a, \(a^2+b^2=\left(a+b\right)^2-2ab\)

Thay a+b=s; ab vào đa thức trên ta được:

\(\left(a+b\right)^2-2ab=s^2-2p\)

b, \(a^3+b^3=\left(a+b\right)^3+3a^2b-3ab^2\)

\(=\left(a+b\right)^3-3ab.\left(a+b\right)\)

Thay a+b=s; ab=p Ta được:

\(\left(a+b\right)^3-3ab.\left(a+b\right)=s^3-3sp\)

c, \(a^4+b^4=\left(x^2+y^2\right)^2-2x^2y^2\)

\(=\left(s^2-2p\right)^2-2p^2=s^4-4s^2p+2p^2\)

CHÚC HỌC TỐT!!

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

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

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

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

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

          \(S=-2\)

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

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

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