1. Tìm x,y biết:
\(\left(x-2\right)^{20}+\left(y+1\right)^{30}\le0\)
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15 tháng 9 2021
a) \(x^2+\left(y-\dfrac{1}{10}\right)^4=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y-\dfrac{1}{10}=0\end{matrix}\right.\)( do \(x^2\ge0,\left(y-\dfrac{1}{10}\right)^4\ge0\))
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{1}{10}\end{matrix}\right.\)
b) \(\left(\dfrac{1}{2}.x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x-5=0\\y^2-\dfrac{1}{4}=0\end{matrix}\right.\)( do \(\left(\dfrac{1}{2}x-5\right)^{20}\ge0,\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\))
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x=5\\y^2=\dfrac{1}{4}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=\pm\dfrac{1}{2}\end{matrix}\right.\)
15 tháng 9 2021
\(a,\Leftrightarrow\left\{{}\begin{matrix}x=0\\y-\dfrac{1}{10}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{1}{10}\end{matrix}\right.\\ b,\left\{{}\begin{matrix}\left(\dfrac{1}{2}x-5\right)^{20}\ge0\\\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\end{matrix}\right.\Leftrightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\)
Mà \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\)
\(\Leftrightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}=0\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x=5\\y^2=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=\pm\dfrac{1}{2}\end{matrix}\right.\)
10 tháng 7 2018
Vì \(\left(\frac{1}{2}x-5\right)^{10}\ge0\)và \(\left(y^2-\frac{1}{4}\right)^{20}\ge0\)
nên \(\left(\frac{1}{2}x-5\right)^{10}+\left(y^2-\frac{1}{4}\right)^{20}=0\)
<=>\(\hept{\begin{cases}\frac{1}{2}x-5=0\\y^2-\frac{1}{4}=0\end{cases}}\)<=>\(\hept{\begin{cases}x=10\\y=\pm\frac{1}{2}\end{cases}}\)
DQ
12 tháng 2 2020
Ta có:\(\hept{\begin{cases}\left\{\frac{1}{2}x-5\right\}^{10}\ge0\forall x\\\left\{y^2-\frac{1}{4}\right\}^{20}\ge0\forall y\end{cases}}\)
Mà \(\left\{\frac{1}{2}x-5\right\}^{10}+\left\{y^2-\frac{1}{4}\right\}^{20}\le0\)
\(\Rightarrow\left\{\frac{1}{2}x-5\right\}^{10}+\left\{y^2-\frac{1}{4}\right\}^{20}=0\)
\(\Leftrightarrow\hept{\begin{cases}\left\{\frac{1}{2}x-5\right\}^{10}=0\\\left\{y^2-\frac{1}{4}\right\}^{20}=0\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{1}{2}x-5=0\\y^2-\frac{1}{4}=0\end{cases}\Leftrightarrow}\hept{\begin{cases}\frac{1}{2}x=5\\y^2=\frac{1}{4}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=10\\y=\pm\frac{1}{2}\end{cases}}}\)
Vậy \(x=10;y=\pm\frac{1}{2}\)
28 tháng 9 2021
a) \(\left|3x-\dfrac{1}{2}\right|+\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|=0\)
Do \(\left|3x-\dfrac{1}{2}\right|,\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|\ge0\forall x,y\)
\(\Rightarrow\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\\dfrac{1}{4}y+\dfrac{3}{5}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-\dfrac{12}{5}\end{matrix}\right.\)
b) \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|+\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\le0\)
Do \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|,\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\ge0\forall x,y\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{9}=0\\\dfrac{5}{7}y-\dfrac{1}{2}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{27}\\y=\dfrac{7}{10}\end{matrix}\right.\)
MA
1
17 tháng 9 2017
Để\(\left(\dfrac{1}{2}\times x-5\right)^{20}+\left(y^2-\dfrac{1}{2}\right)^{20}\le0\) thì \(\left(\dfrac{1}{2}\times x-5\right)\)và \(\left(y^2-\dfrac{1}{2}\right)\le0\) .
Để \(\left(\dfrac{1}{2}\times x-5\right)\)và \(\left(y^2-\dfrac{1}{2}\right)\le0\) thì \(\dfrac{1}{2}\times x\le5\) và \(y^2\le\dfrac{1}{2}\).
Vậy ta có :
\(\dfrac{x}{2}\le5\Rightarrow x\le10\)
\(y^2\le\dfrac{1}{2}\Rightarrow y\le\dfrac{1}{4}\)
\(\left(x-2\right)^{20}\ge0\forall x\\ \left(y+1\right)^{30}\ge0\forall x\\ \Rightarrow\left(x-2\right)^{20}+\left(y+1\right)^{30}\ge0\forall x\)
Mà \(\left(x-2\right)^{20}+\left(y+1\right)^{30}=0\)
Để thỏa mãn điều kiện thì \(\left\{{}\begin{matrix}\left(x+2\right)^{20}=0\\\left(y+1\right)^{30}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x+2=0\\y+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\)
Vậy ...
ta có : \(\left(x-2\right)^{20}\ge0\) với mọi x
và \(\left(y+1\right)^{30}\ge0\) với mọi y
\(\Rightarrow\) \(\left(x-2\right)^{20}+\left(y+1\right)^{30}\ge0\) với mọi giá trị của x ; y
mà \(\left(x-2\right)^{20}+\left(y+1\right)^{30}\le0\)
\(\Rightarrow\) \(\left(x-2\right)^{20}+\left(y+1\right)^{30}=0\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)^{20}=0\\\left(y+1\right)^{30}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\) vậy \(x=2;y=-1\)