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Bài giải
a, \(\left|x+3\right|+\left|y-1\right|=0\)
Mà \(\hept{\begin{cases}\left|x+3\right|\ge0\forall x\\\left|y-1\right|\ge0\forall x\end{cases}}\Rightarrow\hept{\begin{cases}\left|x+3\right|=0\\\left|y-1\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)
Vậy \(\left(x\text{ ; }y\right)=\left(-3\text{ ; }1\right)\)
b, \(\left|x+5\right|+\left|y+1\right|\le0\)
Mà \(\hept{\begin{cases}\left|x+5\right|\ge0\forall x\\\left|y+1\right|\ge0\end{cases}}\Rightarrow\text{ }\left|x+5\right|+\left|y+1\right|=0\)
Dấu " = " xảy ra khi \(\hept{\begin{cases}\left|x+5\right|=0\\\left|y+1\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-5\\y=-1\end{cases}}\)
Vậy \(\left(x\text{ ; }y\right)=\left(-5\text{ ; }-1\right)\)
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A = |x - 1| + |x + 5| + (x - 2)2 + 2017
A = |x - 1| + |x + 5| + |(x - 2)2| + 2017
A = |x - 1| + |x + 5| + |x2 + 4 - 4x| + 2017
Áp dụng bđt |a| + |b| + |c| \(\ge\)|a+b+c| ta có:
A = |x - 1| + |x + 5| + |x2 + 4 - 4x| + 2017 \(\ge\)|x - 1 + x + 5 + x2 + 4 - 4x| + 2017
A\(\ge\) |x2 - 2x + 8| + 2017
A \(\ge\) |x2 - x - x + 1 + 7| + 2017
A\(\ge\) |(x - 1)2 + 7| + 2017
A\(\ge\) (x - 1)2 + 2024
Dấu "=" xảy ra khi x - 1 \(\ge\)0; x + 5 \(\ge\)0
=> x \(\ge\)1; x \(\ge\)-5
=> x \(\ge\)1
Vậy GTNN của A là 2024 khi x = 1
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/x+1/>= 0
/x+3/>=0
=>/x+1/+/x+3/>=0
=>3x>=0
=> x>=0
=> /x+1/=x+1 ;/x+3/=x+3=> x+1+x+3=3x=>2x+4=3x =>x=4
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a: \(\Leftrightarrow\left\{{}\begin{matrix}3x-2>-4\\3x-2< 4\end{matrix}\right.\Leftrightarrow-\dfrac{2}{3}< x< 2\)
c: \(\Leftrightarrow\left[{}\begin{matrix}3x-1>5\\3x-1< -5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>2\\x< -\dfrac{4}{3}\end{matrix}\right.\)
d: \(\Leftrightarrow\left[{}\begin{matrix}3x+1>x-2\\3x+1< -x+2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x>-3\\4x< 1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>-\dfrac{3}{2}\\x< \dfrac{1}{4}\end{matrix}\right.\)
Ta có [x-y+5]=0 và [x-1]=0
ta đc [x-1]=0=>x=1
vậy [x-y+5]=0 <=> [1-y+5]=0 <=> [6-y]=0
=>y=6
Vậy giá trị của y là 6