\(\left(x^2+3x+2\right)\left(x^2+7x+12\right)=24\)
\(\Rightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=24\)
\(\Rightarrow\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)=24\)
\(\Rightarrow\left(x^2+4x+x+4\right)\left(x^2+3x+2x+6\right)=24\)
\(\Rightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)=24\)
\(\Rightarrow\left(x^2+5x+5-1\right)\left(x^2+5x+5+1\right)=24\)
\(\Rightarrow\left(x^2+5x+5\right)^2-1=24\)
\(\Rightarrow\left(x^2+5x+5\right)^2=25\)
\(\Rightarrow\left(x^2+5x+5\right)^2-25=0\)
\(\Rightarrow\left(x^2+5x+5+5\right)\left(x^2+5x+5-5\right)=0\)
\(\Rightarrow\left(x^2+5x+10\right)\left(x^2+5x\right)=0\)
\(\Rightarrow\left(x^2+5x+\dfrac{25}{4}+\dfrac{15}{4}\right)\left(x^2+5x\right)=0\)
\(\Rightarrow\left[\left(x^2+5x+\dfrac{25}{4}\right)+\dfrac{15}{4}\right]\left(x^2+5x\right)=0\)
\(\Rightarrow x\left(x+5\right)\left[\left(x+\dfrac{5}{2}\right)^2+\dfrac{15}{4}\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-5\\\left(x+\dfrac{5}{2}\right)^2+\dfrac{15}{4}>0\end{matrix}\right.\)
hãy c/m tính đúng đắn của câu:'hày yêu quý sách vì đó là nguồn gốc của mọi tri thức'
