a.x^3-1^3

b.x^3-5^3

c)(2x)^3+3^3

d)x^3+1/2^3

a) 3(x - 1)² - 3x(x - 5) = 3(x² - 2x + 1) - 3x² + 15x = 3x² - 6x + 3 - 3x² + 15x = 9x + 3 = 21 => x = (21 - 3) : 9 = 18 : 9 = 2

b) 3(x + 2)² + (2x - 1)² - 7(x + 3)(x - 3) = 3(x² + 4x + 4) + 4x² - 4x + 1 - 7(x² - 9) = 3x² + 12x + 12 + 4x² - 4x + 1 - 7x² + 63

= 8x + 76 = 36 => x = (36 - 76) : 8 = -40 : 8 = -5

Bài 2:

a: Ta có: \(A=\left(x+1\right)^3+\left(x-1\right)^3\)

\(=x^3+3x^2+3x+1+x^3-3x^2+3x-1\)

\(=2x^3+6x\)

b: Ta có: \(B=\left(x-3\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(3x-1\right)\left(3x+1\right)\)

\(=x^3-9x^2+27x-27-x^3-27+9x^2-1\)

\(=27x-55\)

a) Ta có: \(x\left(x-1\right)-x^2+2x=5\)

\(\Leftrightarrow x^2-x-x^2+2x=5\)

hay x=5

b) Ta có: \(2x^2-2x=\left(x-1\right)^2\)

\(\Leftrightarrow2x\left(x-1\right)-\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x-x+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

c) Ta có: \(\left(x+3\right)\cdot\left(x^2-3x+9\right)-x\left(x-2\right)^2=19\)

\(\Leftrightarrow x^3+27-x\left(x^2-4x+4\right)-19=0\)

\(\Leftrightarrow x^3+8-x^3+4x^2-4x=0\)

\(\Leftrightarrow4x^2-4x+8=0\)(Vô lý)

a) 

(x³ – 3x²+3x-1) – (x³+9) +(3x²-12) = 2

3x-22 = 2

3x =24

=> x= 8

b)

x³+6x²+12x+8-6x²-x³-x²+x² = 5

                    12x+8 = 5

                      12x = -3

                       =>x = -1/4

a) ( x - 1 )³ - ( x + 3 )(x² - 3x + 9 ) + 3( x² - 4 ) = 2

(x³-3x²\(\times\)1+3x\(\times\)1²-1³)-(x³+3³)+(3x²-3\(\times\)4)=2

x³-3x²+3x-1-x³-9+3x²-12=2

3x-40=2

3x=42

x=14

b ) ( x + 2 )³ - 6x² - x² ( x + 1 ) + x² = 5

(x³+3\(\times\)2x²+3x\(\times\)2²+2³)-6x²-(x³+x²)+x²=5

x³+6x²+12x+8-6x²-x³-x²+x²=5

12x+8=5

12x=\(-\)3

x=\(-\frac{1}{4}\)

bài dễ mà 

Lời giải:

a. $x(3x+1)+(x-1)^2-(2x+1)(2x-1)=0$

$\Leftrightarrow (3x^2+x)+(x^2-2x+1)-(4x^2-1)=0$

$\Leftrightarrow 3x^2+x+x^2-2x+1-4x^2+1=0$

$\Leftrightarrow (3x^2+x^2-4x^2)+(x-2x)+(1+1)=0$

$\Leftrightarrow -x+2=0$

$\Leftrightarrow x=2$

b.

$(x+1)^3+(2-x)^3-9(x-3)(x+3)=0$

$\Leftrightarrow [(x+1)+(2-x)][(x+1)^2-(x+1)(2-x)+(2-x)^2]-9(x-3)(x+3)=0$

$\Leftrightarrow 3[x^2+2x+1-(x-x^2+2)+(x^2-4x+4)]-9(x-3)(x+3)=0$

$\Leftrightarrow 3(3x^2-3x+3)-9(x^2-9)=0$

$\Leftrightarrow 9(x^2-x+1)-9(x^2-9)=0$

$\Leftrightarrow 9(x^2-x+1-x^2+9)=0$
$\Leftrightarrow 9(-x+10)=0$

$\Leftrightarrow -x+10=0\Leftrightarrow x=10$

c.

$(x-1)^3-(x+3)(x^2-3x+9)+3x^2=25$

$\Leftrightarrow (x^3-3x^2+3x-1)-(x^3+3^3)+3x^2=25$

$\Leftrightarrow x^3-3x^2+3x-1-x^3-27+3x^2=25$
$\Leftrightarrow (x^3-x^3)+(-3x^2+3x^2)+3x-28=25$

$\Leftrightarrow 3x-28=25$

$\Leftrightarrow x=\frac{53}{3}$

d.

$(x+2)^3-(x+1)(x^2-x+1)-6(x-1)^2=23$
$\Leftrightarrow (x^3+6x^2+12x+8)-(x^3+1)-6(x^2-2x+1)=23$

$\Leftrightarrow x^3+6x^2+12x+8-x^3-1-6x^2+12x-6=23$

$\Leftrightarrow (x^3-x^3)+(6x^2-6x^2)+(12x+12x)+(8-1-6)=23$
$\Leftrightarrow 24x+1=23$

$\Leftrgihtarrow 24x=22$

$\Leftrightarrow x=\frac{11}{12}$

e.

$(x+3)(x^2-3x+9)-x(x-2)(x+2)+11=0$

$\Leftrightarrow x^3+3^3-x(x^2-4)+11=0$

$\Leftrightarrow x^3+27-x^3+4x+11=0$

$\Leftrightarrow (x^3-x^3)+4x+(27+11)=0$

$\Leftrightarrow 4x+38=0$

$\Leftrightarrow x=\frac{-19}{2}$

f.

$x(x-3)-x+3=0$

$\Leftrightarrow x(x-3)-(x-3)=0$

$\Leftrightarrow (x-3)(x-1)=0$

$\Leftrightarrow x-3=0$ hoặc $x-1=0$

$\Leftrightarrow x=3$ hoặc $x=1$

\(A=x^2-2x+1-x^2+4=5-2x\)

\(B=27x^3+8-x^2+9=27x^3-x^2+17\)

\(C=3x^2y-6xy^2-2x\left(x^2-2x^2y+x^2y^2\right)=3x^2y-6xy^2-2x^3+4x^3y-2x^3y^2\)

Em chỉ cần nhớ hằng đẳng thức và áp dụng là biến đổi được ^^