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HH
2
30 tháng 7 2017
a) \(\left|x+2\right|+\left|2y-1\right|=0\)
\(\Leftrightarrow\hept{\begin{cases}\left|x+2\right|=0\\\left|2y-1\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x+2=0\\2y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-2\\y=0,5\end{cases}}\)
Vậy (x; y) = (-2; 0,5)
b) \(\left|x-y\right|+\left|2x+3\right|=0\Leftrightarrow\hept{\begin{cases}\left|x-y\right|=0\\\left|2x+3\right|=0\end{cases}}\)
+) |2x + 3| = 0
2x + 3 = 0
2x = -3
x = -1,5
+) |x - y| = 0
x - y = 0
-1,5 - y = 0
y = -1,5
Vậy (x; y) = (-1,5; -1,5)
c, \(\left|2x+y\right|+\left|y+\left(1:4\right)\right|=0\)
\(\left|2x+y\right|+\left|y+\frac{1}{4}\right|=0\)
\(\Leftrightarrow\hept{\begin{cases}\left|2x+y\right|=0\\\left|y+\frac{1}{4}\right|=0\end{cases}}\)
\(\left|y+\frac{1}{4}\right|=0\Leftrightarrow y+\frac{1}{4}=0\Leftrightarrow y=-\frac{1}{4}\)
\(\left|2x+y\right|=0\Leftrightarrow2x+y=0\Leftrightarrow2x-\frac{1}{4}=0\Leftrightarrow2x=\frac{1}{4}\Leftrightarrow x=\frac{1}{8}\)
Vậy \(\left(x;y\right)=\left(\frac{1}{8};-\frac{1}{4}\right)\)
24 tháng 8 2017
a)\(\left|2x-3y\right|+\left|2y-4z\right|=0\)
\(\left\{{}\begin{matrix}\left|2x-3y\right|\ge0\forall x;y\\\left|2y-4z\right|\ge0\forall y;z\end{matrix}\right.\) \(\Rightarrow\left|2x-3y\right|+\left|2y-4z\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|2x-3y\right|=0\\\left|2y-4z\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x=3y\\2y=4z\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{2}\\\dfrac{y}{4}=\dfrac{z}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{6}=\dfrac{y}{4}\\\dfrac{y}{4}=\dfrac{z}{2}\end{matrix}\right.\)
\(\Rightarrow\dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{2}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{2}=\dfrac{x+y+z}{6+4+2}=\dfrac{7}{12}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{7}{12}.6=\dfrac{7}{2}\\y=\dfrac{7}{12}.4=\dfrac{7}{3}\\z=\dfrac{7}{12}.2=\dfrac{7}{6}\end{matrix}\right.\)
b)\(\left|x-2\right|+\left|x-3\right|+\left|x-4\right|=0\)
\(\left\{{}\begin{matrix}\left|x-2\right|\ge0\\\left|x-3\right|\ge0\\\left|x-4\right|\ge0\end{matrix}\right.\) \(\Leftrightarrow\left|x-2\right|+\left|x-3\right|+\left|x-4\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|x-2\right|=0\\\left|x-3\right|=0\\\left|x-4\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=3\\x=4\end{matrix}\right.\)
Vì \(2\ne3\ne4\) nên \(x\in\varnothing\)
c)
\(\left|x+1\right|+\left|x+2\right|+...+\left|x+8\right|+\left|x+9\right|\)
Với mọi \(x\ge0\) ta có:
\(\left\{{}\begin{matrix}\left|x+1\right|=x+1\\\left|x+2\right|=x+2\\\left|x+8\right|=x+8\\\left|x+9\right|=x+9\end{matrix}\right.\)\(\Leftrightarrow x+1+x+2+...+x+8+x+9=x-1\)
