3x² + 3y² + 4xy - 2x + 2y + 2 = 0

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

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

Vì ( x + y )²\(\ge\)0 ; ( x - 1 )² \(\ge\)0 ; ( y + 1 )²\(\ge\)0 (\(\forall\)x;y )

=> 2 ( x + y )² + ( x - 1 )² + ( y + 1 )² \(\ge\)0\(\forall\)x;y

(*) xảy ra <=>\(\hept{\begin{cases}\left(x+y\right)^2=0\\\left(x-1\right)^2=0\\\left(y+1\right)^2=0\end{cases}}\)<=>\(\hept{\begin{cases}x+y=0\\x=1\\y=-1\end{cases}}\left(tm\right)\)

Vậy x = 1 ; y = - 1 tm đề bài

Vì \(ab+bc+ca=1\)

\(\Rightarrow a^2+1=a^2+ab+bc+ca\)

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

Tương tự ta có: \(b^2+1=\left(b+a\right)\left(b+c\right)\); \(c^2+1=\left(c+a\right)\left(c+b\right)\)

\(\Rightarrow A=\left(a+b\right)^2.\left(b+c\right)^2.\left(c+a\right)^2.\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\)

\(=\left(a+b\right)^2.\left(b+c\right)^2.\left(c+a\right)^2.\left(a+b\right)\left(b+c\right)\left(b+a\right)\left(b+c\right)\left(c+b\right)\left(c+a\right)\)

\(=\left(a+b\right)^4.\left(b+c\right)^4.\left(c+a\right)^4\)

mình cần bài 3.2 e ; 3.4 ; 3.5 ; 3.6

Bài 3.5 

Gọi 3 số nguyên liên tiếp lần lượt là: \(a-1\), \(a\), \(a+1\)\(\left(a\inℤ\right)\)

Tổng các lũy thừa bậc 3 của 3 số nguyên liên tiếp là: \(\left(a-1\right)^3+a^3+\left(a+1\right)^3\)

Ta có: \(\left(a-1\right)^3+a^3+\left(a+1\right)^3\)

\(=a^3-3a^2+3a-1+a^3+a^3+3a^2+3a+1\)

\(=3a^3+6a=3a\left(a^2+2\right)=3a\left(a^2-1+3\right)\)

\(=3a\left(a^2-1\right)+9a=3a\left(a-1\right)\left(a+1\right)+9a\)

Vì \(a\), \(a-1\), \(a+1\)là 3 số nguyên liên tiếp 

\(\Rightarrow a\left(a-1\right)\left(a+1\right)⋮3\)\(\Rightarrow3a\left(a-1\right)\left(a+1\right)⋮9\)

mà \(9a⋮9\)\(\Rightarrow3a\left(a-1\right)\left(a+1\right)+9a⋮9\)

hay \(\left(a-1\right)^3+a^3+\left(a+1\right)^3⋮9\)

Vậy tổng các lũy thừa bậc ba của 3 số nguyên liên tiếp chia hết cho 9

ĐK : \(x\ne0;\pm1\)

\(\frac{x+3}{x^2-1}-\frac{x+1}{x^2-x}=\frac{x+3}{\left(x-1\right)\left(x+1\right)}-\frac{x+1}{x\left(x-1\right)}\)

\(=\frac{x\left(x+3\right)}{x\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)\left(x+1\right)}{x\left(x-1\right)\left(x+1\right)}=\frac{x^2+3x-x^2+1}{x\left(x-1\right)\left(x+1\right)}\)

\(=\frac{3x+1}{x\left(x-1\right)\left(x+1\right)}\)