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+ Với x < -5 thì |x + 5| = -(x + 5) = -x - 5
=> -x - 5 = 4x + 1
=> -x - 4x = 1 + 5
=> -5x = 6
=> \(x=-\frac{6}{5}\), không thỏa mãn x < -5
+ Với \(x\ge-5\) thì |x + 5| = x + 5
=> x + 5 = 4x + 1
=> 4x - x = 5 - 1
=> 3x = 4
=> \(x=\frac{4}{3}\), thỏa mãn \(x\ge-5\)
Vậy \(x=\frac{4}{3}\)
\(\left|x+5\right|=4x+1\)
\(=>\left[\begin{array}{nghiempt}x+5=4x+1\\x+5=-\left(4x+1\right)=-4x-1\end{array}\right.\)
\(=>\left[\begin{array}{nghiempt}3x=4\\5x=-6\end{array}\right.\)
\(=>\left[\begin{array}{nghiempt}x=\frac{4}{3}\\x=-\frac{6}{5}\end{array}\right.\)
a) Ta có: \(\left|x+\dfrac{19}{5}\right|\ge0\forall x\in Q\)
\(\left|y+\dfrac{2017}{2018}\right|\ge0\forall y\in Q\)
\(\left|z-2019\right|\ge0\forall x\in Q\)
\(\Rightarrow\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{2017}{2018}\right|+\left|z-2019\right|\ge0\forall x,y,z\in Q\)
Dấu \("="\) xảy ra khi \(\left\{{}\begin{matrix}\left|x+\dfrac{19}{5}\right|=0\\\left|y+\dfrac{2017}{2018}\right|=0\\\left|z-2019\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-19}{5}\\y=\dfrac{-2017}{2018}\\z=2019\end{matrix}\right.\).
b) Lại có:
\(\left|x-\dfrac{9}{5}\right|\ge0\forall x\in Q\)
\(\left|y+\dfrac{3}{4}\right|\ge0\forall y\in Q\)
\(\left|z+\dfrac{7}{2}\right|\ge0\forall z\in Q\)
\(\Rightarrow\left|x-\dfrac{9}{5}\right|+\left|y+\dfrac{3}{4}\right|+\left|z+\dfrac{7}{2}\right|\ge0\forall x,y,zQ\)
Mà theo đề bài:
\(\left|x-\dfrac{9}{5}\right|+\left|y+\dfrac{3}{4}\right|+\left|z+\dfrac{7}{2}\right|\le0\forall\)
\(\Rightarrow\left|x-\dfrac{9}{5}\right|+\left|y+\dfrac{3}{4}\right|+\left|z+\dfrac{7}{2}\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x-\dfrac{9}{5}\right|=0\\\left|y+\dfrac{3}{4}\right|=0\\\left|z+\dfrac{7}{2}\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{9}{5}\\y=\dfrac{-3}{4}\\z=\dfrac{-7}{2}\end{matrix}\right.\)
Vậy .....
a) \(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{2017}{2018}\right|+\left|z-2019\right|=0\)
Ta có: \(\left|x+\dfrac{19}{5}\right|\ge0;\left|y+\dfrac{2017}{2018}\right|\ge0;\left|z-2019\right|\ge0\)
Để \(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{2017}{2018}\right|+\left|z-2019\right|=0\) thì:
\(\left\{{}\begin{matrix}\left|x+\dfrac{19}{5}\right|=0\\\left|y+\dfrac{2017}{2018}\right|=0\\\left|z-2019\right|=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-19}{5}\\y=\dfrac{-2017}{2018}\\z=2019\end{matrix}\right.\)
Vậy............................
b) Ta có: \(\left|x-\dfrac{9}{5}\right|\ge0;\left|y+\dfrac{3}{4}\right|\ge0;\left|z+\dfrac{7}{2}\right|\ge0\)
Mà \(\left|x-\dfrac{9}{5}\right|+\left|y+\dfrac{3}{4}\right|+\left|z+\dfrac{7}{2}\right|\le0\) thì:
\(\left|x-\dfrac{9}{5}\right|=\left|y+\dfrac{3}{4}\right|=\left|z+\dfrac{7}{2}\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{9}{5}\\y=\dfrac{-3}{4}\\z=\dfrac{-7}{2}\end{matrix}\right.\)
Vậy............................
TH1: x<-1
Pt sẽ là \(2-x-2\left(-x-1\right)=x\)
\(\Leftrightarrow2-x+2x+2=x\)
=>x+4=x(loại)
TH2: -1<=x<2
Pt sẽ là \(2-x-2\left(x+1\right)=x\)
\(\Leftrightarrow2-x-2x-2=x\)
=>x=0(nhận)
TH3: x>=2
Pt sẽ là \(x-2-2\left(x+1\right)=x\)
=>x-2-2x-2=x
=>-x-4=x
=>-2x=4
hay x=-2(loại)
\(\left|x+\frac{1}{2}\right|-2x=3\)
<=>\(\left|x+\frac{1}{2}\right|=3+2x\)
<=>\(x+\frac{1}{2}=-\left(3+2x\right)\)hoặc\(3+2x\)
Xét \(x+\frac{1}{2}=-\left(3+2x\right)\)
<=>\(x+\frac{1}{2}=3-2x\)
<=>\(x=\frac{5}{6}\left(Loai\right)\)
Xét \(x+\frac{1}{2}=3+2x\)
<=>\(x=-\frac{7}{6}\left(tm\right)\)
Vậy \(x=-\frac{7}{6}\)
\(\left|x-\frac{1}{2}\right|-2x=3\)
\(\Rightarrow\left[\begin{array}{nghiempt}x-\frac{1}{2}-2x==3\\\frac{1}{2}-x-2x=3\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}-x=\frac{7}{2}\\-3x=\frac{5}{2}\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{7}{2}\\x=-\frac{5}{6}\end{array}\right.\)
Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(B=\left|x-2011\right|+\left|x-2\right|\)
\(=\left|x-2011\right|+\left|2-x\right|\)
\(\ge\left|x-2011+2-x\right|=2009\)
Xảy ra khi \(2\le x\le2011\)
\(\left|x-2011\right|+\left|x-2\right|=\left|x-2011\right|+\left|2-x\right|\)
Áp dụng bđt:
\(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)
\(\Rightarrow\left|x-2011\right|+\left|2-x\right|\ge\left|x-2011+2-x\right|\)
\(\Rightarrow\left|x-2011\right|+\left|2-x\right|\ge2009\)
Dấu "=" xảy ra khi:
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2011\ge0\Rightarrow x\ge2011\\2-x\ge0\Rightarrow x\le2\end{matrix}\right.\\\left\{{}\begin{matrix}x-2011< 0\Rightarrow x< 2011\\2-x< 0\Rightarrow x>2\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow2< x< 2011\)
+ Với \(x< -1\) thì |x + 1| = -(x + 1) = -x - 1; |x - 2| = 2 - x
Ta có:
H = (-x - 1) - (2 - x)
H = -x - 1 - 2 + x
H = -3
+ Với \(-1\le x< 1\) thì |x + 1| = x + 1; |x - 2| = 2 - x
Ta có:
H = (x + 1) - (2 - x)
H = x + 1 - 2 + x
H = 2x - 1