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48.17+48.5-96
=48.(17+5)-48.2
=48.22-48.2
=48.(22-2)
=48.20
=960

\(\frac{6}{15.18}+\frac{6}{18.21}+\frac{6}{21.24}+...+\frac{6}{87.90}\)
\(\rightarrow\frac{6}{3}.\left(\frac{3}{15.18}+\frac{3}{18.21}+...+\frac{3}{87.90}\right)\)
\(\rightarrow2.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)
\(\rightarrow2.\left(\frac{1}{15}-\frac{1}{90}\right)\)
\(\rightarrow\frac{1}{9}\)

A = 6/3 . ( 1/15.18 + 1/18.21 + 1/21/24 + . . . + 1/87.90 )
A = 6/3 . ( 1/15 - 1/18 + 1/18 - 1/21 + 1/21 - 1/24 + . . . + 1/87 - 1/90 )
A = 2 . ( 1/15 - 1/90 )
A = 2. 5/90
A = 10/90 = 1/9
\(\frac{6}{15.18}+\frac{6}{18.21}+\frac{6}{21.24}+...+\frac{6}{84.87}+\frac{6}{87.90}\)
\(=\frac{6}{3}\left(\frac{3}{15.18}+\frac{3}{18.21}+\frac{3}{21.24}+...+\frac{3}{84.87}+\frac{3}{87.90}\right)\)
\(=2\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{84}-\frac{1}{87}+\frac{1}{87}-\frac{1}{90}\right)\)
\(=2\left(\frac{1}{15}-\frac{1}{90}\right)=2\left(\frac{6-1}{90}\right)=2\times\frac{1}{18}=\frac{1}{9}\)

\(A=\frac{5}{18.21}+\frac{5}{21.24}+\frac{5}{24.27}+...+\frac{5}{123.126}\)
\(A=\frac{5}{3}\left(\frac{3}{18.21}+\frac{3}{21.24}+\frac{3}{24.27}+...+\frac{3}{123.126}\right)\)
\(A=\frac{5}{3}\left(\frac{1}{18}-\frac{1}{21}+\frac{1}{21}-\frac{1}{24}+\frac{1}{24}-\frac{1}{27}+...+\frac{1}{123}-\frac{1}{126}\right)\)
\(A=\frac{5}{3}\left(\frac{1}{18}-\frac{1}{126}\right)\)
\(A=\frac{5}{3}.\frac{1}{21}\)
\(A=\frac{5}{63}\)

\(132.9+132=132.\left(9+1\right)=132.10=1320.\)
\(48.17+48.3=48.\left(17+3\right)=48.20=960\)
\(45.48-90.24+145=45.48-45.2.24+145=45.\left(48-2.24\right)+145\)=45.0+145=145
132 . 9 + 132 = 132 . 9 + 132 . 1 = 132 . (9 + 1) = 132 . 10 = 1320
48 . 17 + 48 . 3 = 48 . (17 + 3) = 48 . 20 = 960
45 . 48 - 90 . 24 + 145 = 45 . 48 - 45 . 2 . 24 + 145 = 45 . (48 - 2 . 24) + 145 = 45 . (48 - 48) + 145 = 45 . 0 + 145 = 0 + 145 = 145
754 . 75 - 2262 . 25 + 4560 = 754 . 75 - 754 . 3 . 25 + 4560 = 754 . (75 - 3 . 25) + 4560 = 754 . (75 - 75) + 4560 = 754 . 0 + 4560 = 4560
1050 : 15 - 350 : 5 + 176 = 70 - 70 + 176 = 0 + 176 = 176

=> A = 6/3.( 1/15 - 1/18 + 1/18 - 1/21 + ..... + 1/87 - 1/90 )
=> A = 2.( 1/15 - 1/90 )
=> A = 2.5/90
=> A = 10/90 = 1/9

Ta có : \(\frac{1}{15.18}+\frac{1}{18.21}+\frac{1}{21.24}+...+\frac{1}{87.90}\)
= \(\frac{1}{3}\left(\frac{3}{15.18}+\frac{3}{18.21}+\frac{3}{21.24}+...+\frac{3}{87.90}\right)\)
= \(\frac{1}{3}\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+\frac{1}{21}-\frac{1}{24}+...+\frac{1}{87}-\frac{1}{90}\right)\)
= \(\frac{1}{3}\left(\frac{1}{15}-\frac{1}{90}\right)\)
= \(\frac{1}{3}.\frac{1}{18}\)
= \(\frac{1}{54}\)
\(\frac{1}{15.18}+\frac{1}{18.21}+....+\frac{1}{87.90}\)
\(=\frac{1}{3}.\left(\frac{1}{15}-\frac{1}{90}\right)=\frac{1}{54}\)

Đề sai. câu đầu phải là \(\dfrac{1}{18.21}\) mới đúng.
Nếu câu đầu là \(\dfrac{1}{18.21}\) thì mik có cách làm sau :
\(\dfrac{1}{18.21}+\dfrac{1}{21.24}+\dfrac{1}{24.27}+...+\dfrac{1}{123.126}\)
= \(\dfrac{1}{3}.\left(\dfrac{3}{18.21}+\dfrac{3}{21.24}+\dfrac{3}{24.27}+...+\dfrac{3}{123.126}\right)\)
= \(\dfrac{1}{3}.\left(\dfrac{1}{18}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{24}+\dfrac{1}{24}-\dfrac{1}{27}+...+\dfrac{1}{123}-\dfrac{1}{126}\right)\)
= \(\dfrac{1}{3}.\left(\dfrac{1}{18}-\dfrac{1}{126}\right)\)
= \(\dfrac{1}{3}.\dfrac{1}{21}\)
= \(\dfrac{1}{63}\)
=12.3.17+21.12.2+12.90 =12.(3+17+21+2+90) =12(20+21+2+90) =12.(41+2+90) =12.(43+90) =12.133 =1596
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