xmu2-16=0
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a,\(\left(x-7\right)\left(x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x+7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-7\end{matrix}\right.\)
Vậy...
b,\(\left(x-5\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-9=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=9\end{matrix}\right.\)
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c,\(\left(x-5\right)\left(x^2-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x^2-9=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=3\end{matrix}\right.\)
Vậy...
Câu d bạn viết lại đề nhé
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\(\dfrac{x^2-3x}{2x^2-3x-9}=\dfrac{x^2+3x}{A}\)
\(\Rightarrow A=\dfrac{\left(x^2+3x\right)\left(2x^2-3x-9\right)}{x^2-3x}\)
\(\Rightarrow A=\dfrac{x\left(x+3\right)\left(2x^2-3x-9\right)}{x\left(x-3\right)}\)
\(\Rightarrow A=\dfrac{\left(x+3\right)\left(2x^2-3x-9\right)}{\left(x-3\right)}\)
mà \(x=-\dfrac{3}{2}\)
\(\Rightarrow A=\dfrac{\left(-\dfrac{3}{2}+3\right)\left(2\left(-\dfrac{3}{2}\right)^2-3\left(-\dfrac{3}{2}\right)-9\right)}{\left(-\dfrac{3}{2}-3\right)}\)
\(\Rightarrow A=\dfrac{\dfrac{3}{2}\left(2.\dfrac{9}{4}+\dfrac{9}{2}-9\right)}{-\dfrac{9}{2}}\)
\(\Rightarrow A=\dfrac{\dfrac{3}{2}\left(\dfrac{9}{2}+\dfrac{9}{2}-9\right)}{-\dfrac{9}{2}}\)
\(\Rightarrow A=\dfrac{\dfrac{3}{2}\left(\dfrac{9}{2}+\dfrac{9}{2}-9\right)}{-\dfrac{9}{2}}=0\)
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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: \(x^2-5x=0\)
=>\(x\left(x-5\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
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x^2 -16=0
x^2=0+16
x^2=16
Có 4^2=16
=> x=4
\(x^2-16=0\)
\(x^2=0+16\)
\(x^2=16\)
\(x^2=4^2\)
\(=>x=4\)