Question

Answer 1

Bài 1: 

b: Ta có: \(f\left(x\right)=x^2-4x+9\)

\(=x^2-4x+4+5\)

\(=\left(x-2\right)^2+5\ge5\forall x\)

Dấu '=' xảy ra khi x=2

Answer 2

Bài 3:

a. x² + 3x = 0

<=> x(x + 3) = 0

<=> \(\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)

b. x³ - 4x = 0

<=> x(x² - 4) = 0

<=> x(x - 2)(x + 2) = 0

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

<=> \(\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)

c. x² + 5x = 6

<=> x² + 5x - 6 = 0

<=> x² - x + 6x - 6 = 0

<=> x(x - 1) + 6(x - 1) = 0

<=> (x + 6)(x - 1) = 0

<=> \(\left[{}\begin{matrix}x+6=0\\x-1=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=-6\\x=1\end{matrix}\right.\)

d. x² - 2015x + 2014 = 0

<=> x² - x - 2014x + 2014 = 0

<=> x(x - 1) - 2014(x - 1) = 0

<=> (x - 2014)(x - 1) = 0

<=> \(\left[{}\begin{matrix}x-2014=0\\x-1=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=2014\\x=1\end{matrix}\right.\)