.

a: |x|-|y|=60

=>6|y|-|y|=60

=>5|y|=60

=>|y|=12

=>y=12 hoặc y=-12

y=12

=>\(x=6y=6\cdot12=72\)

y=-12

=>\(x=6y=6\cdot\left(-12\right)=-72\)

b: |x|+|y|<2

mà |x|+|y|>=0 và x,y nguyên

nên |x|+|y|∈{0;1}

=>(|x|;|y|)∈{(0;1);(0;0);(1;0)}

=>(x;y)∈{(0;0);(0;1);(0;-1);(1;0);(-1;0)}

d: xy+5x-7y=35

=>x(y+5)-7y-35=0

=>(x-7)(y+5)=0

=>x=7 và y=-5

e: xy+2x-3y=9

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

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

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

=>(x;y)∈{(4;1);(6;-1);(2;-5);(0;-3)}

f: xy-2x+5y-12=0

=>x(y-2)+5y-10=2

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

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

=>(x;y)∈{(-4;4);(-3;3);(-6;0);(-7;1)}

a) Có \(\left|x-3y\right|^5\ge0\);\(\left|y+4\right|\ge0\)

\(\rightarrow\left|x-3y\right|^5+\left|y+4\right|\ge0\)

mà \(\left|x-3y\right|^5+\left|y+4\right|=0\)

\(\rightarrow\left\{{}\begin{matrix}\left|x-3y\right|^5=0\\\left|y+4\right|=0\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}x=3y\\y=-4\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}x=-12\\y=-4\end{matrix}\right.\)

b) Tương tự câu a, ta có:

\(\left\{{}\begin{matrix}\left|x-y-5\right|=0\\\left(y-3\right)^4=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=y+5\\y=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=3\end{matrix}\right.\)

c. Tương tự, ta có:

\(\left\{{}\begin{matrix}\left|x+3y-1\right|=0\\\left|y+2\right|=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1-3y\\y=-2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=-2\end{matrix}\right.\)

a. \(\left|x-3y\right|^5\ge0,\left|y+4\right|\ge0\Rightarrow\left|x-3y\right|^5+\left|y+4\right|\ge0\) \(\Rightarrow VT\ge VP\)

Dấu bằng xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left|x-3y\right|^5=0\\\left|y+4\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3y\\y=-4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-12\\y=-4\end{matrix}\right.\) Vậy...

b. \(\left|x-y-5\right|\ge0,\left(y-3\right)^4\ge0\Rightarrow\left|x-y-5\right|+\left(y-3\right)^4\ge0\) \(\Rightarrow VT\ge VP\)

Dấu bằng xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left|x-y-5\right|=0\\\left(y-3\right)^4=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=y+5\\y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=3\end{matrix}\right.\) Vậy ...

c. \(\left|x+3y-1\right|\ge0,3\cdot\left|y+2\right|\ge0\Rightarrow\left|x+3y-1\right|+3\left|y+2\right|\ge0\) \(\Rightarrow VT\ge VP\) Dấu bằng xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left|x+3y-1\right|=0\\3\left|y+2\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1-3y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1-\left(-2\right)\cdot3=7\\y=-2\end{matrix}\right.\) Vậy...

a: Ta có: \(\left(x^2+2\right)\left(x-4\right)-\left(x+2\right)^3=-16\)

\(\Leftrightarrow x^3-4x^2+2x-8-x^3-6x^2-12x-8=-16\)

\(\Leftrightarrow-10x^2-10x=0\)

\(\Leftrightarrow-10x\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

c: Ta có: \(x^3+3x^2+3x+28=0\)

\(\Leftrightarrow\left(x+1\right)^3=-27\)

\(\Leftrightarrow x+1=-3\)

hay x=-4

a: |2x+1|+|y-1|=4

mà 2x+1 lẻ

nên (|2x+1|;|y-1|)∈{(1;3);(3;1)}

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

=>(2x;y)∈{(0;4);(2;2);(-2;-2);(-4;0);(0;-2);(-4;2);(-2;4);(2;0)}

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

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

=>\(3-\left|2x-3\right|\ge0\)

=>|2x-3|<=3

mà 2x-3 lẻ

nên |2x-3|∈{1;3}

TH1: |2x-3|=1

=>\(y^2=3-1=2\)

mà y nguyên

nên y∈∅

TH2: |2x-3|=3

=>\(y^2=3-3=0\)

=>y=0(nhận)

|2x-3|=3

=>2x-3=3 hoặc 2x-3=-3

=>2x=6 hoặc 2x=0

=>x=3(nhận) hoặc x=0(nhận)

c: (x-3)(y-5)=-7

=>(x-3;y-5)∈{(1;-7);(-1;7);(7;-1);(-7;1)}

=>(x;y)∈{(4;-2);(2;12);(10;4);(-4;6)}

a: |2x+1|+|y-1|=4

mà 2x+1 lẻ

nên (|2x+1|;|y-1|)∈{(1;3);(3;1)}

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

=>(2x;y)∈{(0;4);(2;2);(-2;-2);(-4;0);(0;-2);(-4;2);(-2;4);(2;0)}

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

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

=>\(3-\left|2x-3\right|\ge0\)

=>|2x-3|<=3

mà 2x-3 lẻ

nên |2x-3|∈{1;3}

TH1: |2x-3|=1

=>\(y^2=3-1=2\)

mà y nguyên

nên y∈∅

TH2: |2x-3|=3

=>\(y^2=3-3=0\)

=>y=0(nhận)

|2x-3|=3

=>2x-3=3 hoặc 2x-3=-3

=>2x=6 hoặc 2x=0

=>x=3(nhận) hoặc x=0(nhận)

c: (x-3)(y-5)=-7

=>(x-3;y-5)∈{(1;-7);(-1;7);(7;-1);(-7;1)}

=>(x;y)∈{(4;-2);(2;12);(10;4);(-4;6)}

\(2x^2+2y^2+3x-6y=5xy-7\)

\(\Leftrightarrow2x^2+2y^2+3x-6y-5xy=-7\)

\(\Leftrightarrow2x^2-4xy+2y^2-xy+3x-6y=-7\)

\(\Leftrightarrow2x\left(x-2y\right)-y\left(x-2y\right)+3\left(x-2y\right)=-7\)

\(\Leftrightarrow\left(2x-y+3\right)\left(x-2y\right)=-7\)

vì x,y nguyên nên \(\hept{\begin{cases}2x-y+3\\x-2y\end{cases}\in Z}\)

Ta có : -7 = ( -7 ) . 1 = (-1 ) . 7

Tới đây bạn tự làm nhé