Để xem ai thông minh mà biết cách làm nha , bài này không khó đâu , cũng khá dễ đấy

Ta có:

65 × 111 - 13 × 15 × 37

= 5 × 13 × 3 × 37 - 13 × 3 × 5 × 37

= 0

Vì 0 nhân với bất kì số nào cũng = 0 nên biểu thức trên = 0

\(\left(1+2+3+...+100\right).\left(1^2+2^2+3^2+...+10^2\right).\left(65.111-13.15.37\right)\)

\(\left(1+2+3+...100\right).\left(1^2+2^2+3^2+...+10^2\right).\left(13.5.111-13.15.37\right)\)

\(\left(1+2+3+...+100\right).\left(1^2+2^2+3^2+...+10^2\right).\left(13.15.37-13.15.37\right)\)

\(=0\)

\(P=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

\(P=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(P=\frac{1}{1}-\frac{1}{100}\)

\(P=\frac{99}{100}\)

\(HT\)

\(P=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.....+\dfrac{1}{99.100}\)

\(P=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{99}-\dfrac{1}{100}\)

\(P=1+\left(\dfrac{-1}{2}+\dfrac{1}{2}\right)+\left(\dfrac{-1}{3}+\dfrac{1}{3}\right)+..+\left(\dfrac{-1}{99}+\dfrac{1}{99}\right)+\dfrac{-1}{100}\)

\(P=1+0+0+....+0+\dfrac{-1}{100}\)

\(P=1+\dfrac{-1}{100}\)

\(P=\dfrac{99}{100}\)

\(A=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{1}{2018}\)

\(A=1+\left(1+\frac{2017}{2}\right)+\left(1+\frac{2016}{3}\right)+...+\left(1+\frac{1}{2018}\right)\)

\(A=\frac{2019}{2019}+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2018}\)

\(A=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}+\frac{1}{2019}\right)\)

Ta có: \(\frac{A}{B}=\frac{2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}+\frac{1}{2019}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}}=2019\)

\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)......\left(1-\frac{1}{2001}\right)\)

\(=\frac{1}{2}.\frac{2}{3}..........\frac{2000}{2001}\)

\(=\frac{1}{2001}\)

\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2003}\right)\left(1-\frac{1}{2004}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2002}{2003}.\frac{2003}{2004}\)

\(=\frac{1.2.3...2002.2003}{2.3.4...2003.2004}\)

\(=\frac{1}{2004}\)

_Chúc bạn học tốt_

\(445-52+155+452\\ =\left(445+155\right)+\left(452-52\right)\\ =600+4000\\ =1000\\ 102\cdot55-102\cdot73+51\cdot36\\ =102\cdot\left(55-73\right)+51\cdot36\\ =102\cdot\left(-18\right)+102\cdot18\\ =0\\ 4\cdot5\cdot2\cdot25\cdot5-1000\\ =5000-1000\\ =4000\\ 8782-291\cdot13\\ =4999\\ 254+\left\{38\cdot\left[\left(42-16\right):2\right]\right\}\\ =254+\left[38\cdot\left(26:2\right)\right]\\ =254+\left(38\cdot13\right)\\ =748\\ \left(2791\cdot34+7882:14\right)\cdot0+1510-510:2\\ =1510-255\\ =1255\)

a. \(445-52+155+452=\left(445+155\right)+\left(452-52\right)=600+400=1000\)

b. \(102\cdot55-102\cdot73+51\cdot36=102\left(55-73+18\right)=102\cdot0=0\)

c.\(4\cdot5\cdot2\cdot25\cdot5-1000=4\cdot25\cdot5\cdot5\cdot2-1000=100\cdot50-1000=5000-1000=4000\)

d. \(8782-291\cdot13=8782-3783=4999\)

e. \(254+\left\{38\left[\left(42-16\right)\div2\right]\right\}=254+\left\{30\cdot13\right\}=254+494=748\)

f. \(\left(2791\cdot34+7882\div14\right)\cdot0+1510-510\div2=0+1510-255=1255\)\

Câu 1

a) A=2018!.(2019 - 1 -2018)

=2018!.0

= 0

vậy A= 0

b)\(B=\left(1-\frac{1}{9}+1-\frac{2}{10}+1+\frac{3}{11}+...+1-\frac{150}{158}\right):\left(\frac{1}{4}.\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{158}\right)\right)\)

\(=\left(\frac{8}{9}+\frac{8}{10}+...+\frac{8}{158}\right):\left(\frac{1}{4}\left(\frac{1}{9}+\frac{1}{10}+...+\frac{1}{158}\right)\right)\)

\(=8.\left(\frac{1}{9}+\frac{1}{10}+...+\frac{1}{158}\right):\left(\frac{1}{4}\left(\frac{1}{9}+\frac{1}{10}+...+\frac{1}{158}\right)\right)\)

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

=32

Vậy B= 32