Toán Lớp 8: Tìm x, biết: a) 9x^2-25=0 b) (x+1)^2-(x-4)(x+4)=8 c) (3x-1)^2+(x+3)^2-10(x+1)(x-1)=0 d) (x+3)^3+(2x+1)(4x^2-2x+1)-9x^2(x+1)=54 e) (x-3)

Question

Toán Lớp 8:  Tìm x, biết: a) 9x^2-25=0 b) (x+1)^2-(x-4)(x+4)=8 c) (3x-1)^2+(x+3)^2-10(x+1)(x-1)=0 d) (x+3)^3+(2x+1)(4x^2-2x+1)-9x^2(x+1)=54 e) (x-3)^3-(x-3)(x^2+3x+9)+9(x+1)^2=-33

tuanh · Answer

a) 9x&sup2; - 25 = 0   &hArr; 9x&sup2; = 25   &hArr; x&sup2; = 25 : 9   &hArr; x = $ frac{5}{3}$    &hArr; x = -$ frac{5}{3}$   b) (x+1)&sup2;-(x-4)(x+4)=8   &hArr; x&sup2; + 2x + 1 - (x&sup2;-16) = 8   &hArr; x&sup2; + 2x + 1 - x&sup2; + 16 = 8   &hArr;&middot; 2x + 17 = 8   &hArr; 2x = 8 - 17   &hArr; 2x = -9   &hArr; x =-$ frac{9}{2}$    c) (3x-1)&sup2;+(x+3)&sup2;-10(x+1)(x-1)   &hArr; 9x&sup2;-6x+1+x&sup2;+6x+9-10(x&sup2;-1)   &hArr; 9x&sup2; + 1 + x&sup2; + 9 - 10x&sup2; + 10  = 0  &hArr; 0 + 20 = 0  &hArr; 20 = 0 ( pt vô nghiệm )  d) (x+3)&sup3;+(2x+1)(4x&sup2;-2x+1)-9x&sup2;(x+1)=54   &hArr; x&sup3;+9x&sup2;+27x+27+8x&sup3;+1-9x&sup3;-9x&sup2;=54   &hArr; 0 + 27x + 28 + 54   &hArr; 27x = 54 - 28   &hArr; 27x = 26   &hArr; x =n$ frac{26}{27}$    e) (x-3)&sup3;-(x-3)(x&sup2;+3x+9)+9(x-1)&sup2;=-33   &hArr; x&sup3;-9x&sup2;+27x-27-(x&sup3;-27)+9(x&sup2;-2x+1)=-33    &hArr; x&sup3; - 9x&sup2; + 27x - 27&sup2; - x&sup3; + 27 + 9x&sup2; - 18x + 9 = -33   &hArr; 9x + 9 = -33   &hArr; 9x = -33-9   &hArr; 9x = -42   &hArr; x = -$ frac{14}{3}$    @giahanchau

huyenmi76 · Answer

Giải đáp + Lời giải và giải thích chi tiết:    a)   $(=) 9x^2 = 25$   $(=) x^2 = 25/9$   $(=) x=5/3 hoac x=-5/3$   b)   $(=) x^2 + 2x + 1 - x^2 + 16 = 8$   $(=) 2x+17=8$   $(=) 2x= 8-17$   $(=) 2x=-9$   $=) x=-9/2$   c)   $(=) 9x^2 + 1 + x^2 + 9 - 10x^2 + 10 = 0$   $(=) 20=0$   Phương tr&igrave;nh v&ocirc; nghiệm   d)   $(=) x^3 + 9x^2 + 27x + 27 + 8x^3 + 1 - 9x^3 - 9x^2 = 54$   $(=) 27x + 28 = 54$   $(=) 27x= 54-28$   $(=) 27x=26$   $(=) x=26/27$   e)   $(=) x^3 - 9x^2 + 27x - 27 - x^3 + 27 + 9x^2 + 18x + 9 = - 33$   $(=) 45x+9=-33$   $(=) 45x=-33-9$   $(=) 45x=-42$   $(=) x=-14/15$