Toán Lớp 8: Tìm x ( đa thức thành nhân tử) A, x(x-3) +5x-15=0 B, 3x²-9x=0 C, 5x(7x-3)-6+14x=0 D,(2x-3)²-x(x+1)=9+17x

Question

Toán Lớp 8:  Tìm x ( đa thức thành nhân tử) A, x(x-3) +5x-15=0 B, 3x²-9x=0 C, 5x(7x-3)-6+14x=0 D,(2x-3)²-x(x+1)=9+17x

truongthanhhung · Answer

chúc bạn học tốt

kimlien · Answer

Hướng dẫn trả lời:     a) xcdot(x - 3) + 5x - 15 = 0   &harr; xcdot(x - 3) + 5cdot(x - 3) = 0   &harr; (x + 5)cdot(x - 3) = 0   &harr;  ( left[  begin{array}{l}x + 5 = 0  x - 3 = 0 end{array}  right. )   &harr;  ( left[  begin{array}{l}x = -5  x = 3 end{array}  right. )   Vậy x = -5 hoặc x = 3   b) 3x^2 - 9x = 0   &harr; 3xcdot(x - 3) = 0   &harr;  ( left[  begin{array}{l}3x = 0  x - 3 = 0 end{array}  right. )   &harr;  ( left[  begin{array}{l}x = 0  x = 3 end{array}  right. )   Vậy x = 0 hoặc x = 3   c) 5xcdot(7x - 3) - 6 + 14x = 0   &harr; 5xcdot(7x - 3) + (14x - 6) = 0   &harr; 5xcdot(7x - 3) + 2cdot(7x - 3) = 0   &harr; (5x + 2)cdot(7x - 3) = 0   &harr;  ( left[  begin{array}{l}5x + 2 = 0  7x - 3 = 0 end{array}  right. )   &harr;  ( left[  begin{array}{l}5x = -2  7x = 3 end{array}  right. )   &harr;  ( left[  begin{array}{l}x =  dfrac{-2}{5}  x =  dfrac{3}{7} end{array}  right. )   Vậy x = -2/5 hoặc x = 3/7   d) (2x - 3)^2 - xcdot(x + 1) = 9 + 17x   &harr; (2x)^2 - 2cdot2xcdot3 + 3^2 - x^2 - x = 9 + 17x   &harr; 4x^2 - 12x + 9 - x^2 - x = 9 + 17x   &harr; 4x^2 - 12x + 9 - x^2 - x - (9 + 17x) = 0   &harr; 4x^2 - 12x + 9 - x^2 - x - 9 - 17x = 0   &harr; (4x^2 - x^2) + (- 12x - x - 17x) + (9 - 9) = 0   &harr; 3x^2 - 30x = 0   &harr; 3xcdot(x - 10) = 0   &harr;  ( left[  begin{array}{l}3x = 0  x - 10 = 0 end{array}  right. )   &harr;  ( left[  begin{array}{l}x = 0  x = 10 end{array}  right. )   Vậy x = 0 hoặc x = 10