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a, xy+2x-y=5

=> x(y+2)-y-2=3

=>x(y+2)-(y+2)=3

=>(x-1)(y+2)=3

=>\(\hept{\begin{cases}x-1=3\Rightarrow x=4\\y+2=1\Rightarrow y=-1\end{cases}}\); \(\hept{\begin{cases}x-1=1\Rightarrow x=2\\y+2=3\Rightarrow y=1\end{cases}}\)

=>\(\hept{\begin{cases}x-1=-1\Rightarrow x=0\\y+2=-3\Rightarrow y=-5\end{cases}}\); \(\hept{\begin{cases}x-1=-3\Rightarrow x=-2\\y+2=-1\Rightarrow y=-3\end{cases}}\)

vậy (x;y)\(\in\)(4,-1);(2,1);(0,-5);(-2.-3)

từ\(\frac{2bz-3cy}{a}\)=\(\frac{3cx-az}{2b}=\frac{ay-2bx}{3c}\)

=>\(\frac{2abz-3acy}{a}\)=\(\frac{6bcx-2abz}{2b}\)=\(\frac{3cay-6cbx}{3c}\)

=\(\frac{2abz-3acy+6bcx-2abz+3cay-6cbx}{2a+4b+6c}\)=0

=>\(\frac{2bz-3cy}{a}=0\)=>2bz=3cy=>\(\frac{z}{3c}\)=\(\frac{y}{2b}\)(1)

=>\(\frac{3cx-az}{2b}\)=0 =>3cx=az =>\(\frac{x}{a}\)=\(\frac{z}{3c}\)(2)

=>\(\frac{ay-2bx}{3c}=0\)=>ay=2bx =>\(\frac{y}{2b}\)=\(\frac{x}{a}\)(3)

Từ (1),(2) và (3) suy ra\(\frac{x}{a}=\frac{y}{2b}=\frac{z}{3c}\)đpcm

giúp mk đi!

the ma van duoc k

a) A + x² - 4xy² + 2xz - 3y² = 0

=> A =  -x² + 4xy² - 2xz + 3y²

b) B + 5x² - 2xy = 6x² + 9xy - y²

=> B = 6x² + 9xy - y² - 5x² + 2xy= x² + 11xy - y²

c) 3xy - 4y² - A = x² - 7xy + 8y²

=> A = 3xy - 4y² - x² + 7xy - 8y² = -12y² + 10xy - x²

Trả lời:

a, A + ( x² - 4xy² + 2xz - 3y² ) = 0 

=> A = - ( x² - 4xy² + 2xz - 3y² ) = - x² + 4xy² - 2xz + 3y²

b, B + ( 5x² - 2xy ) = 6x² + 9xy - y² 

=> B = 6x² + 9xy - y² - ( 5x² - 2xy ) = 6x² + 9xy - y² - 5x² + 2xy = x² + 11xy - y²

c, ( 3xy - 4y² ) - A = x² - 7xy + 8y² 

=> A = 3xy - 4y² - ( x² - 7xy + 8y² ) = 3xy - 4y² - x² + 7xy - 8y² = 10xy - 12y² - x²

d, B + ( 4x²y + 5y² - 3xz + z² ) = x² + 11xy - y² + 4x²y + 5y² - 3xz + z² = x² + 11xy + 4y² + 4x²y - 3xz + z²