Ta có:
\(32x^2-36x+13=0\\ =4\left(8x^2-9x+\dfrac{13}{4}\right)\\ =32\left(x^2-\dfrac{9}{8}x+\dfrac{13}{32}\right)\\ =32\left[\left(x^2-2\cdot x\cdot\dfrac{9}{16}+\dfrac{81}{256}\right)+\dfrac{23}{256}\right]\\ =32\left(x-\dfrac{9}{16}\right)^2+\dfrac{23}{8}\)
Vì: `(x-9/16)^2>=0` với mọi x
`=>32(x-9/16)^2>=0` với mọi x
`=>32(x-9/16)^2+23/8>=23/8>0` với mọi x
`=>` Vô nghiệm
\(32x^2-36x+13=0\)
\(\Leftrightarrow4x^2-\dfrac{9}{2}x+\dfrac{13}{8}=0\)
\(\Leftrightarrow\left(2x\right)^2-2.2x.\dfrac{9}{8}+\left(\dfrac{9}{8}\right)^2-\left(\dfrac{9}{8}\right)^2+\dfrac{13}{8}=0\)
\(\Leftrightarrow\left(2x-\dfrac{9}{8}\right)^2+\dfrac{23}{64}=0\)
\(\Leftrightarrow\left(2x-\dfrac{9}{8}\right)^2=-\dfrac{23}{64}\)
\(\Rightarrow\)Vô lý vì \(\left(2x-\dfrac{9}{8}\right)^2\ge0\forall x\)
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