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\(35-\left[\left(2x-3\right)^2:7\right]=28\)
\(\Rightarrow\left[\left(2x-3\right)^2:7\right]=35-28\)
\(\Rightarrow\left(2x-3\right)^2:7=7\)
\(\Rightarrow\left(2x-3\right)^2=1\)
\(\Rightarrow2x-3=\pm1\)
\(\Rightarrow x=2\) hay \(x=1\)
35 - [(2\(x\) - 3)2:7 ] = 28
(2\(x-3\))2 : 7 = 35 - 28
(2\(x\) - 3)2 : 7 = 7
(2\(x\) - 3)2 = 7 \(\times\) 7
(2\(x-3\))2 = 72
\(\left[{}\begin{matrix}2x-3=-7\\2x-3=7\end{matrix}\right.\)
\(\left[{}\begin{matrix}2x=-7+3\\2x=7+3\end{matrix}\right.\)
\(\left[{}\begin{matrix}2x=-4\\2x=10\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)
Vậy \(x\in\) {-2; 5}
Bài làm
\(\left(2x-1\right)^7=x^7\Leftrightarrow2x-1=x\Leftrightarrow x=1\)
\(\left(2x-1\right)^7=x^7\)
\(2x-1=x\)
2x-x=1
X.(2-1)=1
X.1=1
X=1:1
X=1
Vậy x=1
TH1 \(x\ge0;\left|2x-3\right|=2x-3\)
\(2x-3-5=7x+1\)
\(\Leftrightarrow2x-7x=3+5+1=9\)
\(\Leftrightarrow-5x=9\Rightarrow x=-\frac{9}{5}\left(ktm\right)\)
TH2:\(x< 0;\left|2x-3\right|=-\left(2x-3\right)\)
\(-\left(2x-3\right)-5=7x+1\)
\(\Leftrightarrow-2x+3-5=7x+1\)
\(\Leftrightarrow-2x-7x=-3+5+1=3\)
\(\Leftrightarrow-9x=3\Rightarrow x=-\frac{3}{9}=-\frac{1}{3}\left(tm\right)\)
Vậy \(x=-\frac{1}{3}\)
\(\left|2x-3\right|=x+1\)
\(\Rightarrow\orbr{\begin{cases}2x-3=x+1\\2x-3=-x-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x-x=1+3\\2x+x=-1+3\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=4\\3x=2\Rightarrow x=\frac{2}{3}\end{cases}}\)