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a.2010-|x-2010|=x
=>| x-2010|=2010-x
Ta có: | x- 2010 |= x-2010 hoặc |x-2010|= -(x-2010)
TH1: | x-2010|= x-2010
=>x-2010= 2010 - x
=> x+x= 2010+2010
=> 2x = 4020
=> x = 2010.
TH2: | x-2010|=-( x- 2010)
=> -x+2010= 2010-x
=>-x+x=2010-2010
=> 0=0(luôn đúng).
=>x=0
Vậy x= 2010 hoặc x=0
b. Ta có: \(\left(2x-1\right)^{2010}\) \(\ge0\)
\(\left(y-\dfrac{2}{5}\right)^{2010}\ge0\)
\(\left|x+y-z\right|\ge0\)
=> Để biểu thức trên xảy ra =>\(\left(2x-1\right)^{2010}=0\)
\(\left(y-\dfrac{2}{5}\right)^{2010}=0\)
\(\left|x+y-z\right|=0\)
* Với \(\left(2x-1\right)^{2010}=0\)
=> 2x -1 =0
=> 2x = 1
=> x= \(\dfrac{1}{2}\)
*Với \(\left(y-\dfrac{2}{5}\right)^{2010}=0\)
=> \(y-\dfrac{2}{5}=0\)
=> y= \(\dfrac{2}{5}\)
* Với \(\left|x+y-z\right|=0\)
=> x+y-z=0
=> \(\dfrac{1}{2}+\dfrac{2}{5}-z=0\)
=> \(\dfrac{9}{10}-z=0\)
=> \(z=\dfrac{9}{10}\)
Vậy \(x=\dfrac{1}{2}\); \(y=\dfrac{2}{5}\); \(z=\dfrac{9}{10}\)
VÌ \(\left(x-1\right)^{2012}\ge0\)
\(\left(y-2\right)^{2010}\ge0\)
\(\left(x-z\right)^{2008}\ge0\)
nên dấu \(=\)xảy ra khi \(\hept{\begin{cases}x=z\\x=1\\y=2\end{cases}\Leftrightarrow\hept{\begin{cases}x=z=1\\y=2\end{cases}}}\)
\(a,\Leftrightarrow y^{200}-y=y\left(y^{199}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=0\\y^{199}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}y=0\\y=1\end{matrix}\right.\)
Vậy ..
\(b,\Leftrightarrow y^{2010}-y^{2008}=y^{2008}\left(y^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y^{2008}=0\\y^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}y=0\\y=1\\y=-1\end{matrix}\right.\)
Vậy ...
\(c,\Leftrightarrow\left(2y-1\right)^{50}-\left(2y-1\right)=\left(2y-1\right)\left(\left(2y-1\right)^{49}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2y-1=0\\\left(2y-1\right)^{49}=1\end{matrix}\right.\)
\(\Leftrightarrow y=\dfrac{1}{2}\)
Vậy ..
\(d,\Leftrightarrow\left(\dfrac{y}{3}-5\right)^{2008}\left(\left(\dfrac{y}{3}-5\right)^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(\dfrac{y}{3}-5\right)^{2008}=0\\\left(\dfrac{y}{3}-5\right)^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{y}{3}-5=0\\\dfrac{y}{3}-5=1\\\dfrac{y}{3}-5=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}y=15\\y=18\\y=12\end{matrix}\right.\)
Vậy ..
