giúp mình vs ạ

Theo đề bài ta có :

\(a+b+b+c+c+a=-3-5+10\)

\(\Rightarrow\)\(2a+2b+2c=2\)

\(\Rightarrow\)\(2\left(a+b+c\right)=2\)

\(\Rightarrow\)\(a+b+c=\frac{2}{2}=1\)

Do đó :

\(a=a+b+c-\left(b+c\right)=1-\left(-5\right)=6\)

\(b=a+b+c-\left(c+a\right)=1-10=-9\)

\(c=a+b+c-\left(-3\right)=1+3=4\)

Vậy \(a=6\)\(;\)\(b=-9\)và \(c=4\)

Chúc bạn học tốt 

Giải:

Vì $a\in Z^{+}$

$\Rightarrow 5^b=a^3+3a^2+5>a+3=5^c$

$\Rightarrow 5^b>5^c\Rightarrow b>c$

$\Rightarrow 5^b⋮5^c$

$\Rightarrow a^3+3a^2+5⋮a+3$

$\Rightarrow a^2\left(a+3\right)+5⋮a+3$

Mà $a^2\left(a+3\right)⋮a+3$

$\Rightarrow 5⋮a+3$

$\Rightarrow a+3\in Ư\left(5\right)$

$\Rightarrow a+3\in \left\{\pm 1;\pm 5\right\}\left(1\right)$

Do $a\in Z^{+}\Rightarrow a+3\ge 4\left(2\right)$

Từ $\left(1\right)$ và $\left(2\right)$

$\Rightarrow a+3=5$

$\Rightarrow a=5-3$

$\Rightarrow a=2$$(\ast )$

Thay $(\ast )$ vào biểu thức ta có:

$2^3+{3.2}^2+5=5^b\iff b=2$

$2+3=5^c\iff c=1$

Vậy: $\{\begin{array}{c}a=2\\ b=2\\ c=1\end{array}$

\(\dfrac{a}{b+c}=\dfrac{b}{c+a}=\dfrac{c}{a+b}\\ \Leftrightarrow\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{a+b}{c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\\ \Leftrightarrow\left\{{}\begin{matrix}b+c=2a\left(1\right)\\c+a=2b\left(2\right)\\a+b=2c\left(3\right)\end{matrix}\right.\\ \Leftrightarrow\left(1\right)-\left(2\right)=b-a=2a-2b\Leftrightarrow a-b=0\Leftrightarrow a=b\\ \left(2\right)-\left(3\right)=c-b=2b-2c\Leftrightarrow b-c=0\Leftrightarrow b=c\\ \left(3\right)-\left(1\right)=a-c=2c-2a\Leftrightarrow a-c=0\Leftrightarrow a=c\)

Vậy \(a=b=c\)

abc = 100a + 10b + c = 98a + 2a + 7b + 2b + b + 2c - c = (98a + 7b) + (2a + 2b + 2c) + (b - c) = 7(14a + b) + 2(a + b + c) + (b - c) chia hết cho 7.

Mà 7(14a + b) chia hết cho 7 và 2(a + b + c) chia hết cho 7 

⇒b - c chia hết cho 7 

Mà 0≤b - c < 7 

Vậy b - c = 0

\(a+b+c=7\Leftrightarrow a=7-b-c.\)

\(\Rightarrow abc=bc.\left(7-b-c\right)=7bc-bc\left(b-c\right)⋮7\)

Do 7bc chia hết cho 7 \(\Rightarrow bc\left(b-c\right)⋮7\)

a, b, c là các chữ số \(\Rightarrow1\le a,b,c\le9\left(a,b,c\in N\right)\)

Ta có: \(a+b+c=1\Rightarrow c\le\frac{1}{3}\)

vì vai trò a,b,c như nhau giả sử: \(c\ge a;c\ge b\)

\(\Rightarrow\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1}\ge\frac{a+b+c}{c^2+1}\ge\frac{9}{10}\)

Theo AM GM 3 số ta có:\(a+b+c\ge3\sqrt[3]{abc}\Rightarrow abc\le\frac{1}{27}\Leftrightarrow\frac{1}{9abc}\le3\)

\(\Rightarrow P\ge\frac{9}{10}+3=\frac{39}{10}\) Dấu bằng xảy ra khi \(a=b=c=\frac{1}{3}\)