Question

[Khuất Đăng Mạnh]

Tìm x biết

a.\(2009-\left|x-2009\right|=x\)

b.\(\left(2x+1\right)^{2008}+\left(y-\dfrac{2}{5}\right)^{2008}+\left|x+y-z\right|=0\)

Answer 1

[Hoàng Thị Ngọc Mai]

a)

\(2009-\left|x-2009\right|=x\)

\(\Rightarrow\left|x-2009\right|=-\left(x-2009\right)\)

\(\Rightarrow x-2009\le0\)

\(\Rightarrow x\le2009\)

Vậy \(x\le2009\)

b)

Vì \(\left(2x+1\right)^{2008}\ge0\forall x\)

\(\left(y-\dfrac{2}{5}\right)^{2008}\ge0\forall y\)

\(\left|x+y-z\right|\ge0\forall x,y,z\)

\(\Rightarrow\left(2x+1\right)^{2008}+\left(y-\dfrac{2}{5}\right)^{2008}+\left|x+y-z\right|\ge0\forall x,y,z\)

Mà theo đề bài :

\(\left(2x+1\right)^{2008}+\left(y-\dfrac{2}{5}\right)^{2008}+\left|x+y-z\right|=0\)

\(\Rightarrow\left(2x+1\right)^{2008}=0;\left(y-\dfrac{2}{5}\right)^{2008}=0;\left|x+y-z\right|=0\)

*) Với \(\left(2x+1\right)^{2008}=0\)

\(\Rightarrow2x+1=0\)

\(\Rightarrow2x=-1\)

\(\Rightarrow x=\dfrac{-1}{2}\)

*) Với \(\left(y-\dfrac{2}{5}\right)^{2008}=0\)

\(\Rightarrow y-\dfrac{2}{5}=0\)

\(\Rightarrow y=\dfrac{2}{5}\)

*) Với \(\left|x+y-z\right|=0\)

\(\Rightarrow x+y-z=0\)

\(\Rightarrow\dfrac{-1}{2}+\dfrac{2}{5}-z=0\)

\(\Rightarrow\dfrac{-1}{10}-z=0\)

\(\Rightarrow z=\dfrac{-1}{10}\)

Vậy \(x=\dfrac{-1}{2};y=\dfrac{2}{5};z=\dfrac{-1}{10}\)

Answer 2

[FAIRY TAIL]

a, 2009 - \(\left|x-2009\right|\) = x

=> \(\left|x-2009\right|\) = 2009 - x

=> \(\left[{}\begin{matrix}x-2009=2009-x\\x-2009=-2009-x\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}2x=4018\\2x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2009\\x=0\end{matrix}\right.\)

Vậy x \(\in\)n { 2009 ; 0 }