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\(2.16\ge2^n>4\)
\(2.2^4\ge2^n>2^2\)
\(2^5\ge2^n>2^2\)
=> \(n\in\left\{3,4,5\right\}\)
Vậy: \(n\in\left\{3,4,5\right\}\)
Bài 1:
a, Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) có:
\(A=\left|x-3\right|+\left|x-5\right|=\left|x-3\right|+\left|5-x\right|\ge\left|x-3+5-x\right|=\left|2\right|=2\)
Dấu " = " khi \(\left\{{}\begin{matrix}x-3\ge0\\5-x\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge3\\x\le5\end{matrix}\right.\)
Vậy \(MIN_A=2\) khi \(3\le x\le5\)
b, Ta có: \(\left\{{}\begin{matrix}\left|y^2-25\right|\ge0\\\left|x^2-4\right|\ge0\end{matrix}\right.\Rightarrow\left|y^2-25\right|+\left|x^2-4\right|\ge0\)
\(\Rightarrow B\ge3\)
Dấu " = " khi \(\left\{{}\begin{matrix}y^2-25=0\\x^2-4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=\pm5\\x=\pm2\end{matrix}\right.\)
Vậy \(MIN_B=2\) khi \(\left\{{}\begin{matrix}x=\pm2\\y=\pm5\end{matrix}\right.\)
Bài 2:
a, Xét \(x\ge-2\) có:
\(A=3x-3x-6-12=-18\)
+) Xét x < -2 có:
\(A=3x+3x+6-12=6x-6\)
Vậy...
b, tương tự
\(3-\frac{2}{3}+\frac{3}{5}\cdot\left(-\frac{10}{9}-\frac{25}{3}\right)-\frac{5}{6}\)
\(=3-\frac{2}{3}+\frac{3}{5}\cdot\left(-\frac{10}{9}-\frac{75}{9}\right)-\frac{5}{6}\)
\(=3-\frac{2}{3}+\frac{3}{5}\cdot-\frac{85}{9}-\frac{5}{6}\)
\(=3-\frac{2}{3}+\left(-\frac{17}{3}\right)-\frac{5}{6}\)
\(=\frac{-25}{6}\)
\(3-\frac{2}{3}+\frac{3}{5}.\left(\frac{-10}{9}-\frac{25}{3}\right)-\frac{5}{6}\)
\(=3-\frac{2}{3}+\frac{3}{5}.\left(\frac{-10}{9}-\frac{75}{9}\right)-\frac{5}{6}\)
\(=3-\frac{2}{3}+\frac{3}{5}.\frac{-85}{9}-\frac{5}{6}\)
\(=3-\frac{2}{3}+\frac{3.\left(-85\right)}{5.9}-\frac{5}{6}\)
\(=3-\frac{2}{3}+\frac{1.\left(-17\right)}{1.3}-\frac{5}{6}\)
\(=3-\frac{2}{3}+\frac{-17}{3}-\frac{5}{6}\)
\(=\frac{3}{1}-\frac{2}{3}+\frac{-17}{3}-\frac{5}{6}\)
\(=\frac{18}{6}-\frac{4}{6}+\frac{-34}{6}-\frac{5}{6}\)
\(=\frac{18-4+\left(-34\right)-5}{6}\)
\(=\frac{-25}{6}\)
