-\dfrac{1}{2005}\right)\)

\(\Rightarrow A=1-\dfrac{1}{2005}\)

\(\Rightarrow A=\dfrac{2004}{2005}\)

a) 8; 16; 24; 32; 40

b) 1; 2

\(B\left(8\right)\in\left\{0;8;16;24;32\right\}\)

\(Ư\left(12\right)\in\left\{1;2\right\}\)

5 < 6:...< 7

vì 5 < 6 < 7

⇒ 6:... = 6

 ⇒...= 6: 6 

⇒... = 1

Vậy 5 < 6:1 < 7

1

a, \(x\) - \(\dfrac{2}{7}\) = \(\dfrac{4}{5}\) \(\times\) \(\dfrac{15}{24}\)

    \(x\) - \(\dfrac{2}{7}\) = \(\dfrac{1}{2}\)

    \(x\)        = \(\dfrac{1}{2}\) + \(\dfrac{2}{7}\)

    \(x\)       = \(\dfrac{11}{14}\)

b, \(x\): \(\dfrac{2}{7}\) = \(\dfrac{9}{8}\) - \(\dfrac{15}{6}\)

     \(x\):  \(\dfrac{2}{7}\) = - \(\dfrac{11}{8}\)

     \(x\)      = - \(\dfrac{11}{8}\) \(\times\) \(\dfrac{2}{7}\)

     \(x\)     = - \(\dfrac{11}{28}\)

a) x = 11/14

b) x = -11/28

 a)\(\left(\dfrac{5}{6}+\dfrac{3}{8}\right)\times\dfrac{2}{7}\)

\(=\dfrac{29}{24}\times\dfrac{2}{7}\)

\(=\) \(\dfrac{29}{84}\)

b) \(\dfrac{6}{7}\times\dfrac{2}{3}\div\dfrac{5}{7}\)

= \(\dfrac{4}{7}\div\) \(\dfrac{5}{7}\)

= \(\dfrac{4}{5}\)

c) \(\dfrac{8}{9}+\dfrac{3}{4}\times\dfrac{4}{9}\)

= \(\dfrac{8}{9}+\dfrac{1}{3}\)

= \(\dfrac{5}{9}\)

Cô tích xanh cho lisa blackpink vì câu a, b đúng. Còn câu c thì em làm như sau:

c, \(\dfrac{8}{9}\) + \(\dfrac{3}{4}\) \(\times\) \(\dfrac{4}{9}\) 

  = \(\dfrac{8}{9}\) + \(\dfrac{1}{3}\)

  =  \(\dfrac{8}{9}\) + \(\dfrac{3}{9}\)

  =    \(\dfrac{11}{9}\)