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8/7: x =4/5 
Câu 1: Tìm x:
a) 2/5× x =(thiếu đề)
8/7: x =4/5
x=8/7:4/5
x=10/7
Câu 2: Tính bằng cách thuận tiện nhất
5/9× 8/17 + 4/9 × 8/17
=(5/9+4/9)x8/17
=1x8/17=8/17
A = \(\dfrac{5}{1.6}\)+\(\dfrac{5}{6.11}\)+\(\dfrac{5}{11.16}\)+\(\dfrac{5}{16.21}\)+...+\(\dfrac{5}{101.106}\)
A = \(\dfrac{1}{1}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{101}-\dfrac{1}{106}\)
A = \(\dfrac{1}{1}\) - \(\dfrac{1}{106}\)
A = \(\dfrac{105}{106}\)
B = \(\dfrac{3}{1.4}\) +\(\dfrac{3}{4.7}\)+\(\dfrac{3}{7.10}\)+...+\(\dfrac{3}{97.100}\)
B = \(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\)
B = \(\dfrac{1}{1}\) - \(\dfrac{1}{100}\)
B = \(\dfrac{99}{100}\)
C = \(\dfrac{1}{2.7}+\dfrac{1}{7.12}\) + \(\dfrac{1}{12.17}\)+...+ \(\dfrac{1}{97.102}\)
C= \(\dfrac{1}{5}\) \(\times\)( \(\dfrac{5}{2.7}+\dfrac{5}{7.12}+\dfrac{5}{12.17}+...+\dfrac{5}{97.102}\))
C = \(\dfrac{1}{5}\)\(\times\)(\(\dfrac{1}{2}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{12}\) + \(\dfrac{1}{12}\) - \(\dfrac{1}{17}\)+...+ \(\dfrac{1}{97}\) - \(\dfrac{1}{102}\))
C = \(\dfrac{1}{5}\) \(\times\)( \(\dfrac{1}{2}\) - \(\dfrac{1}{102}\))
C = \(\dfrac{1}{5}\) \(\times\) \(\dfrac{25}{51}\)
C = \(\dfrac{5}{51}\)
D = \(\dfrac{1}{2}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\) + \(\dfrac{1}{30}\) + \(\dfrac{1}{42}\) + \(\dfrac{1}{56}\) + \(\dfrac{1}{72}\)
D = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\)+\(\dfrac{1}{7.8}\)+ \(\dfrac{1}{8.9}\)
D = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\) - \(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)
D = \(\dfrac{1}{1}\) - \(\dfrac{1}{9}\)
D = \(\dfrac{8}{9}\)
E = \(\dfrac{3}{2.4}\)+\(\dfrac{3}{4.6}\)+\(\dfrac{3}{6.8}\)+...+\(\dfrac{3}{98.100}\)
E = \(\dfrac{3}{2}\) \(\times\) ( \(\dfrac{2}{2.4}\) + \(\dfrac{2}{4.6}\)+ \(\dfrac{2}{6.8}\)+...+\(\dfrac{2}{98.100}\))
E = \(\dfrac{3}{2}\)\(\times\)( \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{8}\)+...+\(\dfrac{1}{98}\) - \(\dfrac{1}{100}\))
E = \(\dfrac{3}{2}\) \(\times\) ( \(\dfrac{1}{2}\) - \(\dfrac{1}{100}\))
E = \(\dfrac{3}{2}\) \(\times\) \(\dfrac{49}{100}\)
E = \(\dfrac{147}{200}\)
Viết lại đề bài cho dễ nhìn
\(A=\frac{9}{4}\cdot5+\frac{9}{5}\cdot6+\frac{9}{6}\cdot7+...+\frac{9}{33}\cdot34+\frac{9}{34}\cdot35+\frac{9}{35}\cdot36\)
Dấu \(\cdot\)là dáu nhân nha
\(A=\frac{9.5}{4}+\frac{9.6}{5}+\frac{9.7}{6}+...+\frac{9.34}{33}+\frac{9.35}{34}+\frac{9.36}{35}\)
\(A=\frac{9.\left(4+1\right)}{4}+\frac{9\left(5+1\right)}{5}+\frac{9.\left(6+1\right)}{6}+...+\frac{9.\left(33+1\right)}{33}+\frac{9.\left(34+1\right)}{34}+\frac{9.\left(35+1\right)}{35}\)
\(A=\frac{9.4+9.1}{4}+\frac{9.5+9.1}{5}+\frac{9.6+9.1}{6}+...+\frac{9.33+9.1}{33}+\frac{9.34+9.1}{34}+\frac{9.35+9.1}{35}\)
Giải thích
Nếu có: 4x3+4x5 thì có thừa số chung là 4, viết lại: 4x(3+5) và 2 kết quả của 2 phép tính bằng nhau
Vậy nếu có 9x(4+1) thì phân tích ra được là 9x4 + 9x1 và 2 kết quả của 2 phép tính này chắc chắn bằng nhau
Tiếp tục bài giải:
\(A=\frac{9.4}{4}+\frac{9.1}{4}+\frac{9.5}{5}+\frac{9.1}{5}+\frac{9.6}{6}+\frac{9.1}{6}+...+\frac{9.33}{33}+\frac{9.1}{33}+\frac{9.34}{34}+\frac{9.1}{34}+\frac{9.35}{35}+\frac{9.1}{35}\)
Giải thích
\(\frac{1}{2}+\frac{3}{2}=\frac{1+3}{2}=\frac{4}{2}=2\)phải không?
