[288:(x-3)2-2] . (x2-169)=0
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a: \(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)
b: \(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
a: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\\x=1\end{matrix}\right.\)
d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)
\(\Leftrightarrow x+3=0\)
hay x=-3
Bài 1:
a: Ta có: \(48751-\left(10425+y\right)=3828:12\)
\(\Leftrightarrow y+10425=48751-319=48432\)
hay y=38007
b: Ta có: \(\left(2367-y\right)-\left(2^{10}-7\right)=15^2-20\)
\(\Leftrightarrow2367-y=1222\)
hay y=1145
Bài 2:
Ta có: \(8\cdot6+288:\left(x-3\right)^2=50\)
\(\Leftrightarrow288:\left(x-3\right)^2=2\)
\(\Leftrightarrow\left(x-3\right)^2=144\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)
a) (x - 3)2 - 5.(x - 2) + 5 = 0.
<=> x^2 - 6x + 9 - 5x + 10 + 5 = 0
<=> x^2 - 11x + 24 = 0
<=> (x-3)(x-8)=0
<=> x = 3 hoặc x = 8
\(...\Rightarrow\left[{}\begin{matrix}288:\left(x-3\right)^2-2=0\\x^2-169=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}288:\left(x-3\right)^2=2\\x^2=169\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-3\right)^2=288:2\\x^2=13^2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left(x-3\right)^2=144=12^2\\x^2=13^2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-3=12\\x-3=-12\\x=13\\x=-13\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=15\\x=-9\\x=13\\x=-13\end{matrix}\right.\) \(\Rightarrow x\in\left\{-9;15;\pm13\right\}\)