ta có : x^2 + y^2 +z^2 = xy + yz + xz
=> 2x^2 + 2y^2 +2z^2 = 2xy + 2yz + 2xz
=> ( x^2 -  2xy + y^2) + ( y^2 - 2yz + z^2 ) + ( z^2 -2xz + x^2 ) =0
=> ( x-y )^2 + ( y-z )^2 + ( z -x)^2 =0
=> x =y=z
thay vào .......


 

CM đẳng thức hay tìm x,y vậy 

Mình sẽ làm theo đề bài của mình nếu đúng thì ... nha 

Biến đổi vế phải  ta có :

( x + y) [ ( x - y)^2 + xy ] = ( x + y)( x^2 - 2xy + y^2 + xy)

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

                                       = x^3 + y^3

VẬy VT  = VP đẳng thức được CM 

\(\frac{x^2+y^2}{xy}=\frac{10}{3}\Leftrightarrow 3x^2-10xy+3y^2=0\Leftrightarrow (x-3y)(3x-y)=0\)

Thay trường hợp vòa là xong

a: \(x-2-\frac{x^2-10}{x+2}\)

\(=\frac{\left(x-2\right)\left(x+2\right)-x^2+10}{x+2}\)

\(=\frac{x^2-4-x^2+10}{x+2}=\frac{6}{x+2}\)

b: \(\frac{x}{y^2-xy}-\frac{y}{xy-x^2}\)

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

\(=\frac{-x-y}{xy}\)

c: \(\frac{1}{x\left(x-1\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}\)

\(=-\frac{1}{x}+\frac{1}{x-1}-\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-2}+\frac{1}{x-3}-\frac{1}{x-3}+\frac{1}{x-4}-\frac{1}{x-4}+\frac{1}{x-5}\)

\(=\frac{1}{x-5}-\frac{1}{x}=\frac{x-\left(x-5\right)}{x\left(x-5\right)}=\frac{5}{x\left(x-5\right)}\)

1: 6(x-2)-y(2-x)=10

=>6(x-2)+y(x-2)=10

=>(x-2)(y+6)=10

=>(x-2;y+6)∈{(1;10);(10;1);(-1;-10);(-10;-1);(2;5);(5;2);(-2;-5);(-5;-2)}

=>(x;y)∈{(3;4);(12;-5);(1;-16);(-8;-7);(4;-1);(7;-4);(0;-11);(-3;-8)}

2: 3x-2xy+3y=6

=>x(3-2y)+3y-4,5=6-4,5

=>-x(2y-3)+1,5(2y-3)=1,5

=>(2y-3)(-x+1,5)=1,5

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

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

=>(2x-3;2y-3)∈{(1;-3);(-3;1);(-1;3);(3;-1)}

=>(x;y)∈{(2;0);(0;2);(1;3);(3;1)}

3: 6x-xy+2y=5

=>x(6-y)+2y-12=5-12=-7

=>-x(y-6)+2(y-6)=-7

=>(y-6)(-x+2)=-7

=>(x-2)(y-6)=7

=>(x-2;y-6)∈{(1;7);(7;1);(-1;-7);(-7;-1)}

=>(x;y)∈{(3;13);(9;7);(1;-1);(-5;5)}