Toán Lớp 8: Phân tích đa thức thành nhân tử a) 4x + 8y b)

Question

Toán Lớp 8:  Phân tích đa thức thành nhân tử        a)    4x +  8y                                                                                b)  16  – x2 +2xy – y2        c)   9x2- 9xy – 4x + 4y                                                                 d)  x2 –  5x + 6 Bài 3 Tìm x biết        a)   x( x +1) – x2   = – 2                                                        b) x2 –  4  = 0                                                                                              c)  x(x-2021) - 2022x + 2021.2022=0

diemchau · Answer

1   a 4x+8y=4(x+2y)   b 16-x&sup2;-2xy+y&sup2;   =16-(x-y)&sup2;   =(4-x+y)(4+x-y)   c 9x&sup2;-9xy-4x+4y   =9x(x-y)-4(x-y)   =(x-y)(9x-4)   d x&sup2;-5x+6   =x&sup2;-2x-3x+6   =x(x-2)-3(x-2)   =(x-2)(x-3)   3   a x(x+1)-x&sup2;=-2   x&sup2;+x-x&sup2;=-2   x=-2   b x&sup2;-4=0   (x-2)(x+2)=0   &rArr;$ begin{cases} x-2=0  x+2=0  end{cases}$   &rArr;$ begin{cases} x=2  x=-2  end{cases}$   c  x(x-2021)-2022x+2021.2022=0   x(x-2021)-2022(x-2021)=0   (x-2021)(x-2022)=0   &rArr;&rArr;$ begin{cases} x-2021=0  x-2022=0  end{cases}$   &rArr;$ begin{cases} x=2021  x=2022  end{cases}$

thucquyenlan · Answer

@ B&agrave;i 2:$  $   a)4x+8y$  $   =4.x+4.2.y$  $   =4(x+2y)$  $   b)16-x^2+2xy-y^2$  $   =16-(x^2-2xy+y^2)$  $   =4^2-(x-y)^2$  $   =[4-(x-y)][4+(x-y)]$  $   =(4-x+y)(4+x-y)$  $   c)9x^2-9xy-4x+4y$  $   =(9x^2-9xy)-(4x-4y)$  $   =9x(x-y)-4(x-y)$  $   =(x-y)(9x-4)$  $   d)x^2-5x+6$  $   =x^2-2x-3x+6$  $   =(x^2-2x)-(3x-6)$  $   =x(x-2)-3(x-2)$  $   =(x-2)(x-3)$  $   @ B&agrave;i 3:$  $   a)x(x+1)-x^2=-2$  $   ->x.x+x.1-x^2=-2$  $   ->x^2+x-x^2=-2$  $   ->(x^2-x^2)+x=-2$  $   ->x=-2$  $   Vậy x=-2$  $   b)x^2-4=0$  $   ->x^2-2^2=0$  $   ->(x-2)(x+2)=0   -> ( left[  begin{array}{l}x-2=0  x+2=0 end{array}  right. ) $  $   -> ( left[  begin{array}{l}x=2  x=-2 end{array}  right. ) $  $   Vậy x=2 v&agrave; x=-2$  $   c)x(x-2021)-2022x+2021.2022=0$  $   ->x(x-2021)-(2022x-2021.2022)=0$  $   ->x(x-2021)-2022(x-2021)=0$  $   ->(x-2021)(x-2022)=0$  $   -> ( left[  begin{array}{l}x-2021=0  x-2022=0 end{array}  right. ) $  $   -> ( left[  begin{array}{l}x=2021  x=2022 end{array}  right. ) $  $   Vậy x=2021 v&agrave; x=2022$  $