dễ lắm
1) \(\left|0,5-\dfrac{1}{3}+x\right|=-\left|y\right|-\dfrac{1}{4}\)
\(\Leftrightarrow\left|0,5-\dfrac{1}{3}+x\right|=-\left(\left|y\right|+\dfrac{1}{4}\right)\)
Vì \(\left|y\right|\ge0\forall y\Rightarrow-\left(\left|y\right|+\dfrac{1}{3}\right)< 0\forall y\)
VT>0; VP<0=> PT vô nghiệm
2
Dấu bằng xảy ra \(\Leftrightarrow\)\(\left\{{}\begin{matrix}x+\dfrac{13}{7}=0\\z-2008=0\end{matrix}\right.\)\(\Leftrightarrow z=2008;x=-\dfrac{13}{7}\)
* Trả lời:
\(1.\) \(\left|0,5-\dfrac{1}{3}+x\right|=-\left|y\right|-\dfrac{1}{4}\)
\(-\) Lý luận: Giá trị tuyệt đối luôn luôn lớn hơn hoặc bằng 0 nên:
\(\left|0,5-\dfrac{1}{3}+x\right|\ne-\left|y\right|-\dfrac{1}{4}\)
\(\Rightarrow\left|0,5-\dfrac{1}{3}+x\right|=-\left|y\right|-\dfrac{1}{4}\) không có giá trị \(x\)
\(2.\) \(\left|x+\dfrac{13}{7}\right|+\left|z-2008\right|=0\)
\(\Rightarrow x+\dfrac{13}{7}=0\) hoặc \(z-2008=0\)
TH1: \(x+\dfrac{13}{7}=0\)
\(\Rightarrow x=-\dfrac{13}{7}\)
TH2: \(z-2008=0\)
\(\Rightarrow z=2008\)
Vậy \(x=-\dfrac{13}{7};z=2008\)
