Question

mọi người giúp mình với

Answer 1

\(2,\Leftrightarrow\left(x+1\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-5\end{matrix}\right.\\ 3,\Leftrightarrow x^2\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ 5,\Leftrightarrow x\left(x^2+6x+9-4\right)=0\\ \Leftrightarrow x\left[\left(x+3\right)^2-4\right]=0\\ \Leftrightarrow x\left(x+1\right)\left(x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-7\end{matrix}\right.\)

Answer 2

2: \(\Leftrightarrow\left(x+5\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-3\end{matrix}\right.\)

Answer 3

2)Ta có:  (x+3)² - 4 = 0

<=>(x+3+2)(x+3-2)=0

<=> (x+5)(x+1)=0

<=> \(\left[\begin{array}{} x+5=0\\ x+1=0 \end{array} \right.\)         <=>\(\left[\begin{array}{} x=-5\\ x=-1 \end{array} \right.\)

    vậy, x ∈ {-5; -1}

3)Ta có:  x⁴ - 9x² = 0

<=> (x²- 3x)(x²+3x)= 0

<=> x²(x-3)(x+3)=0

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

    Vậy, x ∈ {0;3;-3}

5) x³ + 6x² + 9x - 4x=0

<=> x(x² + 6x + 9 - 4)=0

<=> x[(x² + 6x + 9) -4] = 0

<=> x[(x+3)² - 2²] = 0

<=> x(x+3-2)(x+3+2) = 0

<=> x(x+1)(x+5)=0

<=> \(\left[\begin{array}{} x=0\\ x+1=0\\x+5=0 \end{array} \right.\)             <=>\(\left[\begin{array}{} x=0\\ x=-1\\x=-5 \end{array} \right.\)

        Vậy, x ∈ {0;-1;-5}