Question

Bài 14 (trang 11 SGK Toán 9 Tập 1)

Phân tích thành nhân tử:

a) $x^2-3$ ;                           b) $x^2-6$  ;

c) $x^2+2\sqrt{3}x+3$ ;           d) $x^2-2\sqrt{5}x+5$.

Hướng dẫn: Dùng kết quả:

                      Với $a \ge 0$ thì $a = (\sqrt{a})^2$.

Answer 1

a, \(x^2-3=\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\)

b, \(x^2-6=\left(x-\sqrt{6}\right)\left(x+\sqrt{6}\right)\)

c, \(x^2+2\sqrt{3}+3=x^2+2\sqrt{3}+\left(\sqrt{3}\right)^2=\left(x+\sqrt{3}\right)^2\)

d, \(x^2-2\sqrt{5}x+5=x^2-2\sqrt{5}x+\left(\sqrt{5}\right)^2=\left(x-\sqrt{5}\right)^2\)

Answer 2

a) \(x^2\) - 3 = (x-\(\sqrt{3}\))(x+\(\sqrt{3}\))

b)\(x^2\)-6=(x-\(\sqrt{6}\))(x+\(\sqrt{6}\))

c) \(x^2+2\sqrt{3}x+3\)= \(\left(x+\sqrt{3}\right)^2\)

d) \(x^2-2\sqrt{5}x+5\)=\(\left(x-\sqrt{5}\right)^2\)

Answer 3

a, ( x^2- √3).(x^2+ √3)

b, (x^2- √6).(x^2+ √6)

c, (√x +  √3)^2

d, ( √x -  √5)^2

Answer 4

a) \(x^2-3=\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\)

b) \(x^2-6=\left(x-\sqrt{6}\right)\left(x+\sqrt{6}\right)\)

c) \(x^2+2\sqrt{3}x+3=\left(x+\sqrt{3}\right)^2\)

d) \(x^2-2\sqrt{5}x+5=\left(x-\sqrt{5}\right)^2\)

Answer 5

Answer 6

Answer 7

 M y   D u y e n   

a) $x^2-3=x^2-(\sqrt{3}{)}^2=(x-\sqrt{3}\left)\right(x+\sqrt{3})$.

b) $x^2-6=x^2-(\sqrt{6}{)}^2=(x-\sqrt{6}\left)\right(x+\sqrt{6})$.

c) $x^2+2\sqrt{3}x+3$

$=x^2+2x\sqrt{3}+(\sqrt{3}{)}^2$

$=(x+\sqrt{3}{)}^2$.

d) $x^2-2\sqrt{5}x+5$

$=x^2-2x\sqrt{5}+(\sqrt{5}{)}^2$

$=(x-\sqrt{5}{)}^2$

Answer 8

Answer 9

Answer 10

a. x² - 3 = x² - \(\left(\sqrt{3}\right)^2\) = (x - \(\sqrt{3}\) ) (x + \(\sqrt{3}\) )
b. x² - 6 = x² - (\(\sqrt{6}\))²  = ( x - \(\sqrt{6}\)) (x + \(\sqrt{6}\))
c. x² + 2\(\sqrt{3}\)x + (\(\sqrt{3}\) )²
= x² + 2x\(\sqrt{3}\) + (\(\sqrt{3}\) )²
= (x+ \(\sqrt{3}\) )² 
d. x² - 2\(\sqrt{5}\)x + 5
= x² - 2x\(\sqrt{5}\)  + (\(\sqrt{5}\) )²
= (x - \(\sqrt{5}\))²

Answer 11

Answer 12

a) $x^2-3=x^2-(\sqrt{3}{)}^2=(x-\sqrt{3}\left)\right(x+\sqrt{3})$

b) $x^2-6=x^2-(\sqrt{6}{)}^2=(x-\sqrt{6}\left)\right(x+\sqrt{6})$

c) $x^2+2\sqrt{3}x+3$

$=x^2+2x\sqrt{3}+(\sqrt{3}{)}^2$

$=(x+\sqrt{3}{)}^2$

d) $x^2-2\sqrt{5}x+5$

$=x^2-2x\sqrt{5}+(\sqrt{5}{)}^2$

$=(x-\sqrt{5}{)}^2$

Answer 13

Answer 14

a) = (x-\(\sqrt{3}\))(x+\(\sqrt{3}\))
b) = (x-\(\sqrt{6}\))(x+\(\sqrt{6}\))
c) = (x+\(\sqrt{3}\))²
d) = (x-\(\sqrt{5}\))²

Answer 15

a) $x^2-3=x^2-(\sqrt{3}{)}^2=(x-\sqrt{3}\left)\right(x+\sqrt{3})$.

b) $x^2-6=x^2-(\sqrt{6}{)}^2=(x-\sqrt{6}\left)\right(x+\sqrt{6})$.

c) $x^2+2\sqrt{3}x+3$

$=x^2+2x\sqrt{3}+(\sqrt{3}{)}^2$

$=(x+\sqrt{3}{)}^2$.

d) $x^2-2\sqrt{5}x+5$

$=x^2-2x\sqrt{5}+(\sqrt{5}{)}^2$

$=(x-\sqrt{5}{)}^2$.

