a) Vì \(\hept{\begin{cases}\left|x-y-2\right|\ge0∀x,y\\\left|y+3\right|\ge0∀y\end{cases}}\)\(\Rightarrow\left|x-y-2\right|+\left|y+3\right|≥0∀x,y\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}\left|x-y-2\right|=0\\\left|y+3\right|=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x-y-2=0\\y+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x-y=2\\y=-3\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\y=-3\end{cases}}\)

b) Vì \(\hept{\begin{cases}\left|x-3y\right|^{2007}\ge0∀x,y\\\left|y+4\right|^{2008}\ge0∀y\end{cases}}\)\(\Rightarrow\left|x-3y\right|^{2007}+\left|y+4\right|^{2008}≥0∀x,y\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}\left|x-3y\right|^{2007}=0\\\left|y+4\right|^{2008}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x-3y\right|=0\\\left|y+4\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x-3y=0\\y+4=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=3y\\y=-4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-12\\y=-4\end{cases}}\)

c) Vì \(\hept{\begin{cases}\left(x+y\right)^{2006}\ge0∀x,y\\2007.\left|y-1\right|\ge0∀y\end{cases}}\)\(\Rightarrow\left(x+y\right)^{2006}+2007.\left|y-1\right|\ge0∀x,y\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}\left(x+y\right)^{2006}=0\\2007.\left|y-1\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y=0\\\left|y-1\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-y\\y-1=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=-y\\y=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\y=1\end{cases}}\)

d) Vì \(\hept{\begin{cases}\left|x-y-5\right|\ge0∀x,y\\2007.\left(y-3\right)^{2008}\ge0∀y\end{cases}}\)\(\Rightarrow\left|x-y-5\right|+2007.\left(y-3\right)^{2008}\ge0\ge0∀x,y\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}\left|x-y-5\right|=0\\2007.\left(y-3\right)^{2008}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x-y-5=0\\\left(y-3\right)^{2008}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x-y=5\\y-3=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x-y=5\\y=3\end{cases}}\Leftrightarrow\hept{\begin{cases}x=8\\y=3\end{cases}}\)

hello

-33/19 < 45/-31

nhớ tít nha

Ta có :

( x - 1 ) . ( x - 1 ) . ( x - 1 ) . ( x - 1 ) = ( x - 1 ) . ( x - 1 )

=> ( x - 1 )⁴ = ( x - 1 )²

=> ( x - 1 )⁴ - ( x - 2 )² = 0

=> ( x - 1 )² . [ ( x - 1 )² - 1 ] = 0

\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^2=0\\\left(x-1\right)^2-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\\left(x-1\right)^2=1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x-1\in\left\{-1;1\right\}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x\in\left\{0;2\right\}\end{cases}}\)

có A=\(|\frac{1}{2}+x|+3\)

\(|\frac{1}{2}+x|\ge0\)\(\Rightarrow|\frac{1}{2}+x|+3\ge3\)

dấu bằng xảy ra khi \(\frac{1}{2}+x=0\)

                                \(x=\frac{-1}{2}\)

vậy GTNN của A = 3 đat khi \(x=\frac{-1}{2}\)

\(A=\left|\frac{1}{2}+x\right|+3\)

Ta có: \(\left|\frac{1}{2}+x\right|\ge0\forall x\Rightarrow\left|\frac{1}{2}+x\right|+3\ge3\forall x\)

Dấu '' = '' xảy ra khi: \(\left|\frac{1}{2}+x\right|=0\Rightarrow\frac{1}{2}+x=0\Rightarrow x=\frac{-1}{2}\)

Vậy \(MinA=3\) khi \(x=\frac{-1}{2}\)

\(\left(-x+0,2\right)^3=0,008\)

\(\Leftrightarrow\left(-x+0,2\right)^3=\left(\frac{2}{10}\right)^3\)

\(\Leftrightarrow-x+\frac{1}{5}=\frac{1}{5}\)

\(\Leftrightarrow x=0\)

vậy có 1 giá trị x thoả mãn(x=0)

Ta có : ( -x + 0,2 )³ = 0,008

\(\Rightarrow\left(-x+\frac{1}{5}\right)^3=\frac{1}{125}\)

\(\Rightarrow\left(-x+\frac{1}{5}\right)^3=\left(\frac{1}{5}\right)^3\)

\(\Rightarrow-x+\frac{1}{5}=\frac{1}{5}\)

\(\Rightarrow-x=\frac{1}{5}-\frac{1}{5}=0\)

\(\Rightarrow x=0\)