1) Ta có \(\hept{\begin{cases}\left|x\right|\ge0\forall x\\\left|y-2\right|\ge0\forall y\end{cases}}\Leftrightarrow\left|x\right|+\left|y-2\right|\ge0\forall x;y\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}x=0\\y-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\y=2\end{cases}}\)
Vậy x = 0 ; y = 2
Thay x = 0 ; y = 2 vào B
=> B = 2.0 - 5.2 + 7.0.2 = -10
Vậy B = -10
Bài 2:
\(a)\)
\(A=\left|x-2021\right|+5\)
Ta có:
\(\left|x-2021\right|\ge0\Rightarrow\left|x-2021\right|+5\ge5\)
Dấu '' = '' xảy ra khi:
\(x-2021=0\)
\(\Leftrightarrow x=2021\)
Vậy \(MinA=5\Leftrightarrow x=2021\)
\(b)\)
\(B=\left|x-2\right|+\left|x-5\right|\)
\(B=\left|x-2\right|+\left|x-5\right|\ge\left|x-2+5-x\right|=\left|3\right|=3\)
Dấu '' = '' xảy ra khi:
\(\left(x-2\right)\left(5-x\right)\ge0\)
\(\Leftrightarrow2\le x\le5\)
Vậy \(MinB=3\Leftrightarrow2\le x\le5\)
2) a Ta có \(\left|x-2021\right|\ge0\forall x\)
=> \(\left|x-2021\right|+5\ge5\)
Dấu "=" xảy ra <=> x - 2021 = 0
=> x = 2021
Vậy Min A = 5 <=> x = 2021
b) Ta có B = |x - 2| + |x - 5| = |x - 2| + |5 - x| \(\ge\left|x-2+5-x\right|=\left|3\right|=3\)
Dấu "=" xảy ra <=> (x - 2)(5 - x) \(\ge0\)
TH1 : \(\hept{\begin{cases}x-2\ge0\\5-x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge2\\x\le5\end{cases}}\Rightarrow2\le x\le5\)
TH2 : \(\hept{\begin{cases}x-2\le0\\5-x\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le2\\x\ge5\end{cases}}\Leftrightarrow x\in\varnothing\)
Vậy Min B = 3 <=> \(2\le x\le5\)
Bài 1:
Ta có:
\(\left|x\right|+\left|y-2\right|=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\y-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\y=2\end{cases}}\)
Nên: \(B=2x-5y+7xy=2.0-5.2+7.0.2=-10\)
Tìm x thỏa mãn: lx - 2022l + lx - 2021l = x - 2023.
