a) (x + 2)(x² - 2x + 4) - -x(x² + 2) = 15

<=> x³ + 8 - x³ - 2x = 15

<=> 2x = -7

<=> x=  -7/2

Vậy S = {-7/2}

b) (x - 2)³ - (x - 3)(x² + 3x + 9) + 6(x + 1)² = 49

<=> x³ - 6x² + 12x - 8 - x³ + 27 + 6x² + 12x + 6 = 49

<=> 24x = 24

<=> x = 1

Vậy S = {1}

hello  các đại ca

đại ca nào ở đây

a) x¹² + 4 = x¹² + 4x⁶ + 4 - 4x⁶ = (x⁶ + 2)² - (2x³)² 

= (x⁶ - 2x³ + 2)(x⁶ + 2x³ + 2)

b) 4x⁸ + 1 = 4x⁸ + 4x⁴  + 1 - 4x⁴ = (2x⁴ + 1)² - (2x²)² 

= (2x⁴ + 2x² + 1)(2x⁴ - 2x²  + 1)

c) x⁷ + x⁵ - 1 = x⁷ - x + x⁵ + x² - (x² - x  + 1) = x(x⁶ - 1) + x²(x³ + 1) - (x² - x + 1)

= x(x³ - 1)(x³ + 1) + x²(x + 1)(x² - x + 1) - (x2 - x + 1)

= (x⁴ - x)(x + 1)(x² - x + 1) + (x³ + x²)(x² - x + 1) - (x² - x + 1)

= (x⁵ + x⁴ - x² - x + x³ + x² - 1)(x² -x + 1)

= (x⁵ + x⁴ + x³ - x - 1)(x² - x + 1)

d) x⁷ + x⁵ + 1 = x⁷ - x + x⁵ - x² + (x² + x + 1)

= x(x³ - 1)((x³ + 1) + x²(x³ - 1) + (x² + x + 1)

= (x⁴ + x)(x  - 1)(x² + x + 1) + x²(x - 1)((x² + x + 1) + (x² + x + 1)

= (x² + x + 1)(x⁵ - x⁴ + x² - x + x³ - x² + 1)

= (x² + x + 1)(x⁵ - x⁴ + x³ - x + 1)

a, A=8x⁶-12x⁴+6x²-1

=( 2x²-1)³

b,B= \(8\left(\frac{x}{2}+y\right)^3-6\left(x+2y\right)^2x+12\left(x+2y\right)x^2-8x^3\)

=\(8\left(\frac{x^3}{8}+3\frac{x^2}{4}y+3\frac{x}{2}y^2+y^3\right)-6x\left(x^2+4xy+4y^2\right)-8x^3\)

=\(\left[2\left(\frac{x}{2}+y\right)\right]^3-3.2\left(\frac{x}{2}+y\right)^2.2x+3.2\left(\frac{x}{2}+y\right).4x^2-\left(2x\right)^3\)

=\(\left[2\left(\frac{x}{2}+y\right)-2x\right]^3\)

c, C=\(\left(x-y\right)^3-\frac{3\left(x-y\right)^2}{2}y+\frac{3\left(x-y\right)}{4}y^2-\frac{y^3}{8}\)

=\(\left[\left(x-y\right)-\frac{y}{2}\right]^3\)

8.

a,A=(x-y)³+3(x-y)

=(x-y)[(x-y)²+3xy]

=(x-y)(x²-2xy+y²+3xy)

=(x-y)(x²+xy+y²)

b,B=(x+y)³+3(x-y)(x+y)²+3(x-y)²(x+y)+(x-y)³

=(x+y)³+3(x+y)²(x-y)+3(x+y)(x-y)²+(x-y)³

=[(x+y)(x-y)]³

=[(x²-y²)]³

\(\left(x+2\right)^2+4x=\left(x+2\right)\left(4x-3\right)\Leftrightarrow x^2+4x+4+4x=4x^2+8x-3x-6\)

\(\Leftrightarrow3x^2-3x-10=0\Leftrightarrow\orbr{\begin{cases}x=\frac{3+\sqrt{129}}{6}\\x=\frac{3-\sqrt{129}}{6}\end{cases}}\)

\(\left(x+2\right)^2+4x=\left(x+2\right)\left(4x-3\right)\)

\(x^2+4x+4+4x=4x^2+8x-3x-6\)

\(3x^2-3x-10=0\)

\(\Delta=\left(-3\right)^2-\left(4.3.-10\right)=129\)

\(\sqrt{\Delta}=\sqrt{129}\)

\(x_1=\frac{3+\sqrt{129}}{6}\)

\(x_2=\frac{3-\sqrt{129}}{6}\)

gấp 9 lần

(x² - 3x)(-5x² + x) - x²(x - 7) + 5x²(x - 2)(x - 4) - (x + 1)(x + 3)

= -5x⁴ + 16x³ - 3x² - x³ + 7x² + 5x²(x² - 6x + 8) - x² - 4x - 3

= -5x⁴ + 15x³ + 3x² - 4x - 3 + 5x⁴ - 30x³ + 40x²

= -15x³ + 43x² - 4x - 3

(-2x² + 5)(x - 6) - 4x(x² - 7x + 2) - 2x(x  + 3)(x - 2) - (x³ - x²)

= -2x³ + 12x² + 5x - 30 - 4x³ + 28x² - 8x - 2x(x² +  x - 6) - x³ + x²

= -8x³ + 41x² - 3x - 30 - 2x³ - 2x² + 12x

= -10x³ + 39x² + 9x - 30

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