Question

1/Phân tích thành nhân tử

a, x² - 7

b, x² - 2√2x + 2

c, x² + 2 √13 x + 13

2/ rút gọn

a, \(\dfrac{x^2-5}{x+5}\) (với x≠ -5)

b, \(\dfrac{x^2+2\sqrt{2}x+2}{x^2-2}\) (với x ≠ ± \(\sqrt{12}\))

Answer 1

1/ 

a, \(x^2-7=\left(x+\sqrt{7}\right)\left(x-\sqrt{7}\right)\)

b, \(x^2-2\sqrt{2}x+2=\left(x-\sqrt{2}\right)^2\)

c, \(x^2+2\sqrt{13}x+13=\left(x+\sqrt{13}\right)^2\)

2/ 

a, \(\dfrac{\left(x+\sqrt{5}\right)\left(x-\sqrt{5}\right)}{x+\sqrt{5}}=x-\sqrt{5}\) ( sửa đề nha )

b, \(\dfrac{\left(x+\sqrt{2}\right)^2}{\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)}=\dfrac{x+\sqrt{2}}{\left(x-\sqrt{2}\right)}\) 

 

Answer 2

Câu `1:`

\(a)\\ x^2-7\\ =\left(x-\sqrt{7}\right)\left(x+\sqrt{7}\right)\\ b)\\ x^2-2\sqrt{2}x+2\\ =x^2-2\sqrt{2}x+\left(\sqrt{2}\right)^2\\ =\left(x-\sqrt{2}\right)^2\\ c)\\ x^2+2\sqrt{13}x+13\\ =x^2+2\sqrt{13}x+\left(\sqrt{13}\right)^2\\ =\left(x-\sqrt{13}\right)^2\)