a, \(\dfrac{25.13}{26.35}=\dfrac{5.1}{2.7}=\dfrac{5}{14}.\)

b, \(\dfrac{6.9-2.17}{63.3-119}=\dfrac{2.3.9-2.17}{9.7.3-17.7}=\dfrac{2\left(3.9-17\right)}{7\left(9.3-17\right)}=\dfrac{2}{7}.\)

c, \(\dfrac{3.13-13.18}{15.40-80}=\dfrac{13\left(3-18\right)}{15.40-40.2}=\dfrac{13.\left(-15\right)}{40\left(15-2\right)}=\dfrac{13.\left(-15\right)}{40.13}=\dfrac{-15}{40}=\dfrac{-3}{8}.\)d, \(\dfrac{2929-101}{2.1919+404}=\dfrac{29.101-101.1}{2.19.101+101.4}=\dfrac{101\left(29-1\right)}{38.101+101.4}=\dfrac{101\left(29-1\right)}{101\left(38+4\right)}=\dfrac{29-1}{38+4}=\dfrac{28}{42}=\dfrac{2}{3}.\)

(còn tiếp!!!)

~ Chúc bn học tốt!!! ~

=-1

=2/5

=38/43

= 3/2

b) \(\dfrac{6\cdot9-2\cdot17}{63\cdot3-119}\)

\(=\dfrac{2\left(3\cdot9-17\right)}{7\cdot\left(3\cdot9-17\right)}\)

\(=\dfrac{2}{7}\)

Trả lời:

a, \(\frac{6\times9-2\times17}{63\times3-119}=\frac{2.3\times9-2\times17}{7.9\times3-7.17}\)

                                     \(=\frac{2\times\left(3\times9-17\right)}{7\times\left(3\times9-17\right)}\)

                                     \(=\frac{2}{7}\)

b, \(\frac{3\times13-13\times18}{15\times40-80}=\frac{13\times\left(3-18\right)}{40\times\left(15-2\right)}\)

                                           \(=\frac{13\times-15}{40\times13}\)

                                           \(=\frac{-3}{8}\)

c, \(\frac{-1997.1996+1}{\left(-1995\right).\left(-1997\right)+1996}=\frac{-1997.1996+1}{\left(1-1996\right).\left(-1997\right)+1996}\)

                                                              \(=\frac{-1997.1996+1}{-1997-1996.\left(-1997\right)+1996}\)

                                                              \(=\frac{-1997.1996+1}{-1996.\left(-1997\right)-1}\)

                                                              \(=\frac{-1997.1996+1}{-\left[1996.\left(-1997\right)+1\right]}\)

                                                               \(=-1\)

d, \(\frac{3.7.13.37.39-10101}{505050-70707}=\frac{10101.39-10101}{50.10101-7.10101}\)

                                                   \(=\frac{10101.\left(39-1\right)}{10101.\left(50-7\right)}\)

                                                   \(=\frac{10101.38}{10101.43}\)

                                                   \(=\frac{38}{43}\)

Phân số a quá dễ, bạn chỉ việc giản ước tử với mẫu sẽ được \(\frac{1}{-1995}\)

b) \(\frac{2.3+4.6+14.21}{3.5+6.10+21.35}\)

Sau khi giản ước ta có: 

\(\frac{2+2+2}{5+5+5}=\frac{8}{15}\)

\(\frac{-1997.1996+1}{\left(1996-1\right).1997+1996}=\frac{-1997.1996+1}{1996.1997-1997+1996}=\frac{-\left(1997.1996-1\right)}{1996.1997-1}=-1\)

\(\frac{2.3.\left(1+2.3+7.7\right)}{3.5.\left(1+3.2+7.7\right)}=\frac{2}{5}\)

\(\frac{3.7.13.37.39-10101}{505050-70707}=\frac{10101.39-10101}{10101.50-10101.7}=\frac{10101.\left(39-1\right)}{10101.\left(50-7\right)}=\frac{38}{43}\)

\(\frac{-5^3.2.2^6}{3^3.5.2^{14}.5}=\frac{-5^3.2^7}{3^3.5^2.2^{14}}=\frac{-5}{3^3.2^7}\)

Công tử họ Nguyễn cấm sủa

Bài 1:

Coi \(A=\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+....+\frac{5}{99.101}\)

\(2A=\left(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+....+\frac{5}{99.101}\right).2\)

\(=5.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{99.101}\right)\)

\(=5.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(=5.\left(1-\frac{1}{101}\right)\)

\(=5.\frac{100}{101}\)

\(=\frac{500}{101}\Rightarrow A=\frac{500}{101}:2=\frac{250}{101}\)

\(\left(\frac{5}{1}-\frac{5}{3}+\frac{5}{3}-\frac{5}{5}....+\frac{5}{99}-\frac{5}{101}\right):\frac{1}{5}\)

\(\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}.....+\frac{1}{99}-\frac{1}{101}\right).5\)

\(\left(\frac{1}{1}-\frac{1}{101}\right).5\)

\(\frac{100}{101}.5\)

\(\frac{500}{101}\)

2,a,\(\frac{2929-101}{3838+404}\)\(=\frac{2828}{4242}=\frac{2}{3}\)

\(b,\frac{54-34}{189-119}=\frac{20}{70}=\frac{2}{7}\)

\(c,d,e,f,f,g,h\)\(tuong\) \(tu\)

\(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...............+\frac{5}{99.101}\)

\(=\frac{5.2}{1.3.2}+\frac{5.2}{3.5.2}+\frac{5.2}{5.7.2}+............+\frac{5.2}{99.101.2}\)

\(=\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+.................+\frac{2}{99.101}\right)\)

\(=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.........+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{5}{2}.\left(1-\frac{1}{101}\right)\)

\(=\frac{5}{2}.\frac{100}{101}\)

\(=\frac{250}{101}\)