\(\left\{{}\begin{matrix}x-y=5\\x^2-y^2=15\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x-y=5\\x+y=3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=-1\end{matrix}\right.\)

\(x^3-y^3=4^3-\left(-1\right)^3=65\)

x²-y²=15 

⇒(x-y)(x+y)=15

⇒ 5(x+y)=15

⇒ x+y=3

x-y=5, x+y=3⇒x-y+x+y=5+3

                    ⇒ 2x=8 

                    ⇒ x=4

x+y=3 ⇒ 4+y=3

          ⇒ y=-1

x³-y³=4³-(-1)³=64+1=65

Có : `x^2-y^2=15`

`->(x-y)(x+y)=15`

`->5(x+y)=15`

`->x+y=3`

Mà `x-y=5`

`->x=4;y=-1`

`->x^3+y^3=4^3-(-1)^3=64+1=65`

`x-y=5`

`<=>(x-y)^2=25`

`<=>x^2-2xy+y^2=25`

`<=>15-2xy=25`

`<=>2xy=-10`

`<=>xy=-5`

`=>x^3-y^3`

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

`=5.(15-5)`

`=5.10=50`

Ta có: \(x-y=5\)

\(\Leftrightarrow x^2+y^2-10xy=25\)

\(\Leftrightarrow15-10xy=25\)

\(\Leftrightarrow10xy=-10\)

hay xy=-1

Ta có: \(x^3-y^3\)

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

\(=5^3+3\cdot5\cdot\left(-1\right)\)

\(=125-15=110\)

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

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

\(\Rightarrow A=2x^2+xy\)

b) \(\left(15x^2-2xy\right)+A=6x^2+9xy-y^2\)

\(\Rightarrow A=6x^2+9xy-y^2-15x^2+2xy\)

\(\Rightarrow A=11xy-9x^2-y^2\)

c) \(\left(x^2-y^2+3y^2-1\right)-A=x^2-2y^2\)

\(\Rightarrow A=x^2-y^2+3y^2-1-x^2+2y^2\)

\(\Rightarrow A=4y^2-1\)

1)\(\frac{10xy^2\left(x+y\right)}{15xy\left(x+y\right)^3}=\frac{2y}{5\left(x+y\right)^2}\)

2) \(\frac{15x\left(x+y\right)^2}{20x^2\left(x+5\right)}=\frac{3\left(x^2+2xy+y^2\right)}{4x\left(x+5\right)}=\frac{3\left(x+y\right)^2}{4x^2+20x}\)

3) \(\frac{15x\left(x-y\right)}{3\left(y-x\right)}=\frac{5x\left(x-y\right)}{-3\left(x-y\right)}=-\frac{5x}{3}\)

4)\(\frac{y^2-x^2}{x^3-3x^2y+3xy^2-y^3}=\frac{\left(y-x\right)\left(x+y\right)}{\left(x-y\right)^3}=\frac{-\left(x-y\right)\left(x+y\right)}{\left(x-y\right)^3}=\frac{-\left(x+y\right)}{\left(x-y\right)^2}\)

đề yêu cầu j vậy em?

\(a,x^2-x=x\left(x-1\right)\\ b,5x^2\left(x-2y\right)-15x\left(x-2y\right)=\left(5x^2-15x\right)\left(x-2y\right)=5x\left(x-3\right)\left(x-2y\right)\\ c,3\left(x-y\right)-5x\left(y-x\right)=3\left(x-y\right)+5x\left(x-y\right)=\left(5x+3\right)\left(x-y\right)\)

Đặt x/2=y/5=k

=>x=2k; y=5k

xy-15x+6y=40

\(\Leftrightarrow10k^2-15\cdot2k+6\cdot5k=40\)

\(\Leftrightarrow10k^2=40\)

\(\Leftrightarrow k^2=4\)

Trường hợp 1: k=2

=>x=4;y=10

TRường hợp 2: k=-2

=>x=-4; y=-10

Đặt `x/2 = y/5 = k`

`=>` `{(x = 2k),(y = 5k):}`

Ta có `: xy - 15x + 6y = 40`

`=> 2k . 5k - 15 . ( 2k ) + 6 . ( 5k ) = 40`

`=> 10k^2 - 30k + 30k = 40`

`=> k^2 = 40 : 10`

`=> k^2 = 4`

`=>` \(\left[ \begin{array}{l}k^2 = 2^2\\k^2 = ( - 2 )^2\end{array} \right.\) 

`=>` \(\left[ \begin{array}{l}k = 2\\k = - 2\end{array} \right.\) 

Xét `k = 2 => {(x = 2 . 2 = 4),(y = 5 . 2 = 10):}`

Xét `k = - 2 => {(x = - 2 . 2 = - 4),(y = - 2 . 5= - 10):}`

Vậy `, ( x ; y ) in { ( 4 ; 10 ) ; ( - 4 ; - 10 ) } .`

` x/2=y/5 => x=2y /5`

thay` x = 2y /5 `vào `xy - 15x + 6y = 40` ta đc

\(\dfrac{2}{5}y^2-6y+6y=40\)

\(=>\dfrac{2}{5}y^2=40\)

\(=>y^2=100\Leftrightarrow\left[{}\begin{matrix}y=10\\y=-10\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{2}{5}.10=4\\x=\dfrac{2}{5}.\left(-10\right)=-4\end{matrix}\right.\)

\(c,=\left(x-y\right)\left(10x^2-15x^3\right)=5x^2\left(2-3x\right)\left(x-y\right)\\ d,=\left(2x-3y\right)\left(5x+15\right)=5\left(x+3\right)\left(2x-3y\right)\)

c: \(10x^2\left(x-y\right)+15x^3\left(y-x\right)\)

\(=10x^2\left(x-y\right)-15x^3\left(x-y\right)\)

\(=5x^2\left(x-y\right)\left(2-3x\right)\)

d: \(5x\left(2x-3y\right)-15\left(3y-2x\right)\)

\(=5x\left(2x-3y\right)+15\left(2x-3y\right)\)

\(=5\left(2x-3y\right)\left(x+3\right)\)

c: \(10x^2\left(x-y\right)+15x^3\left(y-x\right)\)

\(=10x^2\left(x-y\right)-15x^3\left(x-y\right)\)

\(=5x^2\left(2-3x\right)\left(x-y\right)\)

d: \(5x\left(2x-3y\right)-15\left(3y-2x\right)\)

\(=5x\left(2x-3y\right)+15\left(2x-3y\right)\)

\(=5\left(x+3\right)\left(2x-3y\right)\)