Vậy \(\frac{9.4+9.1}{4}=\frac{9.4}{4}+\frac{9.1}{4}\)
Tiếp tục bài giải:
\(A=9+\frac{9}{4}+9+\frac{9}{5}+9+\frac{9}{6}+...+9+\frac{9}{33}+9+\frac{9}{34}+9+\frac{9}{35}\)
Giải thích
Lười quá, tự hiểu nha
Tiếp
Ta thấy cứ 1 số 9 lại đi với 1 phân số \(\frac{9}{n}\)
Mà n bắt đầu từ 4, kết thúc ở 35
=> Số số 9 có trong A: (35-4) +1 = 32 (số 9)
Vậy
\(A=32.9+\frac{9}{4}+\frac{9}{5}+\frac{9}{6}+...+\frac{9}{33}+\frac{9}{34}+\frac{9}{35}\)
\(A=288+\frac{9}{4}+\frac{9}{5}+\frac{9}{6}+...+\frac{9}{33}+\frac{9}{34}+\frac{9}{35}\)
Từ 4 đến 35 có 3 số chia hết ho 9 : 9;18;27
=> \(A=288+\frac{9}{4}+\frac{9}{5}+\frac{9}{6}+...+\frac{9}{9}+...+\frac{9}{18}+...+\frac{9}{27}+...+\frac{9}{33}+\frac{9}{34}+\frac{9}{35}\)
\(A=288+\frac{9}{4}+\frac{9}{5}+\frac{9}{6}+...+1+...+\frac{1}{2}+...+\frac{1}{2}+...+\frac{9}{33}+\frac{9}{34}+\frac{9}{35}\)
Bí...
Giúp mình hoàn thành bài giải nhé
a)\(\left(\frac{25}{49}+\frac{17}{39}+\frac{22}{39}\times\frac{25}{49}\right)\times\frac{41}{25}\)
\(=\left(\frac{25}{49}+\left(\frac{17}{39}+\frac{22}{39}\right)\times\frac{25}{49}\right)\times\frac{41}{25}\)
\(=\left(\frac{25}{49}+\frac{39}{39}\times\frac{25}{49}\right)\times\frac{41}{25}\)
\(=\left(\frac{25}{49}+1\times\frac{25}{49}\right)\times\frac{41}{25}\)
\(=\left(\frac{25}{49}+\frac{25}{49}\right)\times\frac{41}{25}\)
\(=\frac{50}{49}\times\frac{41}{25}\)
\(=\frac{2050}{1225}\)
Đặt \(B=\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+......+\frac{2}{100\cdot103}\)
\(B=\frac{2}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.....+\frac{1}{100}-\frac{1}{103}\right)\)
\(B=\frac{2}{3}\cdot\left(1-\frac{1}{103}\right)\)
\(B=\frac{2}{3}\cdot\frac{102}{103}\)
\(\Rightarrow B=\frac{68}{103}\)
Đặt \(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{100.103}\)
\(A=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(A=\frac{2}{3}\left(1-\frac{1}{103}\right)\)
\(A=\frac{2}{3}\cdot\frac{102}{103}\)
\(A=\frac{68}{103}\)