Tìm x,y biết:
\(x-3|^{2014}+|6+2y|^{2015}\le0.\)
\(-2x+4|-|y+5\ge0\)
Tìm x;y biết \(\text{|}x-3\text{|}^{2014}+\text{|}6+2y\text{|}^{2015}\le0\)
Ta có: \(\left|x-3\right|^{2014}\ge0;\left|6+2y\right|^{2015}\ge0\)
\(\Rightarrow\left|x-3\right|^{2014}+\left|6+2y\right|^{2015}\ge0\)
Mà theo đề: \(\left|x-3\right|^{2014}+\left|6+2y\right|^{2015}\le0\)
=> \(\left|x-3\right|^{2014}+\left|6+2y\right|^{2015}=0\)
=> \(\left|x-3\right|=\left|6+2y\right|=0\)
=> \(x-3=6+2y=0\)
=> \(x=3;y=-3\).
Tìm x,y biết:
\(\left|x-3\right|^{2014}+\left|6+2y\right|^{2015}\le0\)
Trả lời:(x;y)=
CÂU 2 :
a, Không dùng máy tính hãy so sánh : \(\dfrac{2014}{\sqrt{2015}}+\dfrac{2015}{\sqrt{2014}}\) và \(\sqrt{2014}+\sqrt{2015}\)
b, Tìm x, y, z biết : \(4x^2+2y^2+2z^2-4xy-2yz+2y-8z+10\le0\)
c, Giair phương trình : \(\sqrt{\dfrac{1}{x+3}}+\sqrt{\dfrac{5}{x+4}}=4\)
a) Ta có: \(\dfrac{2014}{\sqrt{2015}}+\dfrac{2015}{\sqrt{2014}}=\)
\(\dfrac{2015-1}{\sqrt{2015}}+\dfrac{2014+1}{\sqrt{2014}}=\sqrt{2015}-\dfrac{1}{\sqrt{2015}}+\sqrt{2014}+\dfrac{1}{\sqrt{2014}}\)
\(\left(\dfrac{1}{\sqrt{2014}}-\dfrac{1}{\sqrt{2015}}>0\right)\)\(>\sqrt{2014}+\sqrt{2015}\)
Vậy \(\dfrac{2014}{\sqrt{2015}}+\dfrac{2015}{\sqrt{2014}}>\sqrt{2014}+\sqrt{2015}\)
cho x;y thỏa mãn
\(\hept{\begin{cases}x-2y+4\le0\\3x+2y+12\ge0\\2x+5y-10\le0\end{cases}}\)
\(cmr:x^2+y^2\ge\frac{16}{5}\)
Cho : \(\left\{{}\begin{matrix}x-6y+18\ge0\\2x+4y+4\ge0\\3x-2y+6\le0\end{matrix}\right.\)
Tìm Min,Max, P=y-x.
Xác định miền nghiệm:
a, \(\left\{{}\begin{matrix}x+y+2>0\\2x-3y-6\le0\\x-2y+3\le0\\\left|y\right|>1\end{matrix}\right.\)
b, \(\left\{{}\begin{matrix}x+y-2\ge0\\x-3y+3\le0\\-1\le x\le1\end{matrix}\right.\)
tìm x ; y biết |x-3|^2014+|6+2y|2015<=0
Ta có:/x-3/^2014>=0;/6+2y/^2015>=0
=>/x-3/^2014+/6+2y/^2015>=0
mà theo đề bài, /x-3/^2014+/6+2y/^2015<=0
=>/x-3/^2014=/6+2y/^2015=0
=>/x-3/=0; /6+2y/=0
=>x-3=0 =>6+2y=0
=>x=3 =>2y=-6=>y=-3
vậy x=3; y=-3
3;-3 đúng 1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000%
Tìm Max của biểu thức F(x;y) = x+2y với điều kiện \(\left\{{}\begin{matrix}0\le y\le4\\x\ge0\\x-y-1\le0\\x+2y-10\le0\end{matrix}\right.\)
Tìm x;y biết |x-3|^2014+| 6+2y|^2015<0