Question

Bài 7:Tìm nghiệm của đa thức
a) 4x + 9                     f) x2 – 2x.

b) -5x+6                      g) (x – 4)(x^2 + 1)

c) x2 – 1                      h) 3x2 – 4x

d) x2 – 9.                     i) x^2 + 9
e) x2 – x.

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Answer 1

a) \(4x+9=0\Leftrightarrow4x=-9\Leftrightarrow x=-\dfrac{9}{4}\)

b) \(-5x+6=0\Leftrightarrow5x=6\Leftrightarrow x=\dfrac{6}{5}\)

c) \(x^2-1=0\Leftrightarrow\left(x-1\right)\left(x+1\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

d) \(x^2-9=0\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

e) \(x^2-x=0\Leftrightarrow x\left(x-1\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

f) \(x^2-2x=0\Leftrightarrow x\left(x-2\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

g) \(\left(x-4\right)\left(x^2+1\right)=0\Leftrightarrow x-4=0\Leftrightarrow x=4\)( do \(x^2+1\ge1>0\))

h) \(3x^2-4x=0\Leftrightarrow x\left(3x-4\right)=0\Leftrightarrow\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{3}\end{matrix}\right.\)

i) \(x^2+9=0\Leftrightarrow x^2=-9\)( vô lý do \(x^2\ge0>-9\))

Vậy \(x\in\left\{\varnothing\right\}\)