3(x + 2)^2 + (2x - 1)^2 - 7(x + 3)(x - 3) = 36

=> 3(x^2 + 4x + 4) + 4x^2 - 4x + 1 - 7(x^2 - 9) = 36

=> 3x^2 + 12x + 12 + 4x^2 - 4x + 1 - 7x^2 + 63 = 36

=> 8x + 76 = 36

=> 8x = -40

=> x = -5

= (x²)³ + 1 

= x⁶ + 1

(2a-b)² - 2 x ( 2a-b) x (a+b) + (a + b)²

= [(2a-b) - (a+b)]²

a) \(\left(2x^2-1\right)^2=\left(2x^2\right)^2-2.2x^2.1+1^2\)

\(=4x^4-4x^2+1\).

b) \(\left(\frac{1}{2}x+3y^2\right)^2=\left(\frac{1}{2}x\right)^2+2.\frac{1}{2}x.3y^2+\left(3y^2\right)^2\)

\(=\frac{1}{4}x^2+3y^2x+9y^4\)

Chúc bn hc tốt!

\(a,\left(2x+y+3\right)^2=4x^2+y^2+9+4xy+12x+6y\)

\(b,\left(x-2y+1\right)^2=x^2+4y^2+1-4xy+2x-4y\)

\(c,\left(x^2-2xy^2-3\right)^2=x^4+2x^2y^4+9-4x^3y^2-6x^2+12xy^2\)

\(x^2-y^2=\left(x+y\right)\left(x-y\right)\)

\(\left(a+b\right)^2=a^2+2ab+b^2\)

ai k mk mk k lại

x²-y²=(a-y)(x+y) ;  

a) ( x+y)x(x-y)

b) a^2+2ab+b^2

a) \(\left(3x^2-2y^3\right)^2\)

\(=\left(3x^2\right)^2-2\cdot3x^2\cdot2y^3+\left(2y^3\right)^2\)

\(=9x^4-12x^2y^3+4y^6\)

b) \(\left(-2x^2-3\right)^2\)

\(=\left(-2x^2\right)^2-2\cdot\left(-2x^2\right)\cdot3+3^2\)

\(=4x^4+12x^2+9\)

a) (x² + 2xy + y²) : (x + y) = (x + y)² : (x + y) = x + y.

b) (125x³ + 1) : (5x + 1) = [(5x)³ + 1] : (5x + 1)

= (5x)² – 5x + 1 = 25x² – 5x + 1.

c) (x² – 2xy + y²) : (y – x) = (x – y)² : [-(x – y)] = - (x – y) = y – x

Hoặc (x² – 2xy + y²) : (y – x) = (y² – 2xy + x²) : (y – x)

= (y – x)² : (y – x) = y - x.

Bài giải:

a) (x² + 2xy + y²) : (x + y) = (x + y)² : (x + y) = x + y.

b) (125x³ + 1) : (5x + 1) = [(5x)³ + 1] : (5x + 1)

= (5x)² – 5x + 1 = 25x² – 5x + 1.

c) (x² – 2xy + y²) : (y – x) = (x – y)² : [-(x – y)] = - (x – y) = y – x

Hoặc (x² – 2xy + y²) : (y – x) = (y² – 2xy + x²) : (y – x)

= (y – x)² : (y – x) = y - x.

a) (x² +2xy + y² ) : (x +y)

= (x +y)² : (x +y)

= x + y

b) (125x³ + 1) : (5x + 1)

= (5x + 1)(25x² - 5x + 1) : (5x + 1)

= 25x² - 5x + 1

c) \(\left(x^2-2xy+y^2\right)\left(y-x\right)\)

= \(\left(x-y\right)^2:\left(y-x\right)\)

= \(x-y\)

a) ta có : \(\left(2+i\sqrt{3}\right)^2=2^2+2.2.i\sqrt{3}+\left(i\sqrt{3}\right)^2\)

\(=4+4\sqrt{3}i-3=1+4\sqrt{3}i\)

b) ta có : \(\left(1+2i\right)^3=1^3+3.1^2.2i+3.1.\left(2i\right)^2+\left(2i\right)^3\)

\(=1+6i-6-8i=-5-2i\)

c) \(\left(3-i\sqrt{2}\right)^3=3^3-3.3^2.i\sqrt{2}+3.3.\left(i\sqrt{2}\right)^2+\left(i\sqrt{2}\right)^3\)

\(=27-27\sqrt{2}i-18-2\sqrt{2}i=9-29\sqrt{2}i\)

d) \(\left(2-i\right)^3=2^3-2.2^2.i+2.2.i^2-i^3\)

\(=8-8i-4+i=4-7i\)