your gay

điều kiện ban đầu <=> (x-1)²+(y-2)²+(z-3)² \(\le1\)

áp dụng bdt sau (ax+ by+ cz)²\(\le\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\)(bunhiacopxky với 3 số)

[ x-1 + 2(y-2) + 2(z-3)]² \(\le\left(1^2+2^2+2^2\right)\left[\left(x-1\right)^2+\left(y-2\right)^2+\left(z-2\right)^2\right]\le9.1=9\)

=>\(-3\le\) x-1 +2(y-2) +2(z-3) \(\le3\) <=> 8\(\le x+2y+2z\le14\)

Giả thiết tương đương \(\left(x-1\right)^2+\left(y+2\right)^2+\left(z-3\right)^2=29\).

Áp dụng bđt Cauchy - Schwarz ta có:

\(\left(2x-3y+4z-20\right)^2=\left[2\left(x-1\right)-3\left(y+2\right)+4\left(z-3\right)\right]^2\le\left(2^2+3^2+4^2\right)\left[\left(x-1\right)^2+\left(y+2\right)^2+\left(z-3\right)^2\right]=29^2\Rightarrow\left|2x-3y+4z-20\right|\le29\)

Sửa đề: \(x^2+y^2+z^2-2x-4y+6z=-14\)

Ta có: \(x^2+y^2+z^2-2x-4y+6z=-14\)

=>\(x^2+y^2+z^2-2x-4y+6z+14=0\)

=>\(x^2-2x+1+y^2-4y+4+z^2+6z+9=0\)

=>\(\left(x-1\right)^2+\left(y-2\right)^2+\left(z+3\right)^2=0\)

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

Bài 1:

xy+2x-3y=1

=>x(y+2)-3y-6=1-6

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

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

=>(x-3;y+2)∈{(1;-5);(-5;1);(-1;5);(5;-1)}

=>(x;y)∈{(4;-7);(-2;-1);(2;3);(8;-3)}