Câu 1:
a)A=|x+1|+2016
Vì |x+1|\(\ge\)0
Suy ra:|x+1|+2016\(\ge\)2016
Dấu = xảy ra khi x+1=0
x=-1
Vậy MinA=2016 khi x=-1
b)B=2017-|2x-\(\frac{1}{3}\)|
Vì -|2x-\(\frac{1}{3}\)|\(\le\)0
Suy ra:2017-|2x-\(\frac{1}{3}\)|\(\le\)2017
Dấu = xảy ra khi \(2x-\frac{1}{3}=0\)
\(2x=\frac{1}{3}\)
\(x=\frac{1}{6}\)
Vậy Max B=2017 khi \(x=\frac{1}{6}\)
c)C=|x+1|+|y+2|+2016
Vì |x+1|\(\ge\)0
|y+2|\(\ge\)0
Suy ra:|x+1|+|y+2|+2016\(\ge\)2016
Dấu = xảy ra khi x+1=0;x=-1
y+2=0;y=-2
Vậy MinC=2016 khi x=-1;y=-1
d)D=-|x+\(\frac{1}{2}\)|-|y-1|+10
=10-|x+\(\frac{1}{2}\)|-|y-1|
Vì -|x+\(\frac{1}{2}\)|\(\le\)0
-|y-1| \(\le\)0
Suy ra: 10-|x+\(\frac{1}{2}\)|-|y-1| \(\le\)10
Dấu = xảy ra khi \(x+\frac{1}{2}=0;x=-\frac{1}{2}\)
y-1=0;y=1
Vậy Max D=10 khi x=\(-\frac{1}{2}\);y=1
Bài 1:
a)Ta thấy: \(\left|x+1\right|\ge0\)
\(\Rightarrow\left|x+1\right|+2016\ge0+2016=2016\)
\(\Rightarrow A\ge2016\)
Dấu = khi x=-1
Vậy MinA=2016 khi x=-1
b)Ta thấy:\(\left|2x-\frac{1}{3}\right|\ge0\)
\(\Rightarrow-\left|2x-\frac{1}{3}\right|\le0\)
\(\Rightarrow2017-\left|2x-\frac{1}{3}\right|\le2017-0=2017\)
\(\Rightarrow B\le2017\)
Dấu = khi x=1/6
Vậy Bmin=2017 khi x=1/6
c)Ta thấy:\(\begin{cases}\left|x+1\right|\\\left|y+2\right|\end{cases}\ge0\)
\(\Rightarrow\left|x+1\right|+\left|y+2\right|\ge0\)
\(\Rightarrow\left|x+1\right|+\left|y+2\right|+2016\ge0+2016=2016\)
\(\Rightarrow D\ge2016\)
Dấu = khi x=-1 và y=-2
Vậy MinD=2016 khi x=-1 và y=-2
d)Ta thấy:\(\begin{cases}-\left|x+\frac{1}{2}\right|\\-\left|y-1\right|\end{cases}\le0\)
\(\Rightarrow-\left|x+\frac{1}{2}\right|-\left|y-1\right|\le0\)
\(\Rightarrow-\left|x+\frac{1}{2}\right|-\left|y-1\right|+10\le0+10=10\)
\(\Rightarrow D\le10\)
Dấu = khi x=-1/2 và y=1
Vậy MaxD=10 khi x=-1/2 và y=1
a) ( x + 1 )( y + 2 ) = 0
\(\Rightarrow\) x + 1 = 0 hoặc y + 2 = 0
+) x + 1 = 0 \(\Rightarrow\) x = -1
+) y + 2 = 0 \(\Rightarrow\) y = -2
Vậy x = -1; y = -2
Câu 2:
a)(x+1)(y+2)=0
\(\Rightarrow\left[\begin{array}{nghiempt}x+1=0\\y+2=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=-1\\y=-2\end{array}\right.\)
Vậy x=-1;y=-2
b) ( x + 2)( x - 3) > 0
\(TH1:\left[\begin{array}{nghiempt}x+2< 0\\x-3< 0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x< -2\\x< 3\end{array}\right.\)\(\Rightarrow x< 3\)
\(TH2:\left[\begin{array}{nghiempt}x+2\rightarrow0\\x-3\rightarrow0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x\rightarrow-2\\x\rightarrow3\end{array}\right.\)\(\Rightarrow x\rightarrow-2\)
Vậy x<3 và x >-2
c)\(x+\frac{1}{2}=3\)
\(x=3-\frac{1}{2}\)
\(x=\frac{5}{2}\)
Vậy \(x=\frac{5}{2}\)
d)|x+1|<2016
Cho x+1=2015 vì 2015<2016
x=2014
Do đó x<2014
e) | x - 1/2| > 5
Cứ cho x-1/2=6 vì 6>5
x=13/2
Vậy x>13/2
Bài 2:
a) ( x+1)( y + 2) = 0
\(\Rightarrow\left[\begin{array}{nghiempt}x+1=0\\y+2=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=-1\\y=-2\end{array}\right.\)
b)( x + 2)( x - 3) > 0
=>x+2 và x-3 cùng dấu
Xét \(\begin{cases}x+2>0\\x-3>0\end{cases}\)\(\Rightarrow\begin{cases}x>-2\\x>3\end{cases}\)\(\Rightarrow-2< x>3\)
Xét \(\begin{cases}x+2< 0\\x-3< 0\end{cases}\)\(\Rightarrow\begin{cases}x< -2\\x< 3\end{cases}\)\(\Rightarrow-2>x< 3\)
Kết hợp từ 2 th ta có:\(-2< x< 3\)
c) ( x + 1/2) = 3
\(\Rightarrow x=3-\frac{1}{2}\)
\(\Rightarrow x=\frac{5}{2}\)
d) | x + 1| < 2016
=>x+1<-2016 hoặc 2016
Xét x+1<-2016
=>x<-2017
Xét x+1<2016
=>x<2015
\(\Rightarrow x\in\left(-2017;2015\right)\)
e) | x - 1/2| > 5
=>x-1/2>5 hoặc -5
Xét x-1/2>5
=>x>11/2
Xét x-1/2>-5
=>x>-9/2
\(\Rightarrow x\in\left(-\infty;-\frac{9}{2}\right)\)U\(\left(\frac{11}{2};\infty\right)\)
