a: =-1/5x^5y^2

b: =-9/7xy^3

c: =7/12xy^2z

d: =2x^4

e: =3/4x^5y

f: =11x^2y^5+x^6

1: xy+x+y+1=0

=>x(y+1)+(y+1)=0

=>(x+1)(y+1)=0

=>\(\begin{cases}x+1=0\\ y+1=0\end{cases}\Rightarrow\begin{cases}x=-1\\ y=-1\end{cases}\)

2: xy+x+6=0

=>x(y+1)=-6

=>(x;y+1)∈{(1;-6);(-6;1);(-1;6);(6;-1);(2;-3);(-3;2);(-2;3);(3;-2)}

=>(x;y)∈{(1;-7);(-6;0);(-1;5);(6;-2);(2;-4);(-3;1);(-2;2);(3;-3)}

3: -xy-x-y-1=0

=>xy+x+y+1=0

=>x(y+1)+(y+1)=0

=>(x+1)(y+1)=0

=>\(\begin{cases}x+1=0\\ y+1=0\end{cases}\Rightarrow\begin{cases}x=-1\\ y=-1\end{cases}\)

4: xy-x-y+1=0

=>x(y-1)-(y-1)=0

=>(x-1)(y-1)=0

=>\(\begin{cases}x-1=0\\ y-1=0\end{cases}\Rightarrow\begin{cases}x=1\\ y=1\end{cases}\)

5: xy+2x+y+11=0

=>x(y+2)+y+2+9=0

=>x(y+2)+(y+2)=-9

=>(x+1)(y+2)=-9

=>(x+1;y+2)∈{(1;-9);(-9;1);(-1;9);(9;-1);(3;-3);(-3;3)}

=>(x;y)∈{(0;-11);(-10;-1);(-2;7);(8;-3);(2;-5);(-4;1)}

6: ĐKXĐ: x<>0

\(\frac{5}{x}+\frac{y}{4}=\frac18\)

=>\(\frac{20+xy}{4x}=\frac18\)

=>\(\frac{40+2xy}{8x}=\frac{x}{8x}\)

=>40+2xy=x

=>x-2xy=40

=>x(1-2y)=40

=>x(2y-1)=-40

mà 2y-1 lẻ(do y nguyên)

nên (x;2y-1)∈{(-40;1);(40;-1);(8;-5);(-8;5)}

=>(x;2y)∈{(-40;2);(40;0);(8;-4);(-8;6)}

=>(x;y)∈{(-40;1);(40;0);(8;-2);(-8;3)}

8: (x+2)(y-3)=-3

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

=>(x;y)∈{(-1;0);(-5;4);(-3;6);(1;2)}

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

bài 1: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

\(\dfrac{x}{x+2}-\dfrac{x}{x-2}\)

\(=\dfrac{x\left(x-2\right)-x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{x^2-2x-x^2-2x}{\left(x-2\right)\left(x+2\right)}=-\dfrac{4x}{x^2-4}\)

Bài 2:

1: \(x^2y^2-8-1\)

\(=x^2y^2-9\)

\(=\left(xy-3\right)\left(xy+3\right)\)

2: \(x^3y-2x^2y+xy-xy^3\)

\(=xy\cdot x^2-xy\cdot2x+xy\cdot1-xy\cdot y^2\)

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

\(=xy\left[\left(x-1\right)^2-y^2\right]\)

\(=xy\left(x-1-y\right)\left(x-1+y\right)\)

3: \(x^3-2x^2y+xy^2\)

\(=x\cdot x^2-x\cdot2xy+x\cdot y^2\)

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

4: \(x^2+2x-y^2+1\)

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

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

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

5: \(x^2+2x-4y^2+1\)

\(=\left(x^2+2x+1\right)-4y^2\)

\(=\left(x+1\right)^2-4y^2\)

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

6: \(x^2-6x-y^2+9\)

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

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

x, y, z thuộc gì thế bạn?

À mình quên, x,y,z ∈ Z nhé ! Giúp mình với

a) 6x² - 12x

= 6x(x - 2)

b) x² + 2x + 1 - y²

= (x² + 2x + 1) - y²

= (x + 1)² - y²

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

c) x + y + z + x² + xy + xz

= (x + x²) + (y + xy) + (z + xz)

= x(1 + x) + y(1 + x) + z(1 + x)

= (x + y + z)(x + 1)

d) xy + xz + y² + yz

= (xy + xz) + (y² + yz)

= x(y + z) + y(y + z)

= (x + y)(x + z)

e) x³ + x² + x + 1

= (x³ + x²) + (x + 1)

= x²(x + 1) + (x + 1)

= (x² + 1)(x + 1)

f) xy + y - 2x - 2

= (xy + y) - (2x + 2)

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

= (y - 2)(x + 1)

g) x³ + 3x - 3x² - 9

= (x³ - 3x²) + (3x - 9)

= x²(x - 3) + 3(x - 3)

= (x² + 3)(x - 3)

h) x² - y² - 2x - 2y

= (x² - y²) - (2x + 2y)

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

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

i) 7x² - 7xy - 5x = 5y

mk thấy con này sai sai ý

à câu í là :7x^2-7xy-5x+5y đấy bạn