Tìm x , y , z thuộc Z
l 5 - 2x l = l x + 4 l
l x - 1 l = l 2x + 5 l
l x + 1 l + l x + 2 l + l x + 3 l = 0
( x - 1 )2 + l y - z l200 + l z - 3 l \(\le\)0
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
5 tháng 2 2022
Bài 2:
a: \(f\left(-x\right)=-x+\left|-x\right|=-x+\left|x\right|< >f\left(x\right)\)
Vậy: Hàm số không chẵn cũng không lẻ
b: \(f\left(-x\right)=-x-\left|-x\right|=-x-\left|x\right|< >f\left(x\right)\)
Vậy: Hàm số không chẵn cũng không lẻ
13 tháng 6 2017
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 .....
PT
13 tháng 6 2017
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............................
TP
1
10 tháng 8 2017
\(xy-2x+y+1=0\)
\(\Rightarrow xy-2x+y-2=-1\)
\(\Rightarrow x\left(y-2\right)+1\left(y-2\right)=-1\)
\(\Rightarrow\left(x+1\right)\left(y-2\right)=-1\)
\(\Rightarrow x+1;y-2\inƯ\left(-1\right)\)
\(Ư\left(-1\right)=\left\{\pm1\right\}\)
Xét ước
\(\left|x-3\right|=2x+1\)
\(\Rightarrow\left[{}\begin{matrix}x-3=2x+1\left(đk:x\ge3\right)\\-x+3=2x+1\left(đk:x< 3\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2x+4\\-x=2x-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-x=4\Rightarrow x=-4\left(KTM\right)\\x=\dfrac{2}{3}\left(TM\right)\end{matrix}\right.\)
10 tháng 8 2017
Bạn ơi, cho mk hỏi sao rằng xy- 2x +y - 2 lại bằng -1 vậy bạn
28 tháng 12 2015
a)|2x-5|=13
2x-5=13=>x=9
2x-5=-13=>x=-4
b)3|x+1|+1=28
3|x+1|=28-1
3|x+1|=27
|x+1|=27:3
|x+1|=9
x+1=9=>x=8
x+1=-9=>x=-10
tick nha
28 tháng 12 2015
a)(x+1)+(x+3)+...+(x+97)+(x+99)=0
x.50+2500=0
x.50=0-2500
x.50=-2500
x=-2500:5
x=-500
HA
19 tháng 2 2020
\(\left(n+3\right).\left(n-2\right)< 0\)
=> n+3 và n-2 khác dấu
\(th1\Leftrightarrow\orbr{\begin{cases}n+3>0\\n-2< 0\end{cases}}\Leftrightarrow\orbr{\begin{cases}n>-3\\n< 2\end{cases}\Leftrightarrow-3< n< 2\left(tm\right)}\)
\(th2\Leftrightarrow\orbr{\begin{cases}n+3< 0\\n-2>0\end{cases}}\Leftrightarrow\orbr{\begin{cases}n< -3\\n>2\end{cases}\Leftrightarrow2< n< -3\left(vl\right)}\)
vậy với -3<n<2 thì
\(n\in\left\{-2;-1;0;1\right\}\)
\(1)|5-2x|=|x+4|\)
\(\Leftrightarrow\orbr{\begin{cases}5-2x=x+4\\5-2x=-x-4\end{cases}\Leftrightarrow\orbr{\begin{cases}-2x-x=4-5\\-2x+x=-4-5\end{cases}\Leftrightarrow}\orbr{\begin{cases}-3x=-1\\-x=-9\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x=9\end{cases}}}\)
Vậy \(x=\frac{1}{3};x=9\)
\(2)|x-1|=|2x+5|\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=2x+5\\x-1=-2x-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x-2x=5+1\\x+2x=-5+1\end{cases}\Leftrightarrow}\orbr{\begin{cases}-x=4\\3x=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-4\\x=-\frac{4}{3}\end{cases}}}\)
Vậy \(x=-4;x=-\frac{4}{3}\)
\(3)|x+1|+|x+2|+|x+3|=0\left(1\right)\)
Ta có: \(|x+1|\ge0\forall x;|x+2|\ge0\forall x;|x+3|\ge0\forall x\)
\(\Leftrightarrow|x+1|+|x+2|+|x+3|\ge0\forall x\)
\(\left(1\right)\Leftrightarrow|x+1|+|x+2|+|x+3|=0\)
\(\Leftrightarrow\left(x+1\right)+\left(x+2\right)+\left(x+3\right)=0\)
\(\Leftrightarrow x+1+x+2+x+3=0\)
\(\Leftrightarrow\left(x+x+x\right)+\left(1+2+3\right)=0\)
\(\Leftrightarrow3x+6=0\)
\(\Leftrightarrow3x=-6\)
\(\Leftrightarrow x=-6:3\)
\(\Leftrightarrow x=-2\)
Vậy x=-2
1: x = 1/3 , x=9
2: x = 4 , x = -4/3
3: x=2