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Áp dụng bđt |a|+|b|+|c|+|d| \(\ge\)|a+b+c+d| ta có:
\(\left|x-1\right|+\left|x-3\right|+\left|x-5\right|+\left|x-7\right|\)\(=\left|x-1\right|+\left|x-3\right|+\left|5-x\right|+\left|7-x\right|\)\(\ge\left|x-1+x-3+5-x+7-x\right|=\left|8\right|=8\)
Dấu "=" xảy ra khi \(\left\{\begin{matrix}x-1\ge0\\x-3\ge0\\x-5\le0\\x-7\le0\end{matrix}\right.\)\(\Rightarrow\left\{\begin{matrix}x-3\ge0\\x-5\le0\end{matrix}\right.\)\(\Leftrightarrow3\le x\le5\)
Mà x nguyên nên \(x\in\left\{3;4;5\right\}\)
Vậy \(x\in\left\{3;4;5\right\}\)
a) \(\left|x-1\right|+3x=5\)
\(\Leftrightarrow\left|x-1\right|=5-3x\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=5-3x\\x-1=3x-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=2\end{cases}}\)
b) \(\left|5x-3\right|-x=7\)
\(\Leftrightarrow\left|5x-3\right|=7+x\)
\(\Leftrightarrow\orbr{\begin{cases}5x-3=7+x\\5x-3=-x-7\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{-2}{3}\end{cases}}\)
\(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{4}\right)\)
\(\Leftrightarrow-\frac{1}{12}\le\frac{x}{12}< \frac{7}{12}\)
=> x \(\in\) {-1;0;1;2;3;4;5;6}
\(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{4}\right)\)
\(\Leftrightarrow\)\(\frac{9-10}{12}\le\frac{x}{12}< 1-\left(\frac{8-3}{12}\right)\)
\(\Leftrightarrow\)\(-\frac{1}{12}\le\frac{x}{12}< \frac{7}{12}\)
\(\Leftrightarrow-1\le x< 7\)
Mà x nguyên
=>x={-1;0;1;2;3;4;5;6}
\(\left(x-2\right)^8=\left(x-2\right)^6\)
\(\Leftrightarrow\left(x-2\right)^8-\left(x-2\right)^6=0\)
\(\Leftrightarrow\left(x-2\right)^6\left[\left(x-2\right)^2-1\right]=0\)
\(\Leftrightarrow\left(x-2\right)^6\left(x-3\right)\left(x-1\right)=0\)
\(\Rightarrow x=2;x=3;x=1\)
=>x-2=0 hoặc x-2=1
=>x-2=0=>x=2
=>x-2=1=>x=3