Câu hỏi của Linh

4)\(x^2+3x-16\)\(=\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{73}{4}\)\(=\left[x^2+2.x.\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2\right]-\left(\dfrac{\sqrt{73}}{2}\right)^2\)\(=\left(x+\dfrac{3}{2}\right)^2-\left(\dfrac{\sqrt{73}}{2}\right)^2\)\(=\left(x+\dfrac{3}{2}-\dfrac{\sqrt{73}}{2}\right).\left(x+\dfrac{3}{2}+\dfrac{\sqrt{73}}{2}\right)\)Câu trả lời: 4)\(x^2+3x-16\)\(=\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{73}{4}\)\(=\left[x^2+2.x.\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2\right]-\left(\dfrac{\sqrt{73}}{2}\right)^2\)\(=\left(x+\dfrac{3}{2}\right)^2-\left(\dfrac{\sqrt{73}}{2}\right)^2\)\(=\left(x+\dfrac{3}{2}-\dfrac{\sqrt{73}}{2}\right).\left(x+\dfrac{3}{2}+\dfrac{\sqrt{73}}{2}\right)\)