Cho pt 2x^2 -5x-2+√5=0 có 2 nghiệm x1,x2 Tính giá trị M= x1(x1^2 – 2)+x2(x2^2 – 2) +7/2x1x2 + 2√5

Đáp án:

$M=\dfrac{257+42\sqrt{5}}{8}$

Giải thích các bước giải:

$Vi-et:x_1+x_2=\dfrac{5}{2}\\ x_1x_2=\dfrac{\sqrt{5}-2}{2}\\ M= x_1(x_1^2 – 2)+x_2(x_2^2 – 2) +\dfrac{7}{2x_1x_2} + 2\sqrt{5}\\ =x_1^3-2x_1+x_2^3-2x_2+\dfrac{7}{2x_1x_2} + 2\sqrt{5}\\ =x_1^3+x_2^3-2(x_1+x_2)+\dfrac{7}{2x_1x_2} + 2\sqrt{5}\\ =(x_1+x_2)(x_1^2-x_1x_2+x_2^2)-2(x_1+x_2)+\dfrac{7}{2x_1x_2} + 2\sqrt{5}\\ =(x_1+x_2)(x_1^2+2x_1x_2+x_2^2-3x_1x_2)-2(x_1+x_2)+\dfrac{7}{2x_1x_2} + 2\sqrt{5}\\ =(x_1+x_2)\left((x_1+x_2)^2-3x_1x_2\right)-2(x_1+x_2)+\dfrac{7}{2x_1x_2} + 2\sqrt{5}\\ =\dfrac{5}{2}\left(\dfrac{5^2}{2^2}-\dfrac{3(\sqrt{5}-2)}{2}\right)-2.\dfrac{5}{2}+\dfrac{7}{2.\dfrac{\sqrt{5}-2}{2}} + 2\sqrt{5}\\ =\dfrac{185-30\sqrt{5}}{8}+\dfrac{7\left(\sqrt{5}+2\right)}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)} -5+ 2\sqrt{5}\\ =\dfrac{185-30\sqrt{5}}{8}+7\left(\sqrt{5}+2\right) -5+ 2\sqrt{5}\\ =\dfrac{185-30\sqrt{5}}{8}+9\sqrt{5}+9\\ =\dfrac{257+42\sqrt{5}}{8}$