Với [x>1x<−1] ta có: x3<x3+2x2+3x+2<(x+1)3⇒x3<y3<(x+1)3 (không xảy ra)
Từ đây suy ra −1≤x≤1
Mà x∈Z⇒x∈{−1;0;1}
∙ Với x=−1⇒y=0
∙ Với x=0⇒y=2√3 (không thỏa mãn)
∙ Với x=1⇒y=2
Vậy phương trình có 2 nghiệm nguyên (x;y) là (−1;0) và (1;2) 

x=-1,y=0

a) \(P=3\left(x^2+2xy+y^2\right)-2\left(x+y\right)-100\)

\(P=3\left(x+y\right)^2-2.5-100\)

\(P=3.5^2-110\)

\(P=-35\)

b) \(Q=\left[x^3+y^3+3xy\left(x+y\right)\right]-2\left(x^2+2xy+y^2\right)+3.5+10\)

\(Q=\left(x+y\right)^3-2\left(x+y\right)^2+25\)

\(Q=5^3-2.5^2+25\)

\(Q=100\)

11: \(2x^2-12xy+18y^2\)

\(=2\left(x^2-6xy+9y^2\right)\)

\(=2\left(x-3y\right)^2\)

12: \(\left(x^2+x\right)^2+3\left(x^2+x\right)+2\)

\(=\left(x^2+x+2\right)\left(x^2+x+1\right)\)

\(x^3+y^3-2x^2-2y^2+3xy\left(x+y\right)-4xy+3\left(x+y\right)+10=\left[x^3+y^3+3xy\left(x+y\right)\right]-2\left(x^2+2xy+y^2\right)+3\left(x+y\right)+10=\left(x+y\right)^3-2\left(x+y\right)^2+3\left(x+y\right)+10=5^3-2.5^2+3.5+10=100\)

\(x^3+y^3+3\left(x^2+y^2\right)+4\left(x+y\right)+4=0\)

\(\Leftrightarrow\left(x+y\right)^3-3xy\left(x+y\right)+3\left(x+y\right)^2-6xy+4\left(x+y\right)+4=0\)

\(\Leftrightarrow\left(x+y+2\right)\left(\left(x+y\right)^2+x+y+2\right)-3xy\left(x+y+2\right)=0\)

\(\Leftrightarrow\left(x+y+2\right)\left(x^2+y^2+2xy+x+y+2-3xy\right)=0\)

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

\(\Leftrightarrow x+y+2=0\)

\(\Leftrightarrow x+y=-2\)

\(M=\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}=\frac{4}{-2}=-2\)

Dấu \(=\)khi \(x=y=-1\).

a) \(\left(x-5\right)^2=\left(3+2x\right)^2\)

\(\Rightarrow\left(3+2x\right)^2-\left(x-5\right)^2=0\)

\(\Rightarrow\left(3+2x+x-5\right)\left(3+2x-x+5\right)=0\)

\(\Rightarrow\left(3x-2\right)\left(x+8\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}3x-2=0\\x+8=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-8\end{matrix}\right.\)

b) \(27x^3-54x^2+36x=9\)

\(\Rightarrow27x^3-54x^2+36x-9=0\)

\(\Rightarrow27x^3-54x^2+36x-8+8-9=0\)

\(\Rightarrow\left(3x-2\right)^3-1=0\)

\(\Rightarrow\left(3x-2-1\right)\left[\left(3x-2\right)^2+3x-2+1\right]=0\)

\(\Rightarrow\left(3x-3\right)\left[\left(3x-2\right)^2+3x-2+\dfrac{1}{4}-\dfrac{1}{4}+1\right]=0\)

\(\Rightarrow\left(3x-3\right)\left[\left(3x-2+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]=0\)

\(\Rightarrow\left(3x-3\right)\left[\left(3x-\dfrac{3}{2}\right)^2+\dfrac{3}{4}\right]=0\left(1\right)\)

mà \(\left(3x-\dfrac{3}{2}\right)^2+\dfrac{3}{4}>0,\forall x\)

\(\left(1\right)\Rightarrow3x-3=0\Rightarrow3x=3\Rightarrow x=1\)

(\(x-5\))² = (3 +2\(x\))² ⇒ \(\left[{}\begin{matrix}x-5=3+2x\\x-5=-3-2x\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=-8\\x=\dfrac{2}{3}\end{matrix}\right.\) vậy \(x\in\){-8; \(\dfrac{2}{3}\)}

  27\(x^3\) - 54\(x^2\) + 36\(x\) = 9

27\(x^3\) - 54\(x^2\) + 36\(x\) - 8 = 1

(3\(x\) - 2)³ = 1 ⇒ 3\(x\) - 2 = 1 ⇒ \(x\) = 1

1

`x^2-2y^2+2/3x^2y^3+B=2x^2+y^2+2/3x^2y^3`

`=>B=2x^2+y^2+2/3x^2y^3-x^2+2y^2-2/3x^2y^3`

`=>B=(2x^2-x^2)+(y^2+2y^2)+(2/3x^2y^3-2/3x^2y^3)`

`=>B=x^2+3y^2`

Thay `x=1 ; y=[-1]/3` vào `B` có:

   `B=1^2+3.([-1]/3)^2=1+3 . 1/9=1+1/3=4/3`

`x^2 - 2y^2 + 2/3x^2y^3 + B = 2x^2 + y^2 + 2/3x^2y^3`

`=> B  = 2x^2 + y^2 + 2/3x^2y^3` `- (x^2 - 2y^2 + 2/3x^2y^3)`

         `= 2x^2 + y^2 + 2/3x^2y^3 - x^2 + 2y^2 - 2/3x^2y^3`

         `= ( 2x^2 - x^2 ) + ( y^2 + 2y^2 ) + ( 2/3x^2y^3 - 2/3x^2y^3 )`

         `= x^2 + 3y^2`

Thay `x=1 ; y=-1/3` vào `B` ta có `:`

`B = 1^2 + 3 . ( -1/3 )^2`

   `= 1 + 1/3`

   `= 4/3`