Question

[Trương Tấn Dũng]

(x+2)(x+4)(x+6)(x+8)-105=0

Answer 1

[Toru]

\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)-105=0\)

\(\Leftrightarrow\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]-105=0\)

\(\Leftrightarrow\left(x^2+10x+16\right)\left(x^2+10x+24\right)-105=0\) (1)

Đặt \(x^2+10x+20=t\), khi đó (1) trở thành:

\(\left(t-4\right)\left(t+4\right)-105=0\)

\(\Leftrightarrow t^2-16-105=0\)

\(\Leftrightarrow t^2-11^2=0\)

\(\Leftrightarrow\left(t-11\right)\left(t+11\right)=0\)

\(\Rightarrow\left(x^2+10x+20-11\right)\left(x^2+10x+20+11\right)=0\)

\(\Leftrightarrow\left(x^2+10x+9\right)\left(x^2+10x+31\right)=0\)

\(\Leftrightarrow\left(x^2+9x+x+9\right)\left[\left(x+5\right)^2+6\right]=0\)

\(\Leftrightarrow x\left(x+9\right)+\left(x+9\right)=0\) (vì \(\left(x+5\right)^2+6>0;\forall x\))

\(\Leftrightarrow\left(x+9\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+9=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=-1\end{matrix}\right.\)

Vậy phương trình đã cho có tập nghiệm là $S=\{-9;-1\}$.

$Toru$

Answer 2

[Dang Tung]

\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)-105=0\\ \Leftrightarrow\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]=105\\ \Leftrightarrow\left(x^2+10x+16\right)\left(x^2+10x+24\right)=105\\ \Leftrightarrow\left(x^2+10x+20-4\right)\left(x^2+10x+20+4\right)=105\\ \Leftrightarrow\left(x^2+10x+20\right)^2-4^2=105\\ \Leftrightarrow\left(x^2+10x+20\right)^2=121\\ \)

\(\Rightarrow\left[{}\begin{matrix}x^2+10x+20=11\left(1\right)\\x^2+10x+20=-11\left(2\right)\end{matrix}\right.\)

Giải (1):

\(x^2+10x+9=0\\ \Leftrightarrow\left(x^2+x\right)+\left(9x+9\right)=0\\ \Leftrightarrow x\left(x+1\right)+9\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x+9\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-9\end{matrix}\right.\)

Giải (2):

Nhận thấy: \(x^2+10x+20=\left(x+5\right)^2-5\ge-5\forall x\inℝ\)

Vậy pt (2) vô nghiệm

Vậy tập nghiệm pt là: \(S=\left\{-1;-9\right\}\)