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\(\left|x\right|=2\Rightarrow\left[{}\begin{matrix}x=-2\\x=2\end{matrix}\right.\)
Thay x=-2 vào B ta có:
\(B=4x^3+x-2022=4.\left(-2\right)^3+\left(-2\right)-2022=-32-2-2022=-2056\)
Thay x=2 vào B ta có:
\(B=4x^3+x-2022=4.2^3+2-2022=32+2-2022=-1988\)
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Vì \(x^2+1>0\) nên \(x^2-4=0\)
\(\Leftrightarrow x^2=4\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
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Do (x-5)^2012 >= 0 và /x-2/>=0
=> (x-5)^2012 + /x-2/ +2014 >= 2014
=> min A = 2014
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\(a,1-\left(\dfrac{\dfrac{5}{3}}{8}+x-\dfrac{\dfrac{7}{5}}{24}\right)-\dfrac{\dfrac{16}{2}}{3}=0\\ \Leftrightarrow1-\left(\dfrac{5}{24}+x-\dfrac{7}{120}\right)=\dfrac{8}{3}\\ \Leftrightarrow\dfrac{3}{20}+x=1-\dfrac{8}{3}=-\dfrac{5}{3}\\ \Leftrightarrow x=-\dfrac{5}{3}-\dfrac{3}{20}=-\dfrac{109}{60}\)
\(\left(x-2\right)^{2012}+\left|y^2-9\right|^{2014}=0\)
Vì \(\left(x-2\right)^{2012}\ge0\forall x\); \(\left|y^2-9\right|^{2014}\ge0\forall y\)
\(\Rightarrow\left(x-2\right)^{2012}+\left|y^2-9\right|^{2014}\ge0\forall x,y\)
mà \(\left(x-2\right)^{2012}+\left|y^2-9\right|^{2014}=0\)( giả thiết )
Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x-2=0\\y^2-9=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y^2=9\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=\pm3\end{cases}}\)
Vậy \(x=2\)và \(y=\pm3\)
Ta có: \(\left(x-2\right)^{2012}+\left|y^2-9\right|^{2014}\ge0\left(\forall x,y\right)\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x-2\right)^{2012}=0\\\left|y^2-9\right|^{2014}=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=2\\y=\pm3\end{cases}}\)
Vậy \(\left(x;y\right)\in\left\{\left(2;3\right);\left(2;-3\right)\right\}\)