1 + 2xy - x² - y²

= 1 - ( x² - 2xy + y² )

= 1² - ( x - y )²

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

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

a² + b² - c² - d² - 2ab + 2cd

= ( a² - 2ab + b² ) - ( c² - 2cd + d² )

= ( a - b )² - ( c - d )²

= [ ( a - b ) - ( c - d ) ][ ( a - b ) + ( c - d ) ]

= ( a - b - c + d )( a - b + c - d )

a³b³ - 1

= ( ab )³ - 1³

= ( ab - 1 )[ ( ab )² + ab.1 + 1² ]

= ( ab - 1 )( a²b² + ab + 1 )

x²( y - z ) + y²( z - x ) + z²( x - y )

= z²( x - y ) + x²y - x²z + y²z + y²x

= z²( x - y ) + ( x²y - y²x ) - ( x²z - y²z )

= z²( x - y ) + xy( x - y ) - z( x² - y² )

= z²( x - y ) + xy( x - y ) - z( x + y )( x - y )

= ( x - y )[ z² + xy - z( x + y ) ]

= ( x - y )( z² + xy - zx - zy )

= ( x - y )[ ( z² - zx ) - ( zy - xy ) ]

= ( x - y )[ z( z - x ) - y( z - x ) ]

= ( x - y )( z - x )( z - y )

Bài 1:

a) Ta có: \(2x=5y.\)

=> \(\frac{x}{y}=\frac{5}{2}\)

=> \(\frac{x}{5}=\frac{y}{2}\) và \(x.y=90.\)

Đặt \(\frac{x}{5}=\frac{y}{2}=k\Rightarrow\left\{{}\begin{matrix}x=5k\\y=2k\end{matrix}\right.\)

Có: \(x.y=90\)

=> \(5k.2k=90\)

=> \(10k^2=90\)

=> \(k^2=90:10\)

=> \(k^2=9\)

=> \(k=\pm3.\)

TH1: \(k=3\)

\(\Rightarrow\left\{{}\begin{matrix}x=3.5=15\\y=3.2=6\end{matrix}\right.\)

TH2: \(k=-3\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-3\right).5=-15\\y=\left(-3\right).2=-6\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(15;6\right),\left(-15;-6\right).\)

e) Ta có: \(\frac{x}{y}=\frac{4}{5}.\)

=> \(\frac{x}{4}=\frac{y}{5}\) và \(x.y=20.\)

Đặt \(\frac{x}{4}=\frac{y}{5}=k\Rightarrow\left\{{}\begin{matrix}x=4k\\y=5k\end{matrix}\right.\)

Có: \(x.y=20\)

=> \(4k.5k=20\)

=> \(20k^2=20\)

=> \(k^2=20:20\)

=> \(k^2=1\)

=> \(k=\pm1.\)

TH1: \(k=1\)

\(\Rightarrow\left\{{}\begin{matrix}x=1.4=4\\y=1.5=5\end{matrix}\right.\)

TH2: \(k=-1\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-1\right).4=-4\\y=\left(-1\right).5=-5\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(4;5\right),\left(-4;-5\right).\)

Chúc bạn học tốt!

viết bằng công thức ở chỗ \(\sum\) đi bạn

Bạn bảo cái gì cơ

x/3=y/4=z/5 =>2x²/18=2y²/32=3z²/75=(2x²+2y²-3z²)/(18+32-75)

=-100/-25=4

Vậy x=6 hoặc x=-6;y=8 hoặc y=-8; z=10 hoặc z=-10