Question

[Nguyễn Hoài Phương]

x.(x-1) (x+1) (x+2) = 24
Giúp mik với ạ

Answer 1

[I am➻Minh]

\(x.\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\)

\(\Leftrightarrow x\left(x+1\right)\left(x-1\right)\left(x+2\right)=24\)

\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)

\(\Leftrightarrow\left(x^2+x-1+1\right)\left(x^2+x-1-1\right)=24\)

\(\Leftrightarrow\left(x^2+x-1\right)^2-1-24=0\)

\(\Leftrightarrow\left(x^2+x-1\right)^2=25\)

\(\Leftrightarrow x^2+x-1=\pm5\)

\(TH1:x^2+x-1=5\) 

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)

\(TH2:x^2+x-1=-5\)

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

Vì  \(x^2+x+4=x^2+2\cdot\frac{1}{2}\cdot x+\frac{1}{4}+\frac{15}{4}=\left(x+\frac{1}{2}\right)^2+\frac{15}{4}>0\)

Mà \(x^2+x+4=0\)

=> pt vô nghiệm

Vậy \(\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)

Answer 2

[l҉o҉n҉g҉ d҉z҉]

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

<=> [ x( x + 1 ) ][ ( x - 1 )( x + 2 ) ] = 24

<=> ( x² + x )( x² + x - 2 ) - 24 = 0

Đặt t = x² + x

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

<=> t² - 2t - 24 = 0

<=> t² - 6t + 4t - 24 = 0

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

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

<=> ( x² + x - 6 )( x² + x + 4 ) = 0

<=> ( x² - 2x + 3x - 6 )( x² + x + 4 ) = 0

<=> [ x( x - 2 ) + 3( x - 2 ) ]( x² + x + 4 ) = 0

<=> ( x - 2 )( x + 3 )( x² + x + 4 ) = 0

Vì x² + x + 4 = ( x + 1/2 )² + 15/4 ≥ 15/4 > 0 ∀ x

=> x - 2 = 0 hoặc x + 3 = 0

<=> x = 2 hoặc x = -3

Vậy ...