\(=\dfrac{11}{8}-\dfrac{3}{8}-\dfrac{3}{4}=1-\dfrac{3}{4}=\dfrac{1}{4}\)

=11/8−3/8−3/4=1−3/4=1/4

1/4

\(\frac{99}{72}-\frac{15}{40}-\frac{1212}{1616}\)= \(\frac{11}{8}-\frac{3}{8}-\frac{6}{8}\)

                                       =\(\frac{11-3-6}{8}\)

                                       =\(\frac{2}{8}=\frac{1}{4}\)

Ai thấy đúng thì k nha

Yêu các bạn nhiều

Đáp án là 1/4

+)\(\frac{99}{72}=\frac{11}{8}\)

+)\(\frac{15}{40}=\frac{3}{8}\)

+)\(\frac{1212}{1616}=\frac{12}{16}\)

\(\Rightarrow\frac{99}{72}-\frac{15}{40}-\frac{1212}{1616}=\frac{11}{8}-\frac{3}{8}-\frac{12}{16}=\frac{1}{4}\)

a: 78/42-4/7=13/7-4/7=9/7

100/110-56/88=10/11-7/11=3/11

75/35-11/7+21/49=15/7-11/7+3/7=7/7=1

99/72-15/40-1212/1616=11/8-3/8-3/4=1-3/4=1/4

\(\frac{1}{3}\)

16/24-1/3=2/3-1/3=1/3

16/24-1/3=2/3-1/3=1/3

hỏi chơi à?

4/5*5/24*1 và 1/7

miru

Rút gọn rồi tính: 

\(\frac{28}{35}.\frac{15}{72}.1\frac{7}{49}\)

\(=\frac{4}{5}.\frac{5}{24}.\frac{56}{49}\)

\(=\frac{4.5.8}{5.24.7}=\frac{4}{21}\)

\(\frac{28}{35}.\frac{15}{72}.1\frac{7}{49}\)

\(=\frac{4}{5}.\frac{5}{24}.\frac{7}{49}\)

\(=\frac{4}{21}\)

k mk nha, cảm ơn những người đã k

hơi lệch

\(1,A=\dfrac{2}{3\cdot7}+\dfrac{2}{7\cdot11}+\dfrac{2}{11\cdot15}+...+\dfrac{2}{99\cdot103}\\ 2A=\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{4}{11\cdot15}+...+\dfrac{4}{99\cdot103}\\ 2A=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{99}-\dfrac{1}{103}\\ 2A=\dfrac{1}{3}-\dfrac{1}{103}=\dfrac{100}{309}\\ A=\dfrac{100}{309}\cdot\dfrac{1}{2}=\dfrac{50}{309}\)

\(2,A=\dfrac{7}{2}+\dfrac{7}{6}+\dfrac{7}{12}+\dfrac{7}{20}+\dfrac{7}{30}+\dfrac{7}{42}+\dfrac{7}{56}+\dfrac{7}{72}+\dfrac{7}{90}\\ A=7\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\right)\\ A=7\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ A=7\left(1-\dfrac{1}{10}\right)=7\cdot\dfrac{9}{10}=\dfrac{63}{10}\)

Bài 1: 

Ta có: \(A=\dfrac{2}{3\cdot7}+\dfrac{2}{7\cdot11}+\dfrac{2}{11\cdot15}+...+\dfrac{2}{99\cdot103}\)

\(=\dfrac{1}{2}\left(\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{4}{11\cdot15}+...+\dfrac{4}{99\cdot103}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{100}{309}=\dfrac{50}{309}\)

Bài 2: 

Ta có: \(A=\dfrac{7}{2}+\dfrac{7}{6}+\dfrac{7}{12}+\dfrac{7}{20}+\dfrac{7}{30}+\dfrac{7}{42}+\dfrac{7}{56}+\dfrac{7}{72}+\dfrac{7}{90}\)

\(=7\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\right)\)

\(=7\left(1-\dfrac{1}{10}\right)\)

\(=\dfrac{63}{10}\)

mình trả lời bạn k mình nhé

ok bạn

Câu 1: \(\frac{17}{51}\)x \(\frac{45}{30}\)= \(\frac{1}{3}\)x \(\frac{3}{2}\)

                                   = \(\frac{1}{2}\).

Câu 2: \(\frac{24}{15}\)x \(\frac{56}{72}\)x \(\frac{10}{14}\)= \(\frac{8}{5}\)x \(\frac{7}{9}\)x \(\frac{5}{7}\)

                                                 = \(\frac{56}{45}\)x \(\frac{5}{7}\)

                                                 = \(\frac{8}{9}\)

Câu 3: \(\frac{32}{48}\times\frac{35}{21}\times\frac{63}{14}\times\frac{9}{30}=\frac{2}{3}\times\frac{5}{3}\times\frac{9}{2}\times\frac{3}{10}=\frac{3}{2}\)

k nha

a)17/51x45/40=1/3x3/2=3/6=1/2

b)24/15x56/72x10/14=8/5x7/9x5/7=8/9

c)32/48x35/21x63/14x9/30=2/3x5/3x9/2x3/10=3/2

x x 134 -x x24 -x x 10 = x x ( 134 - 24 - 10)

x = x x 100

x = 1300 : 100

x = 13

\(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)

\(\Rightarrow2A=2^{101}-2^{100}+2^{99}-2^{98}+....+2^3-2^2\)

\(\Rightarrow2A+A=\left(2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\right)+\left(2^{100}-2^{99}+...+2^2-2\right)\)

\(\Rightarrow3A=2^{101}-2\)

\(\Rightarrow A=\frac{2^{101}-2}{3}\)

   Vậy ......

\(2A=2^{101}-2^{100}+2^{99}-...+2^3-2^2\)

\(\Rightarrow2A+A=2^{101}-2\)

\(\Rightarrow3A=2^{101}-2\)

\(\Rightarrow A=\frac{2^{101}-2}{3}\)