`@` `\text {Ans}`

`\downarrow`

\((x+y)(x-y)+(xy^4-x^3y^2) \div (xy^2) \)

`= x(x-y) + y(x-y) + xy^4 \div xy^2 - x^3y^2 \div xy^2`

`= x^2 - xy + xy - y^2 + y^2 - x^2`

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

`= 0`

2)     => X/3 = Y/4

(2X^2 + Y^2)/(2.3^2 + 4^2) =  136/34 = 4

2X^2 = 4.18 = 72 => x  = 6

y^2 = 4.16 = 64 => y = 8

5)  (a+2b-3c)/(2+2.3 - 3.4) =  20/4 = 5

a = 10

2b = 30 => b = 15

3c = 60 => c = 20 

kho hieu qua 

18 voi 16 lay dau ra vay 

a) \(2xy-y^2-6x+4y=7\)

\(\Leftrightarrow2xy-6x-y^2+3y+y-3=4\)

\(\Leftrightarrow\left(2x-y+1\right)\left(y-3\right)=4\)

Tới đây bạn xét bảng giá trị thu được nghiệm \(\left(x,y\right)\).

b) \(x^2+y^2-x⋮xy\Rightarrow x^2+y^2-x⋮x\Rightarrow y^2⋮x\).

Đặt \(y^2=kx,\left(k\inℤ\right),d=\left(x,k\right)\).

\(x^2+\left(kx\right)^2-x⋮xy\Rightarrow x+k^2x-1⋮y\).

suy ra \(x+k^2x-1⋮d\Rightarrow1⋮d\Rightarrow d=1\).

Do đó \(kx=y^2\)mà \(\left(k,x\right)=1\)nên \(x\)là số chính phương. 

 2(x-y)² -y(x-y)² +xy²-x²y= 2(x-y)²-y(x-y)²+(xy^2-x^2y)=2(x-y)²-y(x-y)²+xy(x-y)=(x-y)\(\left[2\left(x-y\right)-y\left(x-y\right)+xy\right]\)=(x-y)(2x-2y-xy+y²+xy)=(x-y)(2x-2y+y²)

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

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

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

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

(x-2)(y+1)=-4

⇔xy+x-2y-2=-4

⇔-31+x-2y-2=-4

⇔x-2y=4+2+31

⇔x-2y=39

⇔x=39+2y

⇔y=x-39 / 2

1)ta có x.y=23=1.23=(-1)(-23)⇒các cặp (x,y)là(1,23);(23,1);(-1,-23);(-23;-1)

vậy......

2) ta có:(x-1 ).(y+2)= -4=-1.4=1.(-4)=-2.2=2.(-2)

⇒th1:x-1=-1                                           y+2=4

            x=-1+1=0                                       y=4-2=2

th2:x-1=1                                           y+2=-4

         x=1+1=2                                       y=-4-2=-6

th3:x-1=-2                                          y+2=2

         x=-2+1=-1                                     y=2-2=0

th4:x-1=2                                         y+2=-2

         x=2+1=3                                     y=-2-2=-4

vậy các cặp (x,y)là(0,2);(2,-6);(-1,0);(3,-4)

(x-2)(y+1)=-4

⇔xy+x-2y-2=-4

⇔-31+x-2y-2=-4

⇔x-2y=4+2+31

⇔x-2y=39

⇔x=39+2y

⇔y=x-39 / 2☺   tick mình với nhé

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

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

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

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

x(x-y)² -y(x-y)²+xy²-x²y=x(x-y)² -y(x-y)²+(xy²-x²y)=x(x-y)² -y(x-y)²+xy(x-y)=\(\left(x-y\right)\left[x\left(x-y\right)-y\left(x-y\right)+xy\right]\)=\(\left(x-y\right)\left[\left(x-y\right)^2+xy\right]\)

giúp mình vs mn ơi