Question

giải các phương trình sau:

a) (x² + 3x+ 2)(x² + 3x+ 3) - 2= 0

b) (x² + x - 2)(x² +x - 3) = 12

c) x(x + 1)(x - 1)(x + 2) = 24

d) (x + 1)(x + 2)(x + 3)(x + 4) - 24 = 0

Mình cần gấp ạ

Answer 1

a) Đặt x² + 3x + 2 = a

<=> a(a + 1) - 2 = 0

<=> a² + a - 2 = 0

<=> a² + 2a - a - 2 = 0

<=> (a - 1)(a + 2) = 0

<=> \(\left[{}\begin{matrix}a-1=0\\a-2=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x^2+3x+2-1=0\\x^2+3x+2-2=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}\left(x^2+3x+\frac{9}{4}\right)=\frac{5}{4}\\x\left(x+3\right)=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}\left(x+\frac{3}{2}\right)^2=\frac{5}{4}\\\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}\left[{}\begin{matrix}x+\frac{3}{2}=\sqrt{\frac{5}{4}}\\x+\frac{3}{2}=-\sqrt{\frac{5}{4}}\end{matrix}\right.\\\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}\left[{}\begin{matrix}x=\frac{\sqrt{5}-3}{2}\\x=\frac{-\sqrt{5}-3}{2}\end{matrix}\right.\\\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\end{matrix}\right.\)

Vậy S = {\(\frac{\sqrt{5}-3}{2}\); \(\frac{-\sqrt{5}-3}{2}\); 0; 3}

b) Đặt x² + x = b

<=> (b - 2)(b - 3) = 12

<=> n² - 3b - 2b + 6 - 12 = 0

<=> b² - 5b - 6 = 0

<=> b² - 6b + b - 6 = 0

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

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

<=> \(\left[{}\begin{matrix}x^2+x-6=0\\x^2+x+1=0\left(vn\right)\end{matrix}\right.\)

<=> x² + 3x - 2x - 6 = 0

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

<=> x = -3 hoặc x = 2

Vậy S = {-3; 2}

Answer 2

c) x(x + 1)(x - 1)(x + 2) = 24

<=> (x² + x)(x² + x - 2) - 24 = 0

Đặt x² + x = t

<=> t(t - 2) - 24 = 0

<=> t² - 2t - 24 = 0

<=> t² - 6t + 4t - 24 = 0

<=> (t + 4)(t - 6) = 0

<=> \(\left[{}\begin{matrix}t+4=0\\t-6=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x^2+x+4=0\left(vn\right)\\x^2+x-6=0\end{matrix}\right.\)(* vn là vô nghiệm)

<=> x² + 3x - 2x - 6 = 0

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

<=> x = -3 hoặc x = 2

Vậy S = {-3; 2}

d) (x + 1)(x + 2)(x + 3)(x + 4) - 24 = 0

<=> (x² + 5x +4)(x² + 5x + 6) - 24 = 0

Đặt x² + 5x = y

<=> (y + 4)(y + 6) - 24 = 0

<=> y² + 10y + 24 - 24 = 0

<=> y(y + 10) = 0

<=> \(\left[{}\begin{matrix}y=0\\y+10=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x^2+5x=0\\x^2+5x+10=0\left(vn\right)\end{matrix}\right.\)

<=> x(x + 5) = 0

<=> x= 0 hoặc x = -5

Vậy S = {0; -5}