a) Gọi ba số lẻ liên tiếp lần lượt là 2k + 1, 2k + 3, 2k + 5 ( k thuộc N )
Theo đề ta có:
\(\left(2k+1\right)\left(2k+3\right)\left(2k+5\right)=105\)
\(\Rightarrow\left(4k^2+8k+3\right)\left(2k+5\right)=105\)
\(\Rightarrow\left(4k^2+8k+3\right)2k+\left(4k^2+8k+3\right)5=105\)
\(\Rightarrow8k^3+16k^2+6k+20k^2+40k+15=105\)
\(\Rightarrow8k^3+36k^2+46k+15=105\)
\(\Rightarrow8k^3+36k^2+46k=90\)
\(\Rightarrow8k^3+36k^2+46k-90=0\)
\(\Rightarrow8k^3-8k^2+44k^2-44k+90k-90=0\)
\(\Rightarrow8k^2\left(k-1\right)+44k\left(k-1\right)+90\left(k-1\right)=0\)
\(\Rightarrow\left(k-1\right)\left(8k^2+44k+90\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}k-1=0\\8k^2+44k+90=0\end{matrix}\right.\)
Mà \(8k^2+44k+90\)
\(=\left(2\sqrt{2}k\right)^2+2.2\sqrt{2}k.\dfrac{11\sqrt{2}}{2}+\left(\dfrac{11\sqrt{2}}{2}\right)^2-\left(\dfrac{11\sqrt{2}}{2}\right)^2+90\)
\(=\left(2\sqrt{2}k+\dfrac{11\sqrt{2}}{2}\right)^2+\dfrac{59}{2}\)
Vì \(\left(2\sqrt{2}k+\dfrac{11\sqrt{2}}{2}\right)^2\ge0\) với mọi k
\(\Rightarrow\left(2\sqrt{2}k+\dfrac{11\sqrt{2}}{2}\right)^2+\dfrac{59}{2}\ge\dfrac{59}{2}\)
=> 8k2 + 44k + 90 vô nghiệm
=> k - 1 = 0
=> k = 1
=> 2k + 1 = 3
Vậy ba số tự nhiên lẻ liên tiếp là 3 ; 5 ; 7