Answer 16

a) $x^2-3=x^2-(\sqrt{3}{)}^2=(x-\sqrt{3}\left)\right(x+\sqrt{3})$.

b) $x^2-6=x^2-(\sqrt{6}{)}^2=(x-\sqrt{6}\left)\right(x+\sqrt{6})$.

c) $x^2+2\sqrt{3}x+3$

$=x^2+2x\sqrt{3}+(\sqrt{3}{)}^2$

$=(x+\sqrt{3}{)}^2$.

d) $x^2-2\sqrt{5}x+5$

$=x^2-2x\sqrt{5}+(\sqrt{5}{)}^2$

$=(x-\sqrt{5}{)}^2$.

Answer 17

Answer 18

Answer 19

Answer 20

a) Ta có: 

$x^2-3=x^2-(\sqrt{3}{)}^2$

            $=(x-\sqrt{3})(x+\sqrt{3})$  

b) Ta có: 

$x^2-6=x^2-(\sqrt{6}{)}^2$

             $=(x-\sqrt{6})(x+\sqrt{6})$  

c) Ta có: 

$x^2+2\sqrt{3}x+3=x^2+2.x.\sqrt{3}+(\sqrt{3}{)}^2$

                           $=(x+\sqrt{3}{)}^2$

d) Ta có:

$x^2-2\sqrt{5}x+5=x^2-2.x.\sqrt{5}+(\sqrt{5}{)}^2$

                            $=(x-\sqrt{5}{)}^2$  

 

Answer 21

a) Ta có: 

$x^2-3=x^2-(\sqrt{3}{)}^2$

          $=(x-\sqrt{3})(x+\sqrt{3})$

b) Ta có: 

$x^2-6=x^2-(\sqrt{6}{)}^2$

          $=(x-\sqrt{6})(x+\sqrt{6})$

c) Ta có: 

$x^2+2\sqrt{3}x+3=x^2+2.x.\sqrt{3}+(\sqrt{3}{)}^2$

                      $=(x+\sqrt{3}{)}^2$

d) Ta có:

$x^2-2\sqrt{5}x+5=x^2-2.x.\sqrt{5}+(\sqrt{5}{)}^2$

                      $=(x-\sqrt{5}{)}^2$ 

 

Answer 22

Answer 23

Answer 24

Answer 25

a, x=7; x=-7

b, x=8; x=-8

c, x=3; x=-3

d, x=4; x=-4

Answer 26

Answer 27

a) x² - 3 = x² - ( \(\sqrt{3}\) )² = ( x - \(\sqrt{3}\) )² = ( x - \(\sqrt{3}\) )( x + \(\sqrt{3}\) )

b) x² - 6 = x² - ( \(\sqrt{6}\) )² = ( x - \(\sqrt{6}\) )( x + \(\sqrt{6}\) )

c) x² + 2.\(\sqrt{3}\).x + 3 = x² + 2.\(\sqrt{3}\).x  + \(\sqrt{3^2}\) = ( x + \(\sqrt{3}\) )²

d) x² - 2\(\sqrt{5}\)x + 5 = x² - 2\(\sqrt{5}\)x + \(\sqrt{5^2}\) = ( x - \(\sqrt{5}\) )²

Answer 28

a) x\(^2\) - 3 = x\(^2\) - (\(\sqrt{3}\) )\(^2\) = ( x - \(\sqrt{3}\) )( x + \(\sqrt{3}\) )

b) x\(^2\) - 6 = x\(^2\) - (\(\sqrt{6}\) )\(^2\) = ( x -\(\sqrt{6}\) )(x + \(\sqrt{6}\) )

c) x\(^2\) + 2\(\sqrt{3x}\) +3 = x\(^2\) + 2x\(\sqrt{3}\) + (\(\sqrt{3}\) )\(^2\) = ( x - \(\sqrt{5}\) )\(^2\)

d) x\(^2\) - 2\(\sqrt{5x}\) + 5 = x\(^2\) - 2\(\sqrt{5x}\) + (\(\sqrt{5}\) )\(^2\) = ( x - \(\sqrt{5}\) )\(^2\)

 

 

Answer 29

a) ( x- căn3)*(x+ căn 3)

b) ( x- căn 6)*(x+ căn 6)

c)(x+ căn 3)^2

d)(x- căn 5)^2

Answer 30

a, =x²-(\(\sqrt{3}\))²=(x-\(\sqrt{3}\))(x+\(\sqrt{3}\))

b,=x²-(\(\sqrt{6}\))²=(x-\(\sqrt{6}\))(x+\(\sqrt{6}\))

c,=(x+\(\sqrt{3}\))²

d,= (x+\(\sqrt{5}\))²