hello ecverone

a) 

\(2.x.y-y+6.x=18\)

\(\text{⇒}y\left(2x-1\right)=18-6x\)

Vì \(x,y\text{∈}Z\text{⇒}2x-1\text{≠}0\)

Chia 2 vế cho \(2x-1\)

\(y=\frac{18-6x}{2x-1}\)

\(\text{⇒}y=\frac{-3.\left(2x-1\right)+15}{2x-1}\)

\(\text{⇒}y=-3+\frac{15}{2x-1}\)

\(\text{⇒}2x-1\text{∈}Ư\left(15\right)=\left\{\text{±}1;\text{±}3;\text{±}5;\text{±}15\right\}\)

\(2x-1=1\text{⇒}x=1\text{⇒}y=12\left(tm\right)\)

\(2x-1=1\text{⇒}x=0\text{⇒}y=-18\left(tm\right)\)

\(2x-1=-1\text{⇒}x=2\text{⇒}y=2\left(tm\right)\)

\(2x-1=3\text{⇒}x=2\text{⇒}y=2\left(tm\right)\)

\(2x-1=-3\text{⇒}x=-1\text{⇒}y=-8\left(tm\right)\)

\(2x-1=5\text{⇒}x=3\text{⇒}y=0\left(tm\right)\)

\(2x-1=-5\text{⇒}x=-2\text{⇒}y=-6\left(tm\right)\)

\(2x-1=15\text{⇒}x=8\text{⇒}y=-2\left(tm\right)\)

\(2x-1=-15\text{⇒}x=-7\text{⇒}y=-4\left(tm\right)\)

Vậy .................

b) 

vì 452 chia x dư 52 suy ra ( 452 - 52 ) = 400 chia hết cho x

và 321 chia x dư 21 suy ra ( 321 - 21 ) = 300 chia hết cho x 

sau đó bn tìm ưc ( 300 ; 400 ) = { 1 ; 2 ; 4 ; 5 ; 10 ; 20 ; 25 ; 50 ; 100 } 

\(a,-215-358+215+\left(-42\right)\\ =\left(-215+215\right)+\left(-358-42\right)\\ =0-400\\ =-400\\ b,13\cdot\left(23+22\right)-3\cdot\left(17+28\right)\\ =13\cdot45-3\cdot45\\ =45\cdot\left(13-3\right)\\ =45\cdot10\\ =450\)

Gọi x là số phần thưởng có thể chia được

ta có; 72 ( dấu chia hết ) x

90( dấu chia hết) x

vậy x thuộc ưcln(72,90)

72= 2^3.3^2

90=2.3^2.5

ư cln( 72, 90)= 2.3^2= 18

vậy có thể chia nhiều nhất 18 phần thưởng

TL :

\(\left\{-3;-9;-27;3;9;27\right\}\)

HT

@@@@@@@@@@@@

Tìm ước của - 27 

Các ước của - 27 là:  \(\text{Ư}\left(-27\right)=\left\{\pm1;\pm3;\pm9;\pm27\right\}\)

\(S=1.2.3+2.3.4+3.4.5+...+9.10.11\)

\(4S=1.2.3.4+2.3.4.\left(5-1\right)+3.4.5.\left(6-2\right)+...+9.10.11.\left(12-8\right)\)

\(=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+...+9.10.11.12-8.9.10.11\)

\(=9.10.11.12\)

\(4S+1=9.10.11.12+1=\left(9.12\right).\left(10.11\right)+1=108.110+1\)

\(=\left(109-1\right)\left(109+1\right)+1=109^2-1+1=109^2\)

Ta có đpcm. 

Ta có \(k\left(k+1\right)\left(k+2\right)=\dfrac{1}{4}k\left(k+1\right)\left(k+2\right)\cdot4\)

\(=\dfrac{1}{4}k\left(k+1\right)\left(k+2\right)\left[\left(k+3\right)-\left(k-1\right)\right]\\ =\dfrac{1}{4}k\left(k+1\right)\left(k+2\right)\left(k+3\right)-\dfrac{1}{4}\left(k-1\right)k\left(k+1\right)\left(k+2\right)\)

Từ đó ta được \(S=\dfrac{1}{4}\cdot1\cdot2\cdot3\cdot4-\dfrac{1}{4}\cdot0\cdot1\cdot2\cdot3+...+\dfrac{1}{4}\cdot9\cdot10\cdot11\cdot12-\dfrac{1}{4}\cdot8\cdot9\cdot10\cdot11\\ \Leftrightarrow S=\dfrac{1}{4}\cdot9\cdot10\cdot11\cdot12\\ \Leftrightarrow4S+1=9\cdot10\cdot11\cdot12+1=11881=109^2\left(đpcm\right)\)