Question

[Mạch Vy Khánh]

Tìm x, y, z biết

4x² + 9y² + 16z² - 4x - 6y - 8z + 3 = 0

Answer 1

[yennhi tran]

\(4x^2-4x+1+9y^2-6y+1+16z^2-8z+1=0\)

\(\Leftrightarrow\left(2x-1\right)^2+\left(3y-1\right)^2+\left(4z-1\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}2x-1=0\\3y-1=0\\4z-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{1}{3}\\x=\frac{1}{4}\end{cases}}\)

vay ................................................

Answer 2

[trang]

Ta có : 

4x² + 9y² + 16z² - 4x - 6y - 8z + 3 = 0

( 2x ) ²  + ( 3y)² + ( 4z)² - 4x - 6y - 8z + 3 = 0

\([\left(2x\right)^2-2.2x+1]+[\left(3y\right)^2-2.3y+1]+[\left(4z\right)^2-2.4z+1]=0\)=0

( 2x-1)²  + ( 3y -1 )² + ( 4z - 1) ² = 0

Mà ( 2x-1)² \(\ge\)0 với mọi x

     ( 3y-1 )² \(\ge0\)với mọi y

      ( 4z - 1) ^{2 \(\ge0\)với mọi z} 

 nên \(\hept{\begin{cases}2x-1=0\\3y-1=0\\4z-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{1}{3}\\z=\frac{1}{4}\end{cases}}}\)

 Vậy x = 1/2 ; y = 1/3 ; z = 1/4