Bài 55 trang 25 sgk toán 8 tập 1

Bài 55 trang 25 sgk toán 8 tập 1

Tìm x, biết:

55. Tìm (x), biết:

a)  ({x^3} - {1 over 4}x = 0);                     

b) ({(2x - 1)^2} - {(x + 3)^2} = 0);

c) ({x^2}(x - 3) + 12 - 4x = 0).

Bài giải:

a) 

(eqalign{
& {x^3} - {1 over 4}x = 0 Rightarrow xleft( {{x^2} - {1 over 4}} ight) = 0 cr
& Rightarrow xleft( {{x^2} - {{left( {{1 over 2}} ight)}^2}} ight) = 0 cr
& Rightarrow xleft( {x - {1 over 2}} ight)left( {x + {1 over 2}} ight) = 0 cr
& Rightarrow left[ matrix{
x = 0 hfill cr
left( {x - {1 over 2}} ight) = 0 Rightarrow x = {1 over 2} hfill cr
left( {x + {1 over 2}} ight) = 0 Rightarrow x = - {1 over 2} hfill cr} ight. cr} )

Vậy (x=0,x={1over 2},x=-{1over2})

b) 

(eqalign{
& {(2x - 1)^2} - {(x + 3)^2} = 0 cr
& Rightarrow left[ {(2x - 1) - (x + 3)} ight].left[ {(2x - 1) + (x + 3)} ight] = 0 cr
& Rightarrow (2x - 1 - x - 3).(2x - 1 + x + 3) = 0 cr
& Rightarrow (x - 4).(3x + 2) = 0 cr
& Rightarrow left[ matrix{
x - 4 = 0 hfill cr
3x + 2 = 0 hfill cr} ight. Rightarrow left[ matrix{
x = 4 hfill cr
x = - {2 over 3} hfill cr} ight. cr} )

Vậy (x=4,x=-{2over 3})

c) 

(eqalign{
& {x^2}(x - 3) + 12 - 4x = 0 cr
& Rightarrow {x^2}(x - 3) - 4(x - 3) = 0 cr
& Rightarrow (x - 3)({x^2} - 4) = 0 cr
& Rightarrow (x - 3)(x - 2)(x + 2) = 0 cr
& Rightarrow left[ matrix{
x = 3 hfill cr
x = 2 hfill cr
x = - 2 hfill cr} ight. cr} )

Vậy ( x=3,x=2,x=-2)

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