xmu2/6= 24/25
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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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\(\Leftrightarrow B=-\left(x^2-4x-1\right)\)
\(\Leftrightarrow B=-\left(x^2-4x+4-5\right)\)
\(\Leftrightarrow B=-\left(x-2\right)^2+5\)
Ta có \(\left(x-2\right)^2\ge0\)với mọi x
\(\Leftrightarrow-\left(x-2\right)^2\le0\)
\(\Leftrightarrow-\left(x-2\right)^2+5\le0+5\)
hay \(B\) \(\le5\)
Dấu "=" xảy ra khi \(\left(x-2\right)^2=0\)
. \(\Leftrightarrow x-2\)\(=0\)
\(\Leftrightarrow\)\(x\) \(=2\)
Vậy min B=5 tại x=2
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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,\(\left(x-5\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-9=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=9\end{matrix}\right.\)
Vậy...
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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\(A=x^2-4x-50=\left(x-2\right)^2-54\ge-54\)
Dấu "=" xảy ra <=> \(x=2\)
Vậy MIN \(A=-54\)khi \(x=2\)