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15 tháng 8 2017

I don't know

Tôi chưa học lớp 6 nên ko biết bài này

Lêu lêu ko làm được bài

15 tháng 8 2017

sô 10 với sô 5 nhé

ƯCLN(a,b)=5

=>\(a=5\cdot x;b=5y\), với điều kiện là ƯCLN(x;y)=1

\(a\cdot b=5\cdot150=750\)

=>\(x\cdot y=30\)

Ta sẽ có bảng sau:

a1235
x5101525
b3015106
y150755030

=>Các cặp số (a;b) cần tìm sẽ là (5;150); (150;5); (10;75); (75;10); (25;30); (30;25)

d) Ta có: \(n^2+5n+9⋮n+3\)

\(\Leftrightarrow n^2+3n+2n+6+3⋮n+3\)

\(\Leftrightarrow n\left(n+3\right)+2\left(n+3\right)+3⋮n+3\)

mà \(n\left(n+3\right)+2\left(n+3\right)⋮n+3\)

nên \(3⋮n+3\)

\(\Leftrightarrow n+3\inƯ\left(3\right)\)

\(\Leftrightarrow n+3\in\left\{1;-1;3;-3\right\}\)

hay \(n\in\left\{-2;-4;0;-6\right\}\)

Vậy: \(n\in\left\{-2;-4;0;-6\right\}\)

8 tháng 3 2021

d) Ta có: n2+5n+9⋮n+3n2+5n+9⋮n+3

⇔n2+3n+2n+6+3⋮n+3⇔n2+3n+2n+6+3⋮n+3

⇔n(n+3)+2(n+3)+3⋮n+3⇔n(n+3)+2(n+3)+3⋮n+3

mà n(n+3)+2(n+3)⋮n+3n(n+3)+2(n+3)⋮n+3

nên 3⋮n+33⋮n+3

⇔n+3∈Ư(3)⇔n+3∈Ư(3)

⇔n+3∈{1;−1;3;−3}

28 tháng 11 2017

Vì  ƯCLN ( a ; b ) = 5 và BCNN ( a ; b ) = 60

=>  ƯCLN ( a ; b ) . BCNN ( a; b ) = 5 . 60 = 300 = a.b

Có ƯCLN ( a ; b ) = 5

=> a chia hết cho 5 = 5k

     b chia hết cho 5 = 5q ( k ; q ) = 1

=> a.b = 5k . 5q

           =( 5 . 5 ) .( k . q )

           =  25 . kq = 300

=> kq = 300 : 25 = 12

Ta có bảng

k12134
q11243
a3002575100
b2530010075

Vậy ( a ; b ) = ( 300 ; 25 ) ; ( 25 ; 300 ) ; ( 75 ; 100 ) ; (  100 ; 75 )

           

18 tháng 12 2019

Thank dat dang

31 tháng 1 2022

UKM

^6^7g^7*(KHV C GTGFCCGttedx

8 tháng 12 2015

a) goi hai so la a ; b va a >b

vi UCLN(a,b)=18=>a=18k            ;       b=18q       (trong do UCLN (k,q)=1 va k>q)

=>a+b=162

18k+18q =162

18(k+q)=162

k+q=9

ta co bang sau   

 

k1234
q8765
a18365472
b14412610890

vay ...........

   
    
    

 

29 tháng 10 2016

21453 

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