Bài 1:\(\left(x^2+3x+2\right)\left(x^2+7x+12\right)=24\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=24\)
\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)
Đặt \(t=x^2+5x+4\) thì ta có:
\(\Leftrightarrow t\left(t+2\right)-24=0\)\(\Leftrightarrow\left(t-4\right)\left(t+6\right)=0\)
\(\Leftrightarrow\)\(\left(x^2+5x+4-4\right)\left(x^2+5x+4+6\right)=0\)
\(\Leftrightarrow\left(x^2+5x\right)\left(x^2+5x+10\right)=0\)
\(\Leftrightarrow x\left(x+5\right)\left[\left(x+\dfrac{5}{2}\right)^2+\dfrac{15}{4}\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Bài 2:
\(\left\{{}\begin{matrix}\left(x-y+1\right)^3+x=2014\\\left(x-y+1\right)^3+y=2013\end{matrix}\right.\)
Trừ theo của \(pt\left(1\right)\) cho \(pt\left(2\right)\) có:
\(\left(x-y+1\right)^3+x-\left(x-y+1\right)^3-y=1\)
\(\Leftrightarrow x-y=1\). Thay vào \(pt\left(1\right)\) ta có:
\(2+x=2014\Leftrightarrow x=2012\Rightarrow y=2011\)
Bài 1 biết làm không giúp với Ace Legona
