x+Y= 15
X=?
Y=?
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\(\left\{{}\begin{matrix}x-y=5\\x^2-y^2=15\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x-y=5\\x+y=3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=-1\end{matrix}\right.\)
\(x^3-y^3=4^3-\left(-1\right)^3=65\)
`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)\(\frac{y^2-x^2}{x^3-3x^2y+3xy^2-y^3}=\frac{\left(y-x\right)\left(x+y\right)}{\left(x-y\right)^3}=\frac{-\left(x-y\right)\left(x+y\right)}{\left(x-y\right)^3}=\frac{-\left(x+y\right)}{\left(x-y\right)^2}\)
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 ) } .`
\(c,=\left(x-y\right)\left(10x^2-15x^3\right)=5x^2\left(2-3x\right)\left(x-y\right)\\ d,=\left(2x-3y\right)\left(5x+15\right)=5\left(x+3\right)\left(2x-3y\right)\)
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: \(5x\left(2x-3y\right)-15\left(3y-2x\right)\)
\(=5x\left(2x-3y\right)+15\left(2x-3y\right)\)
\(=5\left(x+3\right)\left(2x-3y\right)\)
\(x;y\in Z\)
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