\(\left(x+\frac{2}{3}\right)^{2012}+\left|y-\frac{1}{4}\right|^{2000}+\left(x-y-z\right)^{2014}=0\)

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

\(\Leftrightarrow\hept{\begin{cases}x=-\frac{2}{3}\\y=\frac{1}{4}\\z=-\frac{11}{12}\end{cases}}\).

1 chia 2 bằng bao nhiêu các bạn chỉ giúp mình với

\(|x-2|+\left(x^2-2x\right)^{2014}=0\)

Ta thấy \(|x-2|\ge0\forall x;\left(x^2-2x\right)^{2014}\ge0\forall x\)

Mà I x-2 I +\(\left(x^2-2x\right)^{2014}=0\)

\(\Rightarrow\hept{\begin{cases}|x-2|=0\\x^2-2x=0\end{cases}\Rightarrow\hept{\begin{cases}x-2=0\\x\left(x-2\right)=0\end{cases}}}\Rightarrow\hept{\begin{cases}x=2\\x=0\end{cases}}\)(vô lí)
Vậy PT vô nghiệm

cho minh hoi cai la co ai lop 6 o

\(3^x+4^2=19^6:\left(19^3.19^2\right)-2.1^{2014}\)

\(\Rightarrow\) \(3^x+16=19^6:19^5-2\)

\(\Rightarrow\) \(3^x+16=19-2\)

\(\Rightarrow\) \(3^x+16=17\)

\(\Rightarrow\) \(3^x=1\)

\(\Rightarrow\) \(3^x=3^0\)

\(\Rightarrow\) \(x=0\)

\(2^{x+1}.2^{2014}=2^{2016}\Leftrightarrow2^{x+1+.2014}=2^{2016}\Rightarrow x+2015=2016\Rightarrow x=1\)