\(\left(2x-1\right)^{2016}\) + \(\left|3y-1\right|\)\(\le\) 0
Tìm x , y
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4 tháng 1 2021
Ta có: \(3x^2+3y^2+4xy+2x-2y+2=0\)
\(\Leftrightarrow x^2+2x+1+y^2-2y+1+2x^2+4xy+2y^2=0\)
\(\Leftrightarrow\left(x+1\right)^2+\left(y-1\right)^2+2\left(x^2+2xy+y^2\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2+\left(y-1\right)^2+2\left(x+y\right)^2=0\)
Ta có: \(\left(x+1\right)^2\ge0\forall x\)
\(\left(y-1\right)^2\ge0\forall y\)
\(2\left(x+y\right)^2\ge0\forall x,y\)
Do đó: \(\left(x+1\right)^2+\left(y-1\right)^2+2\left(x+y\right)^2\ge0\forall x,y\)
Dấu '=' xảy ra khi
\(\left\{{}\begin{matrix}x+1=0\\y-1=0\\x+y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\\-1+1=0\left(đúng\right)\end{matrix}\right.\)
Thay x=-1 và y=1 vào biểu thức \(M=\left(x+y\right)^{2016}+\left(x+2\right)^{2017}+\left(y-1\right)^{2018}\), ta được:
\(M=\left(-1+1\right)^{2016}+\left(-1+2\right)^{2017}+\left(1-1\right)^{2018}\)
\(=0^{2016}+1^{2017}+0^{2018}=1\)
Vậy: M=1
9 tháng 3 2021
\(C=2\left(x-y\right)+13x^3y^2\left(x-y\right)-15xy\left(x-y\right)+1=1\)
Vậy C=1
9 tháng 3 2021
\(C=2x-2y+13x^3y^2\left(x-y\right)+15\left(y^2x-x^2y\right)+\left(\dfrac{2015}{2016}\right)^0\)
\(C=2\left(x+y\right)+13x^3y^2\left(x-y\right)+15xy\left(x-y\right)+1\)
Mà x - y = 0 (bài cho)
\(\Rightarrow C=2.0+13x^3y^2.0+15xy.0+1\)
\(C=1\)
Vậy C=1
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 7 2021
a.
\(\left\{{}\begin{matrix}\left(x-1\right)^2-\left(y+1\right)^2=0\\x+3y-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1-y-1\right)\left(x-1+y+1\right)=0\\x+3y-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y-2\right)\left(x+y\right)=0\\x+3y-5=0\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x-y-2=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{4}\\y=\dfrac{3}{4}\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x+y=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=\dfrac{5}{2}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 7 2021
b.
\(\left\{{}\begin{matrix}xy-2x-y+2=0\\3x+y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(y-2\right)-\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)
TH1:
\(\left\{{}\begin{matrix}x-1=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\)
TH2:
\(\left\{{}\begin{matrix}y-2=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
LC
3
17 tháng 3 2017
Vì \(\left(2x-1\right)^{2016}\ge0;\left(3y+6\right)^{2014}\ge0;\left(z-1\right)^{2012}\ge0\)
\(\Rightarrow\left(2x-1\right)^{2016}+\left(3y+6\right)^{2014}+\left(z-1\right)^{2012}\ge0\)
Để \(\left(2x-1\right)^{2016}+\left(3y+6\right)^{2014}+\left(z-1\right)^{2012}=0\)\(\Leftrightarrow\left(2x-1\right)^{2016}=0;\left(3y+6\right)^{2014}=0;\left(z-1\right)^{2012}=0\)
\(\Leftrightarrow2x-1=0;3y+6=0;z-1=0\)
\(\Rightarrow x=\dfrac{1}{2};y=-2;z=1\)
\(\Rightarrow4x+y-3z=4.\dfrac{1}{2}+\left(-2\right)-3.1=2-2-3=-3\)
(2x-1)2016>(=)0
l3y-1l>(=)0
=>(2x-1)2016+l3y-1l>(=)0
mà (2x-1)2016+l3y-1l<(=)0
=>(2x-1)2016=l3y-1l=0
=>2x-1=3y-1=0
=>x=1/2;y=1/3
vậy x=1/2;y=1/3