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

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

Ta có: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2020\cdot2021}+\dfrac{1}{2021\cdot2022}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2020}-\dfrac{1}{2021}+\dfrac{1}{2021}-\dfrac{1}{2022}\)

\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)

1/1x2+1/2x3+1/3x4+...+1/2020x2021+1/2021x2022

=1/1-1/2+1/2-1/3+1/3-1/4+...+1/2020-1/2021+1/2021-1/2022.

=1/1-1/2022

=2021/2022

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

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

\(A=-1.2018=-2018\)

\(725:25.10-432:18+127=\left(725:25\right).10-\left(432:18\right)+127=29.10-24+127=290-24+127=393\)

\(\frac{27.13-12}{24+37.12}\)

\(=\frac{351-12}{24+444}\)

\(=\frac{113}{156}\)

đặt S=1.2.3+2.3.4+....+47.48.49

4S=1.2.3.(4-0)+2.3.4.(5-1)+...+47.48.49.(50-46)

4S=1.2.3.4-1.2.3+2.3.4.5-1.2.3.4+....+47.48.49.50-46.47.48.49

4S=47.48.49.50-1.2.3

S=(47.48.49.50-1.2.3):4

cool queen đúng rồi

???????????

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+.......+\dfrac{1}{x\cdot\left(x+1\right)}=\dfrac{122}{123}\)

\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+......+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{122}{123}\)

\(\Leftrightarrow1-\dfrac{1}{x+1}=\dfrac{122}{123}\)

\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{123}\)

\(\Leftrightarrow x=122\)

17,58 x 43 + 57 x 17,58

= 17,58 x ( 43 + 57 )

= 17,58 x 100

= 1758