\(\sqrt{7-2\sqrt{12}}=\sqrt{\left(2-\sqrt{3}\right)^2}=\left|2-\sqrt{3}\right|=2-\sqrt{3}\)

=> Chọn C

2/7+4/5-7/12+12/7+21/5-17/12

=38/35-7/12+12/7+21/5-17/12

= 211/420+12/7+21/5-17/12

=133/60+21/5-17/12

=42/12-17/12

=23/12

\(\frac{2}{2\times7}+\frac{2}{7\times12}+..+\frac{2}{47\times52}=\frac{2}{5}\left(\frac{1}{2}-\frac{1}{7}\right)+\frac{2}{5}\left(\frac{1}{7}-\frac{1}{12}\right)+..+\frac{2}{5}\left(\frac{1}{47}-\frac{1}{52}\right)\)

\(=\frac{2}{5}\left(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+..+\frac{1}{47}-\frac{1}{52}\right)=\frac{2}{5}\times\left(\frac{1}{2}-\frac{1}{52}\right)=\frac{5}{26}\)

a] 4/12 ; 5/12 ; 11/2 ; 1/4

b] 1 ; 9/12; 12/5 ; 11/3

c] 8/21; 6/11;8/7

d] 2/3 ;2/7 15/2

a]1/3  5/12     11/2       1/4

b]1      3/4         12/5         33/9

c]8/21      6/11        8/7

d]2/3        2/7          15/2

a) 1/3 ; 5/12 ; 11/2 ; 1/14

b) 1 ; 9/12 ; 12/5 ; 11/3

c) 8/21 ; 6/11 ; 8/7

d) 2/3 ; 2/7 ; 15/2

\(a,=\dfrac{5}{7}-\dfrac{33}{8}=-\dfrac{191}{56}\\ b,=\left(\dfrac{12}{17}+\dfrac{5}{17}\right)+\left(\dfrac{19}{7}+\dfrac{3}{7}\right)=1+3=4\\ c,=\left(0,125\cdot8\right)^{12}-\left(\dfrac{45}{15}\right)^3=1-3^3=-26\\ d,=\left(-\dfrac{1}{3}\right)\left(5\dfrac{2}{7}-2\dfrac{2}{7}\right)=-\dfrac{1}{3}\cdot3=-1\\ e,=\dfrac{3^4\cdot3^6}{3^9}=3\)

7/12.(2/3-5/3+3)

=7/12.2

=7/6

giải:  12/15(2/7+3/7+1/7+1/7)=12/15.7/7=12/15=4/5

\(=\frac{12}{15}.\left(\frac{2}{7}+\frac{3}{7}+\frac{1}{7}+\frac{1}{7}\right)\)

\(=\frac{12}{15}.1\)

\(=\frac{12}{15}=\frac{4}{5}\)

\(7,\)

\(A=7+\left(\dfrac{7}{12}-\dfrac{1}{2}+3\right)-\left(\dfrac{1}{12}+5\right)\)

\(=7+\dfrac{7}{12}-\dfrac{1}{2}+3-\dfrac{1}{12}-5\)

\(=\left(\dfrac{7}{12}-\dfrac{1}{12}\right)+\left(7+3-5\right)-\dfrac{1}{2}\)

\(=\dfrac{6}{12}+5-\dfrac{1}{2}\)

\(=\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+5\)

\(=5\)

\(8,\)

\(A=-\dfrac{1}{4}+\dfrac{7}{33}-\dfrac{5}{3}-\left(-\dfrac{15}{12}+\dfrac{6}{11}-\dfrac{68}{49}\right)\)

\(=-\dfrac{1}{4}+\dfrac{7}{33}-\dfrac{5}{3}+\dfrac{15}{12}-\dfrac{6}{11}+\dfrac{68}{49}\)

\(=\left(-\dfrac{1}{4}+\dfrac{15}{12}\right)+\left(\dfrac{7}{33}-\dfrac{5}{3}-\dfrac{6}{11}\right)+\dfrac{68}{49}\)

\(=\left(-\dfrac{3}{12}+\dfrac{15}{12}\right)+\left(\dfrac{7}{33}-\dfrac{55}{33}-\dfrac{18}{33}\right)+\dfrac{68}{49}\)

\(=\dfrac{12}{12}-\dfrac{66}{33}+\dfrac{68}{49}\)

\(=1-2+\dfrac{68}{49}\)

\(=-1+\dfrac{68}{49}\)

\(=\dfrac{19}{49}\)

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

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

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

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