Question

Giải phương trình: x(x+1)(x+2)(x+3)=24

Answer 1

\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=24\)

\(\Leftrightarrow\left[x\left(x+3\right)\right]\left[\left(x+1\right)\left(x+2\right)\right]=24\)

\(\Leftrightarrow\left(x^2+3x\right)\left(x^2+3x+2\right)=24\)

Dat \(x^2+3x+2=a\left(a>0\right)\)

\(\Leftrightarrow\left(a-2\right)a=24\)

\(\Leftrightarrow a^2-2a-24=0\)

\(\Leftrightarrow a^2-6a+4a-24=0\)

\(\Leftrightarrow\left(a-6\right)\left(a+4\right)=0\\ \left[{}\begin{matrix}a=6\\a=-4\left(Loai\right)\end{matrix}\right.\)

Thay a=6:

\(x^2-3x+4=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)

Vayy...

 

 

Answer 2

\(x(x+1)(x+2)(x+3)=24\)

\(\Leftrightarrow[x(x+3)][(x+1)(x+2)]-24=0 \)

\(\Leftrightarrow(x^2+3x)(x^2+3x+2)-24=0\)

\(\Leftrightarrow[(x^2+3x+1)-1][(x^2+3x+1)+1]-24=0\)

Đặt \(a=x^2+3x+1\)

\(\Leftrightarrow(a-1)(a+1)-24=0\)

\(\Leftrightarrow (a^2-1)-24=0\)

\(\Leftrightarrow a^2-1-24=0\)

\(\Leftrightarrow a^2-25=0\)

\(\Leftrightarrow(a-5)(a+5)=0\)

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

Vậy tập nghiệm pt \(S=\{-1;4\}\).

 

Answer 3

x(x+1)(x+2)(x+3)=24

<=>(x²+3x)(x²+3x+2)=24

<=>(x²+3x)²+2(x²+3x)+1=25

<=>(x²+3x+1)²=5²=(-5)²

<=> x²+3x+1=5 hoặc x²+3x+1=-5

<=>x²+3x-4=0 hoặc x²+3x+6=0(vô nghiệm)

<=>x=1 hoặc x=-4

Vậy S = {1;-4}