a) Ta có: \(\left(a+b\right)^2=a^2+2ab+b^2=a^2-2ab+b^2+4ab=\left(a-b\right)^2+4ab^{\left(đpcm\right)}\)

b)Từ kết quá câu a),ta suy ra: \(\left(a-b\right)^2=\left(a+b\right)^2-4ab=9^2-4.20=81-80=1\)

\(\Rightarrow a-b=1\Rightarrow\left(a-b\right)^{2015}=1^{2015}=1\)

Vậy \(\left(a-b\right)^{2015}=1\)

(a+b)^2=(a-b)^2+4ab

(a+b)^2=a^2-2ab+b^2+4ab

(a+b)^2=a^2+2ab+b^2

(a+b)^2=(a+b)^2

b,(a+b)=81

suy ra (a+b)^2=81

(a-b)^2+4ab=81

(a-b)^2=81-4*20

(a-b)^2=81-80

(a-b)^2=1

suy ra (a-b)=1hoac (a-b)=-1

a<b suy ra a-b<0

suy ra a-b=-1

(a-b)^2015=(-1)^2015=-1

Ta có:

\(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

\(\Leftrightarrow a^2+b^2+c^2=ab+bc+ca\)

Ta lại có: 

\(a^2+b^2+c^2\ge ab+bc+ca\)

Dấu = xảy ra khi \(a=b=c\)

Thế vào N ta được

\(N=\frac{a^{2015}+b^{2015}+c^{2015}}{\left(a+b+c\right)^{2015}}=\frac{3a^{2015}}{3^{2015}.a^{2015}}=\frac{1}{a^{2014}}\)

\(A=-2\)

\(\Leftrightarrow5x^2+y^2+4xy-6x-2y=-2\)

\(\Leftrightarrow4x^2+x^2+y^2+4xy-4x-2x-2y+1+1=0\)

\(\Leftrightarrow\left(4x^2+4xy+y^2\right)-2\left(2x+y\right)+1+\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow\left(2x+y\right)^2-2\left(2x+y\right)+1+\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(2x+y-1\right)^2+\left(x-1\right)^2=0\)(1) 

Mà \(\left(2x+y-1\right)^2+\left(x-1\right)^2\ge0\)nên (1) xảy ra

\(\Leftrightarrow\hept{\begin{cases}2x+y-1=0\\x-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-1\\x=1\end{cases}}\)

\(\Rightarrow B=1^{2015}.\left(-1\right)^{2016}-1^{2016}.\left(-1\right)^{2017}+2014\)

\(=1+1+2014=2016\)

Giúp mình với đang cần gấp!!

Ta có: A = -2

=> 5x² + y² + 4xy - 6x - 2y = -2

=> 5x² + y² + 4xy - 6x - 2y + 2 = 0

=> (4x² + 4xy + y²) - 2(2x + y) + 1 + (x² - 2x + 1) = 0

=> (2x + y)² - 2(2x + y) + 1 + (x - 1)² = 0

=> (2x + y - 1)² + (x - 1)² = 0

       <=> \(\hept{\begin{cases}2x+y-1=0\\x-1=0\end{cases}}\)

        <=> \(\hept{\begin{cases}y=1-2x\\x=1\end{cases}}\)

        <=> \(\hept{\begin{cases}y=1-2.1=-1\\x=1\end{cases}}\)

Với x = 1; y = -1 => B = 1²⁰¹⁵.(-1)²⁰¹⁶ - 1²⁰¹⁶.(-1)²⁰¹⁷ + 2014

                                    = 1 + 1 + 2014 = 2016