Question

[Hoàng Bin]

Giải các phương trình sau:

a, (x-2)³+(3x-1)(3x+1)=(x+1)³

b,(x-1)³-x(x+1)²=5x(2-x)-11(x+2)

c,(x+1)(2x-3)=(2x-1)(x-5)

d,(x-1)-(2x-1)=9-x

e,(x-3)(x+4)-2(3x-2)=(x-4)²

f,x(x+3)²-3x=(x+2)+1

g,(x+1)(x²-x+1)-2x=x(x+1)(x-1)

Answer 1

[Edogawa Conan]

a) (x - 2)³ + (3x - 1)(3x + 1) = (x + 1)³

<=> x³ - 6x² + 12x - 8 + 9x² - 1 = x³ + 3x² + 3x + 1

<=> x³ + 3x² + 12x - x³ - 3x² - 3x = 1 + 9

<=> 9x = 10

<=> x = 10/9

vậy S = {10/9}

b) (x - 1)³ - x(x + 1)² = 5x(2 - x) - 11(x + 2)

 <=> x³ - 3x² + 3x  - 1 - x³ - 2x² - x = 10x - 5x² - 11x - 22

<=> -5x² + 2x - 10x + 5x² + 11x = -22 + 1

<=> 3x = -21

<=> x = -7

Vậy S = {-7}

c) (x + 1)(2x - 3) = (2x - 1)(x + 5)

<=> 2x² - x - 3 = 2x² + 9x - 5

<=> 2x² -x - 2x² - 9x = -5 + 3

<=>-10x = -2

<=> x = 1/5 Vậy S = {1/5}

Answer 2

[Edogawa Conan]

d) (x - 1) - (2x - 1) = 9 - x

<=> x - 1 - 2x + 1 = 9 - x

<=> -x + x = 9

<=> 0x = 9 (vô nghiệm)

=> pt vô nghiệm

e) (x - 3)(x + 4) - 2(3x - 2) = (x - 4)²

<=> x² + x - 12 - 6x + 4 = x² - 8x + 16

<=> x² - 5x - x² + 8x = 16 + 8

<=> 3x = 24

<=> x = 8

Vậy S = {8}

g) (x + 1)(x2 - x + 1) - 2x = x(x + 1)(x - 1)

<=> x³ + 1 - 2x = x³ - x

<=> x³ - 2x - x³ + x = -1

<=> -x = -1 <=> x = 1

Vậy S = {1}