a)|x - y -5| + 2019|y - 3|\(^{^{2020}}\) = 0b)2(x - 5)\(^4\) + 5|2y - 7|\(^5\) = 0c)(x + 3y - 1)\(^2\) + (2y - \(\frac{1}{2}\))\(^{^{2020}}\) = 0d)3|x - y|\(^5\) + 10|y + \(\frac{2}{3}\)|\(^7\) _< 0e)\(\frac{1}{2}\)(\(\frac{3}{4}\)x - \(\frac{1}{2}\))\(^{2006}\)+ |\(\frac{4}{5}\)y + \(\frac{6}{25}\)| =...
Đọc tiếp
a)|x - y -5| + 2019|y - 3|\(^{^{2020}}\) = 0
b)2(x - 5)\(^4\) + 5|2y - 7|\(^5\) = 0
c)(x + 3y - 1)\(^2\) + (2y - \(\frac{1}{2}\))\(^{^{2020}}\) = 0
d)3|x - y|\(^5\) + 10|y + \(\frac{2}{3}\)|\(^7\) _< 0
e)\(\frac{1}{2}\)(\(\frac{3}{4}\)x - \(\frac{1}{2}\))\(^{2006}\)+ |\(\frac{4}{5}\)y + \(\frac{6}{25}\)| = 0
khó quá
a. Vì \(\left|x-y-5\right|\ge0\forall x;y;2019\left|y-3\right|^{2020}\ge0\forall y\)
\(\Rightarrow\left|x-y-5\right|+2019\left|y-3\right|^{2020}\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\orbr{\begin{cases}\left|x-y-5\right|=0\\2019\left|y-3\right|^{2020}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-y-5=0\\y-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-y=5\\y=3\end{cases}}\)
b. \(2\left(x-5\right)^4\ge0\forall x;5\left|2y-7\right|^5\ge0\forall y\)
\(\Rightarrow2\left(x-5\right)^4+5\left|2y-7\right|^5\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\orbr{\begin{cases}2\left(x-5\right)^4=0\\5\left|2y-7\right|^5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-5=0\\2y-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\y=\frac{7}{2}\end{cases}}\)