\(\Leftrightarrow9x+90=x-1\)
\(\Leftrightarrow9x=x-89\)
\(\Leftrightarrow-8x=89\)
\(\Leftrightarrow x=\dfrac{89}{-8}\left(KTM\right)\)
Với mọi \(x< 0\) ta có:
\(\left\{{}\begin{matrix}x+1=-x-1\\x+2=-x-2\\x+8=-x-8\\x+9=-x-9\end{matrix}\right.\) \(\Leftrightarrow\left(-x-1\right)+\left(-x-2\right)+...+\left(-x-8\right)+\left(-x-9\right)=x-1\)
\(\Leftrightarrow-9x-90=x-1\)
\(\Leftrightarrow-9x=x+89\)
\(\Leftrightarrow-10x=89\)
\(\Leftrightarrow x=\dfrac{89}{-10}\left(TM\right)\)
d)\(\left|2x-3y\right|+\left|5y-2z\right|+\left|2z-6\right|=0\)
\(\left\{{}\begin{matrix}\left|2x-3y\right|\ge0\\ \left|5y-2z\right|\ge0\\ \left|2z-6\right|\ge0\end{matrix}\right.\) \(\Leftrightarrow\left|2x-3y\right|+\left|5y-2z\right|+\left|2z-6\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|2x-3y\right|=0\\\left|5y-2z\right|=0\\\left|2z-6\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}z=3\\y=\dfrac{6}{5}\\x=\dfrac{9}{5}\end{matrix}\right.\)
5 tháng 12 2017
a/ | x-2011y | + ( y-1)2017=0
Câu này có gì đó nhầm lẫn rồi
b/ (2x -1)2 + | 2y - x | - 8 = 12 - 5.22
=> (2x -1)2 + | 2y - x | - 8 = 12 - 20
=> (2x -1)2 + | 2y - x | = 0
=> (2x -1)2 + | 2y - x | = 0
Ta thấy (2x -1)2 và | 2y - x | luôn lớn hơn hoặc bằng 0
=> (2x -1)2 + | 2y - x | = 0
<=> (2x -1)2 = 0 và | 2y - x | = 0
=> 2x -1 = 0 2y - x = 0
=> x = 1/2 y = x/2 = 1/4
c/ | x - 2014y | + | x - 2015 | = 0
Tương tự b nhé bạn
LN
3
24 tháng 9 2018
a,Vì: \(\left(x-1\right)^2\ge0\forall x\)
\(\left(2y-5\right)^4\ge0\forall y\)
\(\Rightarrow\left(x-1\right)^2+\left(2y-5\right)^2\ge0\forall x,y\)
Dấu = xảy ra khi: \(\hept{\begin{cases}\left(x-1\right)^2=0\\\left(2y-5\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=\frac{5}{2}\end{cases}}}\)
=.= hok tốt!!
24 tháng 9 2018
b, Vì: \(\left(2x+3\right)^2\ge0\forall x\)
\(\left(x+2y-3\right)^2\ge0\forall x,y\)
\(\Rightarrow\left(2x+3\right)^2+\left(x+2y-3\right)^2\ge0\forall x,y\)
Mà: \(\left(2x+3\right)^2+\left(x+2y-3\right)^2< 0\)
=> Ko có giá trị của x , y thỏa mãn
=.= hok tốt!!
a,Ta có : \(\left\{{}\begin{matrix}\left|x-3\right|\ge0\\\left|2x-y+1\right|\ge0\end{matrix}\right.\\ \Rightarrow\left|x-3\right|+\left|2x-y+1\right|\ge0\\ \Leftrightarrow\left|x-3\right|+\left|2x-y+1\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-3=0\\2x-y+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\y=7\end{matrix}\right.\)
Vậy x = 3 ; y = 7
b, \(\left\{{}\begin{matrix}\left|x-2y+1\right|\ge0\\\left|x-y-2\right|\ge0\end{matrix}\right.\\ \Rightarrow\left|x-2y+1\right|+\left|x-y-2\right|\ge0\\ \Leftrightarrow\left|x-2y+1\right|+\left|x-y-2\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-2y+1=0\\x-y-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=5\\y=3\end{matrix}\right.\)
Vậy x = 5; y = 3