Question

Giải các phương trình sau:

a) \(x^2+x-6=0\)

b) \(x^2+x+4=0\)

c) \(t^2+6t-72=0\)

d) \(x^2-9x+8=0\)

e) \(x^2-9x+26=0\)

Answer 1

a) PT <=> \(\left(x^2-2x\right)+\left(3x-6\right)=0\)

<=> \(x\left(x-2\right)+3\left(x-2\right)=0\)

<=> \(\left(x-2\right)\left(x+3\right)=0\)

<=> \(\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

KL: ...

b) \(PT< =>\left(x^2+x+\frac{1}{4}\right)+\frac{15}{4}=0\)

<=> \(\left(x+\frac{1}{2}\right)^2=\frac{-15}{4}\)

<=> x = \(\varnothing\)

c) PT <=> \(\left(t^2-6t\right)+\left(12t-72\right)=0\)

<=> \(t\left(t-6\right)+12\left(t-6\right)=0\)

<=> \(\left(t+12\right)\left(t-6\right)=0\)

<=> \(\left[{}\begin{matrix}t=-12\\t=6\end{matrix}\right.\)

d) PT <=> \(\left(x^2-x\right)-\left(8x-8\right)=0\)

<=> \(x\left(x-1\right)-8\left(x-1\right)=0\)

<=> \(\left(x-1\right)\left(x-8\right)=0\)

<=> \(\left[{}\begin{matrix}x=1\\x=8\end{matrix}\right.\)

e) PT <=> \(\left(x^2-9x+\frac{81}{4}\right)+\frac{23}{4}\)

<=> \(\left(x-\frac{9}{2}\right)^2=\frac{-23}{4}\)

<=> x = \(\varnothing\)

Answer 2

a) \(x^2+x-6=0\)

\(\Leftrightarrow x^2+3x-2x-6=0\)

\(\Leftrightarrow x\left(x+3\right)-2\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)

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

Vậy : \(S=\left\{2;-3\right\}\)