a) |x + 25| + |-y + 5| =0

=> |x + 25| = 0 hoặc |-y + 5| = 0

Từ đó bạn cứ bỏ giá trị tuyệt đối rồi tính nha! Mấy bài khác cũng vậy

a) Để y nguyên thì \(6x-4⋮2x+3\)

\(\Leftrightarrow-13⋮2x+3\)

\(\Leftrightarrow2x+3\in\left\{1;-1;13;-13\right\}\)

\(\Leftrightarrow2x\in\left\{-2;-4;10;-16\right\}\)

hay \(x\in\left\{-1;-2;5;-8\right\}\)

\(a,\left\{{}\begin{matrix}\left|x-3y\right|\ge0\\\left|y+4\right|\ge0\end{matrix}\right.\Rightarrow VT\ge0\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-3y=0\\y+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3y=-12\\y=-4\end{matrix}\right.\)

\(b,Sửa:\left|x-y-5\right|+\left(y+3\right)^2=0\\ \left\{{}\begin{matrix}\left|x-y-5\right|\ge0\\\left(y+3\right)^2\ge0\end{matrix}\right.\Rightarrow VT\ge0\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-y-5=0\\y+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y+5=2\\y=-3\end{matrix}\right.\)

\(c,\left\{{}\begin{matrix}\left|x+y-1\right|\ge0\\\left(y-2\right)^4\ge0\end{matrix}\right.\Rightarrow VT\ge0\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x+y-1=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1-y=-1\\y=2\end{matrix}\right.\)

\(d,\left\{{}\begin{matrix}\left|x+3y-1\right|\ge0\\3\left|y+2\right|\ge0\end{matrix}\right.\Rightarrow VT\ge0\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x+3y-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1-3y=7\\y=-2\end{matrix}\right.\)

\(e,Sửa:\left|2021-x\right|+\left|2y-2022\right|=0\\ \left\{{}\begin{matrix}\left|2021-x\right|\ge0\\\left|2y-2022\right|\ge0\end{matrix}\right.\Rightarrow VT\ge0\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}2021-x=0\\2y-2022=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2021\\y=1011\end{matrix}\right.\)

a: (x-2)(y-3)=5

=>\(\left(x-2\right)\cdot\left(y-3\right)=1\cdot5=5\cdot1=\left(-1\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-1\right)\)

=>\(\left(x-2;y-3\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(3;8\right);\left(7;4\right);\left(1;-2\right);\left(-3;2\right)\right\}\)

b: (2x-1)*(y-4)=-11

=>\(\left(2x-1\right)\cdot\left(y-4\right)=1\cdot\left(-11\right)=\left(-11\right)\cdot1=\left(-1\right)\cdot11=11\cdot\left(-1\right)\)

=>\(\left(2x-1;y-4\right)\in\left\{\left(1;-11\right);\left(-11;1\right);\left(-1;11\right);\left(11;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(1;-7\right);\left(-5;5\right);\left(0;15\right);\left(6;3\right)\right\}\)

c: xy-2x+y=3

=>\(x\left(y-2\right)+y-2=1\)

=>\(\left(x+1\right)\left(y-2\right)=1\)

=>\(\left(x+1\right)\cdot\left(y-2\right)=1\cdot1=\left(-1\right)\cdot\left(-1\right)\)

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

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

a) ( x - 1 ) . ( y + 2 ) = 7

Lập bảng ta có :

b) x . ( y - 3 ) = -12

Lập bảng ta có :

c) xy - 3x - y = 0

x . ( y - 3 ) - y = 0

x . ( y - 3 ) - y + 3 = 3

x . ( y - 3 ) - ( y - 3 ) = 3

( x - 1 ) . ( y - 3 ) = 3

Lập bảng ta có :

d) xy + 2x + 2y = -16

x . ( y + 2 ) + 2y = -16

x . ( y + 2 ) + 2y + 4 = -12

x . ( y + 2 ) + 2 . ( y + 2 ) = -12

( x + 2 ) . ( y + 2 ) = -12

Lập bảng ta có :

Ta có : (x - 1).(y + 2) = 7

=> (x - 1) và y + 2 thuộc Ư(7) = {-7;-1;1;7}

Ta có bảng : 

Vậy có 4 cặp x;y thoả mãn : (-6,-3) ; (0 , -9) ; (2 , 5) ; (8, -1) 

Ta có : x(y - 3) = -12 

=> x ; y - 3 thuộc Ư(-12) = {-12;-6;-4;-3;-2;-1;1;2;3;4;6;12}

Ta có bảng tương tự câu vừa nãy nhé  

a.

$xy=-21=7.(-3)=(-7).3=3.(-7)=(-3).7=21.(-1)=(-21).1=(-1).21=1(-21)$

Do đó $(x,y)=(7,-3); (-7,3); (3,-7); (-3,7); (21,-1); (-21,1); (-1,21); (1,-21)$

b.

$(x+5)(y-3)=14=1.14=14.1=(-14)(-1)=(-1)(-14)=2.7=7.2=(-2)(-7)=(-7)(-2)$

Do đó:

$(x+5,y-3)=(1,14); (14,1); (-14,-1); (-1,-14); (2,7); (7,2); (-2,-7); (-7,-2)$

Đến đây thì đơn giản rồi.

c.

$x(y-2)=-19$, bạn làm tương tự

d. Tương tự

5x²+2y+y²-4x-40=0

△=(-4)²-4.5.(2y+y²-40)

△=16-40y-20y²+800

△=-(784+40y+20y²)

△=-(32y+8y+16y²+4y²+16+4+764)

△=-[(4y+4)²+(2y+2)²+764]<0

=>PHƯƠNG TRÌNH VÔ NGHIỆM.

\(\Leftrightarrow2x^2-x+1=xy+2y\)

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

\(\Leftrightarrow y=\dfrac{2x^2-x+1}{x+2}=2x-5+\dfrac{11}{x+2}\)

Do y nguyên \(\Rightarrow\dfrac{11}{x+2}\) nguyên \(\Rightarrow x+2=Ư\left(11\right)\)

Mà x nguyên dương \(\Rightarrow x+2\ge3\Rightarrow x+2=11\Rightarrow x=9\)

\(\Rightarrow y=14\)

Vậy \(\left(x;y\right)=\left(9;14\right)\)