Cho a,b,c là ba số tự nhiên.Biết a + 5 = b + 7 = c + 10.Hãy so sánh a,b,c.
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6 tháng 8 2021
đáp án: a < b < c
vd:
a => 1 + 5 = 6
b => 1 + 7 = 8
c => 1 + 10 = 11
dễ thấy : a < b < c
LL
2
9 tháng 7 2018
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9 tháng 8 2020
Đề bài: Cho 3 số \(a+b+c=0\)..........
Vì \(a+b+c=0\)\(\Rightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)
\(\Rightarrow A=a\left(a+b\right)\left(c+a\right)=a.\left(-c\right).\left(-b\right)=abc\)(1)
\(B=b\left(b+c\right)\left(a+b\right)=b.\left(-a\right).\left(-c\right)=abc\)(2)
\(C=c\left(c+a\right)\left(b+c\right)=c.\left(-b\right).\left(-a\right)=abc\)(3)
Từ (1) , (2) và (3) \(\Rightarrow A=B=C\)
PN
9 tháng 8 2020
3 số mà thêm d vô mần chi rứa:v
Ta có : \(a+b+c=0< =>\hept{\begin{cases}a+b=-c\\a+c=-b\\b+c=-a\end{cases}}\)
Thay vào các biểu thức A,B,C ta có :
\(\hept{\begin{cases}A=a.\left(-c\right).\left(-b\right)=abc\\B=b.\left(-a\right).\left(-c\right)=abc\\C=c.\left(-b\right).\left(-a\right)=abc\end{cases}}\)
Suy ra \(A=B=C\)
26 tháng 6 2015
Vì a,b,c là các số tự nhiên khác 0 nên a,b,c > 0.
Do vậy a < a + b < a + b + c
b < b + c < a + b + c
c < c + a < a + b + c
\(\Rightarrow\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1\)
ND
1
3 tháng 7 2017
Ta có: a<b
\(\Rightarrow a+c< b+c\)
\(\Rightarrow a.\left(a+c\right)< b.\left(b+c\right)\)
\(\Rightarrow\frac{a}{b}< \frac{a+c}{b+c}\)\(\left(a,b\in Z+;c\in N\right)\)
*) a + 5 = b + 7
⇒ a = b + 7 - 5
⇒ a = b + 2
⇒ a > b (1)
*) b + 7 = c + 10
⇒ b = c + 10 - 7
⇒ b = c + 3
⇒ b > c (2)
Từ (1) và (2) ⇒ a > b > c
a+5=b+7=c+10
=>a+5=b+7 và b+7=c+10
=>a-b=2>0 và b-c=10-7=3>0
=>a>b và b>c
=>a>b>c