a=4

b=6

~~ chúc bạn học tốt ~~

bố mày cần cm

+)Theo bài:(3a+2b).(2a+3b)\(⋮\)5

=>[(3a+2b).(2a+3b)]²\(⋮\)5²

=>[(3a+2b).(2a+3b)].[(3a+2b).(2a+3b)]\(⋮\)25

Mà[(3a+2b).(2a+3b)].[(3a+2b).(2a+3b)]\(⋮\)25

=>[(3a+2b).(2a+3b)]\(⋮\)25 hoặc [(3a+2b).(2a+3b)]\(⋮\)25

Mà [(3a+2b).(2a+3b)]=[(3a+2b).(2a+3b)]

\(ab+2a-3b=11\)

\(\Rightarrow a\left(b+2\right)-3b-6=5\)

\(\Rightarrow a\left(b+2\right)-3\left(b+2\right)=5\)

\(\Rightarrow\left(a-3\right)\left(b+2\right)=5\)

Ta có bảng sau:

Vậy cặp số \(\left(a;b\right)\) là \(\left(4;3\right);\left(2;-7\right);\left(8;-1\right);\left(-2;-3\right)\)

giúp mik vs ạ

các số nguyên đó là 1,3,5,7,9

k mình nha cậuHoàng Phan

Đáp án:

 (b,a)∈{(3,8),(7,4),(1,−2),(−3,2)}(b,a)∈{(3,8),(7,4),(1,−2),(−3,2)} 

Các bước giải:

ab+1=2a+3bab+1=2a+3b

→(ab−3b)+1=(2a−6)+6→(ab−3b)+1=(2a−6)+6 

→b(a−3)=2(a−3)+5→b(a−3)=2(a−3)+5 

→b(a−3)−2(a−3)=5→b(a−3)−2(a−3)=5 

→(b−2)(a−3)=5→(b−2)(a−3)=5 

→(b−2,a−3)∈U(5)={(1,5),(5,1),(−1,−5),(−5,−1)}→(b−2,a−3)∈U(5)={(1,5),(5,1),(−1,−5),(−5,−1)} 

→(b,a)∈{(3,8),(7,4),(1,−2),(−3,2)}→(b,a)∈{(3,8),(7,4),(1,−2),(−3,2)} 

#Châu's ngốc

\(ab+bc+ca=3abc\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\) (do a,b,c là các số dương)

Áp dụng BĐT Bunhiacopxki dạng phân thức:

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{c}\ge\dfrac{6^2}{a+2b+3c}\)

\(\Rightarrow\dfrac{36}{a+2b+3c}\le\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{3}{c}\left(1\right)\)

Tương tự: \(\left\{{}\begin{matrix}\dfrac{36}{b+2c+3a}\le\dfrac{1}{b}+\dfrac{2}{c}+\dfrac{3}{a}\left(2\right)\\\dfrac{36}{c+2a+3b}\le\dfrac{1}{c}+\dfrac{2}{a}+\dfrac{3}{b}\left(3\right)\end{matrix}\right.\)

Lấy (1) + (2) + (3) ta được:

\(36F\le6\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=6.3=18\)

\(\Rightarrow F\le\dfrac{1}{2}\)

MaxF=1/2 khi \(a=b=c=1\)