Ta có (1)  ⇔ x 4 + x 2 + 20 = y 2 + y

Ta thấy:  x 4 + x 2 < x 4 + x 2 + 20 ≤ x 4 + x 2 + 20 + 8 x 2 ⇔ x 2 ( x 2 + 1 ) < y ( y + 1 ) ≤ ( x 2 + 4 ) ( x 2 + 5 )

Vì x, y ∈ Z nên ta xét các trường hợp sau

+ TH1.  y ( y + 1 ) = ( x 2 + 1 ) ( x 2 + 2 ) ⇔ x 4 + x 2 + 20 = x 4 + 3 x 2 + 2 ⇔ 2 x 2 = 18 ⇔ x 2 = 9 ⇔ x = ± 3

Với  x 2 = 9   ⇒ y 2 + y = 9 2 + 9 + 20 ⇔ y 2 + y − 110 = 0 ⇔ y = 10 ; y = − 11 ( t . m )

+ TH2  y ( y + 1 ) = ( x 2 + 2 ) ( x 2 + 3 ) ⇔ x 4 + x 2 + 20 = x 4 + 5 x 2 + 6 ⇔ 4 x 2 = 14 ⇔ x 2 = 7 2   ( l o ạ i )

+ TH3: y ( y + 1 ) = ( x 2 + 3 ) ( x 2 + 4 ) ⇔ 6 x 2 = 8 ⇔ x 2 = 4 3   ( l o ạ i )

+ TH4:  y ( y + 1 ) = ( x 2 + 4 ) ( x 2 + 5 ) ⇔ 8 x 2 = 0 ⇔ x 2 = 0 ⇔ x = 0

Với  x 2 = 0  ta có  y 2 + y = 20 ⇔ y 2 + y − 20 = 0 ⇔ y = − 5 ; y = 4

Vậy PT đã cho có nghiệm nguyên (x;y) là :

(3;10), (3;-11), (-3; 10), (-3;-11), (0; -5), (0;4).

\(\Leftrightarrow\)\(4y^2+12y=4x^4+4x^2+72\)

\(\Leftrightarrow\left(2y+3\right)^2=\left(2x^2+1\right)^2+80\)

\(\Leftrightarrow\left(2y+3\right)^2-\left(2x^2+1\right)^2=80\)

\(\Leftrightarrow\left(2y+3-2x^2-1\right)\left(2y+3+2x^2+1\right)=80\)

\(\Leftrightarrow\left(y-x^2+1\right)\left(y+x^2+2\right)=20\)

Do \(x,y\in Z\) => \(y+1-x^2;y+x^2+2\in Z\)

=>\(y+1-x^2;y+x^2+2\inƯ\left(20\right)\)

Kẻ bảng làm nốt nha.

Bài 1:

xy+2x-3y=1

=>x(y+2)-3y-6=1-6

=>x(y+2)-3(y+2)=-5

=>(x-3)(y+2)=-5

=>(x-3;y+2)∈{(1;-5);(-5;1);(-1;5);(5;-1)}

=>(x;y)∈{(4;-7);(-2;-1);(2;3);(8;-3)}

Bài 1:

xy+2x-3y=1

=>x(y+2)-3y-6=1-6

=>x(y+2)-3(y+2)=-5

=>(x-3)(y+2)=-5

=>(x-3;y+2)∈{(1;-5);(-5;1);(-1;5);(5;-1)}

=>(x;y)∈{(4;-7);(-2;-1);(2;3);(8;-3)}

4:

(x+1)(y-2)=5

=>\(\left(x+1;y-2\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(0;7\right);\left(4;3\right);\left(-2;-3\right);\left(-6;1\right)\right\}\)

1×1=3

4: \(\left(x^2-1\right)\left(x^2-6\right)<0\)

TH1: \(x^2-1>0\) và \(x^2-6<0\)

=>1<\(x^2\) <6

mà x là số nguyên

nên x∈{2;-2}

TH2: \(x^2-1<0\) và \(x^2-6>0\)

