Tìm x,y biết |x-3|^2014+|6+2y|^2015<hoặc=0 Trả lời:( x,y ) (Nhập các giá trị theo thứ tự,cách nhau bởi dấu ";")
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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
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Xem thêm câu trả lời
Tìm x;y biết |x-3|^2014+| 6+2y|^2015<0
Tìm x;y biết
/x-3/ ^ 2014 + /6+2y/^2015 lớn hơn hoặc bằng 0
x=3;y=-3
chưa tìm ra cách giải nhưng thử vào đúng !!
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Tìm x;y biết|x-3|2014+|6+2y|2015 <=0
Tìm x;y biết |X-3|2014+|6+2y|2015<hoặc bằng 0
Vì \(\left|x-3\right|^{2014}\ge0;\left|6+2y\right|^{2015}\ge0\)
\(\Rightarrow\left|x-3\right|^{2014}+\left|6+2y\right|^{2015}\ge0\)
Mà đề lại cho : \(\left|x-3\right|^{2014}+\left|6+2y\right|^{2015}\le0\Rightarrow\left|x-3\right|^{2014}=0;\left|6+2y\right|^{2015}=0\)
\(\Rightarrow x-3=0;6+2y=0\Rightarrow x=3;y=-3\)
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Tìm x , y biết :| x - 3|2014 + |6 + 2y| 2015 < 0
Tìm x;y biết | x - 3 |2014+| 6+2y |2015 nhỏ hơn hoặc bằng 0
Ta có:\(\hept{\begin{cases}\left|x-3\right|\ge0\\\left|6+2y\right|\ge0\end{cases}\Rightarrow\hept{\begin{cases}\left|x-3\right|^{2014}\ge0\\\left|6+2y\right|^{2015}\ge0\end{cases}\Rightarrow}\left|x-3\right|^{2014}+\left|6+2y\right|^{2015}\ge0}\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}\left|x-3\right|^{2014}=0\\\left|6-2y\right|^{2015}=0\end{cases}\Rightarrow\hept{\begin{cases}x=3\\y=3\end{cases}}}\)
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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: \(\left|x-3\right|^{2014}+\left|6+2y\right|^{2015}\le0\)
=> \(\left|x-3\right|^{2014}+\left|6+2y\right|^{2015}=0\)
=> \(\left|x-3\right|=\left|6+2y\right|=0\)
=> \(x-3=6+2y=0\)
=> \(x=3;y=-3\).
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Tìm x; y biết |x - 3|2014 + | 6 + 2y| 2015 \(\le\) 0. Trả lời (x; y)= (...)
Vì |x - 3|2014 ≥ 0 ; |6 + 2y|2015 ≥ 0
=> |x - 3|2014 + |6 + 2y|2015 ≥ 0
Mà để |x - 3|2014 + |6 + 2y|2015 ≤ 0 <=> |x - 3|2014 = 0 ; |6 + 2y|2015 = 0
=> x = 3 và y = - 3
Vậy x = 3 và y = - 3
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