\(x^2-11x+2\)
\(\text{Sử dụng biệt thức( cách này lớp 9 kì 2 hok nha)}\)
\(\text{denta}=b^2-4ac=11^2-2.1.4=113>0\)
=> pt có 2 No là:
\(x_1=\frac{11+\sqrt{113}}{2};x_2=\frac{11-\sqrt{113}}{2}\)
\(x^2-11x+2\)
\(=\left[x^2-2.x.\frac{11}{2}+\left(\frac{11}{2}\right)^2\right]-\frac{7}{2}\)
\(=\left(x-\frac{11}{2}\right)^2-\left(\sqrt{\frac{7}{2}}\right)^2\)
\(=\left(x-\frac{11}{2}+\sqrt{\frac{7}{2}}\right)\left(x-\frac{11}{2}-\sqrt{\frac{7}{2}}\right)\)
\(=\left(x-\frac{11}{2}+\frac{\sqrt{14}}{2}\right)\left(x-\frac{11}{2}-\frac{\sqrt{14}}{2}\right)\)
\(=\left(x+\frac{\sqrt{14}-11}{2}\right)\left(x-\frac{\sqrt{14}+11}{2}\right)\)
Tham khảo nhé~
