Vì \(a+b+c=2016\Rightarrow a=2016-\left(b+c\right);b=2016-\left(a+c\right);c=2016-\left(a+b\right)\)

Ta có:\(S=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\) 

           \(S=\frac{2016-\left(b+c\right)}{b+c}+\frac{2016-\left(a+c\right)}{a+c}+\frac{2016-\left(a+b\right)}{a+b}\)

           \(S=\frac{2016}{b+c}-1+\frac{2016}{a+c}-1+\frac{2016}{a+b}-1\)

           \(S=2016.\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)-3\)

           \(S=2016.\frac{1}{2016}-3\)

          \(S=-2\)

a) a² + b² = a² + b² + 2ab - 2ab = (a + b)² - 2ab = s² - 2p

b) a³ + b³ = (a + b)(a² - ab + b²) = (a + b)(a² + 2ab + b² - 3ab) = (a + b).[(a + b)² - 3ab] = s.(s² - 3p) = s³ - 3ps

c) a⁴ + b⁴ = a⁴ + b⁴ + 4a²b² - 4a²b² = (a² + b²)² - 4(ab)² = (s² - 2p)² - 4p²

= (s² - 2p - 2p)(s² - 2p + 2p) = s².(s² - 4p) = s⁴ - 4ps²

bn ơi a²+b² ko = a²+2ab+b²

\(1,\\ a,X=\left\{3;4\right\};\left\{2;3;4\right\};\left\{1;2;3;4\right\}\\ b,X=\left\{2;4\right\}\\ X=\left\{2\right\}\\ X=\left\{4\right\}\\ X=\varnothing\)

\(2,\\ a,A=\left\{-3;-2;0;1;2;3;4\right\}\\ B=\left\{0;1;2;3;4;6;9;10\right\}\\ b,A=\left\{1;2;3;4;5\right\}\\ B=\left\{1;2;3;6;9\right\}\)

TC: a/b=b/c

=>

(a+b+c)/(a+b)+(a+b+c)/(b+c)+(a+b+c)/(c+a)=(a+b+c)/90

1+c/(a+b)+1+a/(b+c)+1+b/(c+a)=2007/90

a/b+c+b/a+c+c/a+b=2007/90-3

                             =19,3

Câu 2: 

a: \(\Leftrightarrow x+2\in\left\{3;9\right\}\)

hay \(x\in\left\{1;7\right\}\)

Thi à :)?

\(a+b=132\)\(\left(1\right)\)

\(a-b=4\) \(\left(2\right)\)

lấy \(\left(1\right)-\left(2\right)\)ta có 

\(a+b-a+b=132-4\)

<=> \(2b=128\)

<=> \(b=64\)

=> \(a=4+b=4+64=68\)

\(a+b=132\)

\(a-b=4\)

\(\Rightarrow a=\left(132+4\right)\text{ : }2\)

\(\Rightarrow a=136\text{ : }2\)

\(\Rightarrow a=68\)

\(\Rightarrow b=68-4\)

\(\Rightarrow b=64\)

\(\text{Vậy : }a=68\)

            \(b=64\)

a+b=132

a−b=4

⇒a=(132+4) : 2

⇒a=136 : 2

⇒a=68

⇒b=68−4

⇒b=64

Vậy : a=68

            b=64