$Tìm $ $x$ $biết$ $(k∈Z)$ $:$ $x^{2}$ $+$ $2x$ = $5k$ 27/10/2021 Bởi Madelyn $Tìm $ $x$ $biết$ $(k∈Z)$ $:$ $x^{2}$ $+$ $2x$ = $5k$
$x^2+2x+1=5k\\ \Leftrightarrow (x+1)^2=5k\\ \Leftrightarrow |x+1|=\sqrt{5k}(k\ge0)\\ \Rightarrow \left[\begin{array}{l} x+1=\sqrt{5k}\\ x+1=-\sqrt{5k}\end{array} \right.\\ \Leftrightarrow\left[\begin{array}{l} x=\sqrt{5k}-1\\ x=-\sqrt{5k}-1\end{array} \right.$ $x^2+2x=5k+1\\ \Leftrightarrow x^2+2x+1=5k+2\\ \Leftrightarrow (x+1)^2=5k+2\\ \Leftrightarrow |x+1|=\sqrt{5k+2}\left(k\ge-\dfrac{2}{5}\right)\\ \Rightarrow \left[\begin{array}{l} x+1=\sqrt{5k+2}\\ x+1=-\sqrt{5k+2}\end{array} \right.\\ \Leftrightarrow\left[\begin{array}{l} x=\sqrt{5k+2}-1\\ x=-\sqrt{5k+2}-1\end{array} \right.$ Bình luận
$x^2+2x+1=5k\\ \Leftrightarrow (x+1)^2=5k\\ \Leftrightarrow |x+1|=\sqrt{5k}(k\ge0)\\ \Rightarrow \left[\begin{array}{l} x+1=\sqrt{5k}\\ x+1=-\sqrt{5k}\end{array} \right.\\ \Leftrightarrow\left[\begin{array}{l} x=\sqrt{5k}-1\\ x=-\sqrt{5k}-1\end{array} \right.$
$x^2+2x=5k+1\\ \Leftrightarrow x^2+2x+1=5k+2\\ \Leftrightarrow (x+1)^2=5k+2\\ \Leftrightarrow |x+1|=\sqrt{5k+2}\left(k\ge-\dfrac{2}{5}\right)\\ \Rightarrow \left[\begin{array}{l} x+1=\sqrt{5k+2}\\ x+1=-\sqrt{5k+2}\end{array} \right.\\ \Leftrightarrow\left[\begin{array}{l} x=\sqrt{5k+2}-1\\ x=-\sqrt{5k+2}-1\end{array} \right.$