Tên bạn là gì

Tên mình là Phạm Trung Hải

⇒" role="presentation" style="border:0px; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax">≥0" role="presentation" style="border:0px; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax">≥0 ⇒" role="presentation" style="border:0px; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax">a≤b≤c" role="presentation" style="border:0px; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax">a≤b≤c ≤" role="presentation" style="border:0px; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax">⇒" role="presentation" style="border:0px; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax">≤" role="presentation" style="border:0px; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax">a≤b" role="presentation" style="border:0px; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax">a≤ba2≤ab≤3" role="presentation" style="border:0px; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax">a2≤ab≤3⇒" role="presentation" style="border:0px; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax">⇔" role="presentation" style="border:0px; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax">0≤b−1≤c−1" role="presentation" style="border:0px; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax">0≤b−1≤c−1 nên b-1 = 1, c-1 =2 ⇒" role="presentation" style="border:0px; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax">(a, b, c) = (0, 0, 0) , (1, 2 , 3) , (1, 3, 2), ( 2, 1, 3) , ( 2, 3, 1 ) , ( 3, 1, 2 ) , ( 3, 2, 1 )

Ghi dấu bé hơn hoặc bằng mà nó không hiện lên =))

TH1 : a.b.c = 0 thì a + b + c = 0 => a = b = c = 0

TH2 : Bài toán đưa về dạng: a + (a+1) + (a+2) = a * (a+1) * (a+2)

Khai triển 2 vế trên ta được 3a + 3 = a^3 + 3a^2 + 2a

Chuyển vế a^3 +3a^2 - a - 3 = 0 (a+3) (a^2-1) = 0

PT trên có 3 nghiệm a = 1 a = -1 a = -3

Vậy ta có các cặp: a = 1 -> 1, 2, 3: tổng = tích = 6 a= -1 -> -1, 0, 1: tổng = tích = 0 a= -3 -> -3, -2 , -1: tổng = tích = -6 Vì là số nguyên dương -> cả 3 số đều phải >=0

Vậy chọn 1,2,3

Vậy (a; b; c) \(\in\) {(0;0;0); (1;2;3)}

a + x = a 

x = a - a 

x = 0 

a + x > a 

x > a - a 

x > 0 

a + x < a 

x < a - a 

x < 0 

a) a + x = a ó x = 0. Vậy x ∈ {0}

b) a + x > a ó x>0. Vậy x ∈ N*

c) a + x < a ó x<0. Vì x ∈ ¥ nên không có giá trị nào của x thỏa mãn. Vậy x =  ∅

a + x = a 

x = a - a 

x = 0 

a + x > a 

x > a - a 

x > 0 

a + x < a 

x < a - a 

x < 0 

ai tick mik cái cho đủ 30 điểm đe

thành thật xin lỗi mình không biết giải

Giúp với bà con ơi. Khó quá trời lun !!!!!!!!!!!!!!!!!!!!!!!!!

goi 3 do can tim la a , b ,c ( a,b,c la so tu nhien )
the de bai ta co : 1/a +1/b+1/c la so tu nhien
vi 1/a , 1/b ,1/c <=1 vay 1/a +1/b+1/c <=3
xet cac th :
th1 : 1/a +1/b+1/c =3 => a=b=xc=1 la nghiem
th2: 1/a +1/b+1/c=2 => a*b+b*c+a*c=2*a*b*c ( 1 )
gia su a = min (a,b,c ) => b*c= max ( a*b ,b*c ,a*c )
neu a=> 2 vay 2*a*b*c => 4*b*c > a*b+b*c+a*c vay a=1 hoac 2
+) voi a=1 ( 1 ) <=> 1+1/b+1/c =2
=> 1/b+1/c = 1 => b+c =b*c => b=c = 2
+) voi a=1 (1) 1/2+1/b+1/c =2
=> 1/b+1/c = 3/2 => b=1 x=2 hoac b=2 c=1
th3: 1/a +1/b+1/c=1 => a*b+b*c+a*c=a*b*c ( 2 )
gia su a = min (a,b,c ) => b*c= max ( a*b ,b*c ,a*c )
neu a=> 4 vay a*b*c => 4*b*c > a*b+b*c+a*c vay a=1,2 hoac 3

oh no