Question

Tìm các số x,y,z trong các biểu thức sau:                                                          a) |2x-4| + |3y+21|  =0                                                         b) |2x-12|+|3y+9|=-|x+y+z|

Answer 1

a; |2\(x\) - 4| + |3y + 21| = 0

   Vì |2\(x\) - 4| ≥ 0 ∀ \(x\); |3y + 21| ≥ 0 ∀ \(x\) 

     vậy |2\(x\) - 4| + |3y + 21| = 0

      ⇔ \(\left\{{}\begin{matrix}2x-4=0\\3y+21=0\end{matrix}\right.\)

      ⇔  \(\left\{{}\begin{matrix}x=2\\y=-7\end{matrix}\right.\)

Answer 2

a)

\(\left|2x-4\right|+\left|3y+21\right|=0\)

Ta thấy:\(\left|2x-4\right|\ge0\forall x;\left|3y+21\right|\ge0\forall y\)

Để \(\left|2x-4\right|+\left|3y+21\right|=0\)

\(\Rightarrow\left[{}\begin{matrix}2x-4=0\\3y+21=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=4\\3y=-21\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\y=-7\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(2;-7\right)\) b) \(\left|2x-12\right|+\left|3y+9\right|=-\left|x+y+z\right|\)  Vì \(\left|2x-12\right|\ge0;\left|3y+9\right|\ge0;-\left|x+y+z\right|\le0\) \(\Rightarrow\left[{}\begin{matrix}2x-12=0\\3y+9=0\\x+y+z=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=6\\y=-3\\x+y+z=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=6\\y=-3\\z=-3\end{matrix}\right.\) Vậy \(\left(x;y;z\right)=\left(6;-3;-3\right)\)    

Answer 3

b; |2\(x\) - 12| + |3y + 9| = - |\(x\) + y + z|

    |2\(x\) - 12| + |3y + 9| + |\(x\) + y + z| = 0

    Vì |2\(x\) - 12| ≥ 0; |3y + 9| ≥ 0; |\(x\) + y + z| ≥ 0

  ⇒      |2\(x\) - 12| + |3y + 9| + |\(x\) + y + z| = 0

 ⇔ \(\left\{{}\begin{matrix}2x-12=0\\3y+9=0\\x+y+z=0\end{matrix}\right.\) ⇔  \(\left\{{}\begin{matrix}x=6\\y=-3\\z=-3\end{matrix}\right.\)