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\(|x^2-5x+4|=4x-4\)
\(\Rightarrow|\left(x-1\right)\left(x-4\right)|=4\left(x-1\right)\)
\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)\left(x-4\right)=4\left(x-1\right)\\\left(x-1\right)\left(x-4\right)=-4\left(x-1\right)\end{cases}}\)
TH1 :
\(\left(x-1\right)\left(x-4\right)=4\left(x-1\right)\)
\(\Rightarrow\left(x-1\right)\left(x-4\right)-4\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x-8\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=8\end{cases}}\)
Th2:
\(\left(x-1\right)\left(x-4\right)=-4\left(x-1\right)\)
\(\Rightarrow\left(x-1\right)\left(x-4\right)+4\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)x=0\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=0\end{cases}}\)
\(\Rightarrow S=\left\{0;1;8\right\}\)
\(\left(x+3\right)\left(5x-20\right)=4-x\)
\(\Leftrightarrow5x^2-5x-60=4-x\)
\(\Leftrightarrow5x^2-4x-64=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=-\frac{16}{5}\end{cases}}\)
P/s haphuong
b: \(\left(2x+1\right)^2=25\)
=>\(\left[{}\begin{matrix}2x+1=5\\2x+1=-5\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2x=4\\2x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
c: \(\left(1-3x\right)^3=64\)
=>\(\left(1-3x\right)^3=4^3\)
=>1-3x=4
=>3x=1-4=-3
=>x=-3/3=-1
d: \(\left(4-x\right)^3=-27\)
=>\(\left(4-x\right)^3=\left(-3\right)^3\)
=>4-x=-3
=>x=4+3=7
e: \(x^2-5x=0\)
=>\(x\left(x-5\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
ngốc quá câu dễ thế mà cũng đăng ak