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b) \(x^2-7x=0\)
\(\Rightarrow x\left(x-7\right)=0\)
\(\Rightarrow\left\{\begin{matrix}x=0\\x-7=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=0\\x=7\end{matrix}\right.\)
Vậy \(x\in\left\{0;7\right\}\)
c) \(x^2=-5x\)
\(\Rightarrow x^2+5x=0\)
\(\Rightarrow x\left(x+5\right)=0\)
\(\Rightarrow\left\{\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Vậy \(x\in\left\{0;-5\right\}\)
a) \(\left(x-5\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x-5=0\\x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{\begin{matrix}x=5\\x=-4\end{matrix}\right.\)
Vậy...
b) \(x^2-7x=0\)
\(\Leftrightarrow x\left(x-7\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x=0\\x-7=0\Leftrightarrow x=7\end{matrix}\right.\)
Vậy...
c) \(x^2=-5x\)
\(\Leftrightarrow x=-5\)
d) \(x^3=x\)
\(\Leftrightarrow\left[\begin{matrix}x=1\\x=-1\\x=0\end{matrix}\right.\)

a) \(\left(4x-13\right)^4+4^3=145\)
\(\Rightarrow\left(4x-13\right)^4+64=145\)
\(\Rightarrow\left(4x-13\right)^4=81\)
\(\Rightarrow4x-13=\pm3\)
+) \(4x-13=3\)
\(\Rightarrow4x=16\)
\(\Rightarrow x=4\)
+) \(4x-13=-3\)
\(\Rightarrow4x=10\)
\(\Rightarrow x=\frac{5}{2}\)
Vậy \(x=4\) hoặc \(x=\frac{5}{2}\)
b) \(3^{x+2}-3^x=72\)
\(\Rightarrow3^x.3^2-3^x=72\)
\(\Rightarrow3^x.\left(3^2-1\right)=72\)
\(\Rightarrow3^x.8=72\)
\(\Rightarrow3^x=9\)
\(\Rightarrow3^x=3^2\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
c) \(2^{x+2}-2^{x-1}=224\)
\(\Rightarrow2^{x-1+3}-2^{x-1}=224\)
\(\Rightarrow2^{x-1}.2^3-2^{x-1}=224\)
\(\Rightarrow2^{x-1}.\left(2^3-1\right)=224\)
\(\Rightarrow2^{x-1}.7=224\)
\(\Rightarrow2^{x-1}=32\)
\(\Rightarrow2^{x-1}=2^5\)
\(\Rightarrow x-1=5\)
\(\Rightarrow x=6\)
Vậy x = 6


\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
\(\Rightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\1-\left(x-7\right)^{10}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\\left[{}\begin{matrix}x-7=-1\\x-7=1\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=6\\x=8\end{matrix}\right.\)
Vậy x = 7 hoặc x = 6 hoặc x = 8
\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\\ \left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\1-\left(x-7\right)^{10}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-7=0\\x-7=\pm1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=7\\x=\pm1+7\end{matrix}\right.\)
vậy x={6;7;8}
x2 - x = 0
<=> x(x-1) = 0
<=> \(\left[\begin{array}{nghiempt}x=0\\x-1=0\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}x=0\\x=1\end{array}\right.\)
bài này mk cx cần nà , tks ông nhá!!!!!!!