=>\(x^2<1\) và \(x^2>6\)

mà 6>1

nên x∈∅

Bài 2:Sửa đề: Tìm x,y nguyên

(2x+1)(5-y)=6

=>(2x+1)(y-5)=-6

mà 2x+1 lẻ

nên (2x+1;y-5)∈{(1;-6);(-1;6);(3;-2);(-3;2)}

=>(2x;y)∈{(0;-1);(-2;11);(2;3);(-4;7)}

=>(x;y)∈{(0;-1);(-1;11);(1;3);(-2;7)}

a) x = 1                

b) x = 5

a: \(x^2+y^2(x-y+1)-\left(x-1\right)\cdot y=22\)

=>\(x^2 + y^2(x - y + 1) - xy + y = 22\)

=>\(x^2-xy+y^2(x-y+1)+y=22\)

=>\(x(x-y+1)-x+y^2(x-y+1)+y=22\)

\(\iff(x-y+1)(x+y^2)-(x-y)=22\)

\(\iff(x-y+1)(x+y^2)-(x-y+1)+1=22\)

=>\((x-y+1)(x+y^2-1)=21\)

TH1: \(\begin{cases}x-y+1=1\\ x+y^2-1=21\end{cases}\)

=>\(\begin{cases}x=y\\ y+y^2-1=21\end{cases}\iff y^2+y-22=0\)

=>\(y^2+y+\frac14-\frac{89}{4}=0\)

=>\(\left(y+\frac12\right)^2=\frac{89}{4}\)

=>\(\left[\begin{array}{l}y+\frac12=\frac{\sqrt{89}}{2}\\ y+\frac12=-\frac{\sqrt{89}}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}y=\frac{\sqrt{89}-1}{2}\left(loại\right)\\ y=\frac{-\sqrt{89}-1}{2}\left(loại\right)\end{array}\right.\)

=>Loại

TH2: \(\begin{cases}x-y+1=21\\ x+y^2-1=1\end{cases}\iff\begin{cases}x=y+20\\ (y+20)+y^2-1=1\end{cases}\iff y^2+y+18=0\)

=>\(y^2+y+\frac14+\frac{71}{4}=0\)

=>\(\left(y+\frac12\right)^2+\frac{71}{4}=0\) (vô lý)

=>Loại

TH3: \(\begin{cases}x-y+1=3\\ x+y^2-1=7\end{cases}\iff\begin{cases}x=y+2\\ (y+2)+y^2-1=7\end{cases}\iff y^2+y-6=0\)

=>(y+3)(y-2)=0

=>y=-3 hoặc y=2

Nếu y=-3 thì x-y+1=3

=>x-y=2

=>x-(-3)=2

=>x+3=2

=>x=-1

Nếu y=2 thì x-y+1=3

=>x-2+1=3

=>x-1=3

=>x=4

TH4: \(\begin{cases}x-y+1=7\\ x+y^2-1=3\end{cases}\iff\begin{cases}x=y+6\\ (y+6)+y^2-1=3\end{cases}\iff y^2+y+2=0\)

=>y∈∅

=>Loại

a) \(6xy+4x-9y-7=0\)

  \(\Leftrightarrow2x.\left(3y+2\right)-9y-6-1=0\)

\(\Leftrightarrow2x.\left(3y+x\right)-3.\left(3y+2\right)=1\)

\(\Leftrightarrow\left(2x-3\right).\left(3y+2\right)=1\)

Mà \(x,y\in Z\Rightarrow2x-3;3y+2\in Z\)

Tự làm típ

\(A=x^3+y^3+xy\)

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

\(A=x^2-xy+y^2+xy\)( vì \(x+y=1\))

\(A=x^2+y^2\)

Áp dụng bất đẳng thức Bunhiakovxky ta có :

\(\left(1^2+1^2\right)\left(x^2+y^2\right)\ge\left(x\cdot1+y\cdot1\right)^2=\left(x+y\right)^2=1\)

\(\Leftrightarrow2\left(x^2+y^2\right)\ge1\)

\(\Leftrightarrow x^2+y^2\ge\frac{1}{2}\)

Hay \(x^3+y^3+xy\ge\frac{1}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=\frac{1}{2}\)