\(a,=\left(x+1\right)^2\\ b,=\left(3x-y\right)^2\\ c,=\left(x-3\right)\left(x+3\right)\\ d,=\left(x+4\right)^3\\ e,=\left(x-2\right)^3\\ f,=\left(x+2\right)\left(x^2-2x+4\right)\\ g,=\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)

a,= (x+4)\(^3\)

b,= (x-2)\(^3\)

c,= x\(^3\)+8

d,=x\(^3\)-27

lời giải đâu ạ

1:

a: A=(x+4)^3=10^3=1000

b: B=(x-2)^3=20^3=8000

1

a) \(A=x^3+3.x^2.4+3x.4^2+4^3=\left(x+4\right)^3=\left(6+4\right)^3=10^3=1000\)

b) \(B=x^3-3.x^2.2+3.x.2^2-2^3=\left(x-2\right)^3=\left(22-2\right)^3=20^3=8000\)

2

\(VT=\left(x-y\right)^2+4xy=x^2-2xy+y^2+4xy=x^2+2xy+y^2=\left(x+y\right)^2=VP\)

a: \(\left(x-2\right)^2-\left(2x+3\right)^2=0\)

=>(x-2-2x-3)(x-2+2x+3)=0

=>(-x-5)(3x+1)=0

=>(x+5)(3x+1)=0

=>\(\left[\begin{array}{l}x+5=0\\ 3x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-5\\ x=-\frac13\end{array}\right.\)

b: \(9\left(2x+1\right)^2-4\left(x+1\right)^2=0\)

=>\(\left\lbrack3\left(2x+1\right)\right\rbrack^2-\left\lbrack2\left(x+1\right)\right\rbrack^2=0\)

=>\(\left(6x+3\right)^2-\left(2x+2\right)^2=0\)

=>(6x+3+2x+2)(6x+3-2x-2)=0

=>(8x+5)(4x+1)=0

=>\(\left[\begin{array}{l}8x+5=0\\ 4x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac58\\ x=-\frac14\end{array}\right.\)

c: \(x^3-6x^2+9x=0\)

=>\(x\left(x^2-6x+9\right)=0\)

=>\(x\left(x-3\right)^2=0\)

=>\(\left[\begin{array}{l}x=0\\ \left(x-3\right)^2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x-3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=3\end{array}\right.\)

d: \(x^2\left(x+1\right)-x\left(x+1\right)+x\left(x-1\right)=0\)

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

=>x(x+1)(x-1)+x(x-1)=0

=>x(x-1)(x+1+1)=0

=>x(x-1)(x+2)=0

=>\(\left[\begin{array}{l}x=0\\ x-1=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=1\\ x=-2\end{array}\right.\)

e: \(\left(x-2\right)^2-\left(x-2\right)\left(x+2\right)=0\)

=>(x-2)(x-2-x-2)=0

=>-4(x-2)=0

=>x-2=0

=>x=2

g: \(x^4-2x^2+1=0\)

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

=>\(x^2-1=0\)

=>\(x^2=1\)

=>\(\left[\begin{array}{l}x=1\\ x=-1\end{array}\right.\)

h: \(4x^2+y^2-20x-2y+26=0\)

=>\(4x^2-20x+25+y^2-2y+1=0\)

=>\(\left(2x-5\right)^2+\left(y-1\right)^2=0\)

=>\(\begin{cases}2x-5=0\\ y-1=0\end{cases}\Rightarrow\begin{cases}x=\frac52\\ y=1\end{cases}\)

i: \(x^2-2x+5+y^2-4y=0\)

=>\(x^2-2x+1+y^2-4y+4=0\)

=>\(\left(x-1\right)^2+\left(y-2\right)^2=0\)

=>\(\begin{cases}x-1=0\\ y-2=0\end{cases}\Rightarrow\begin{cases}x=1\\ y=2\end{cases}\)

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

\(=\left(2x+y\right).3y\)

b) \(\left(x+1\right)^3+\left(x-1\right)^3\)

\(=\left(x+1+x-1\right)\left[\left(x+1\right)^2-\left(x+1\right)\left(x-1\right)+\left(x-1\right)^2\right]\)

\(=2x\left[\left(x+1\right)^2-\left(x^2-1\right)+\left(x-1\right)^2\right]\)

c) \(9x^2-3x+2y-4y^2\)

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

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

\(=\left(3x-2y\right)\left[3x+2y-1\right]\)

d) \(4x^2-4xy+2x-y+y^2\)

\(=4x^2-4xy+y^2+2x-y\)

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

\(=\left(2x-y\right)\left(2x-y+1\right)\)

e) \(x^3+3x^2+3x+1-y^3\)

\(=\left(x+1\right)^3-y^3\)

\(=\left(x+1-y\right)\left[\left(x+1\right)^2+y\left(x+1\right)+y^2\right]\)

g) \(x^3-2x^2y+xy^2-4x\)

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

\(=x\left(x-y\right)^2-4x\)

\(=x\left[\left(x-y\right)^2-4\right]\)

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

a) (x + 2y)² - (x - y)²

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

= 3y(2x + y)

b) (x + 1)³ + (x - 1)³

= (x + 1 + x - 1)[(x + 1)² - (x + 1)(x - 1) + (x - 1)²]

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

= 2x(x² + 3)

c) 9x² - 3x + 2y - 4y²

= (9x² - 4y²) - (3x - 2y)

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

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

d) 4x² - 4xy + 2x - y + y²

= (4x² - 4xy + y²) + (2x - y)

= (2x - y)² + (2x - y)

= (2x - y)(2x - y + 1)

e) x³ + 3x² + 3x + 1 - y³

= (x³ + 3x² + 3x + 1) - y³

= (x + 1)³ - y³

= (x + 1 - y)[(x + 1)² + (x + 1)y + y²]

= (x - y + 1)(x² + 2x + 1 + xy + y + y²)

g) x³ - 2x²y + xy² - 4x

= x(x² - 2xy + y² - 4)

= x[(x² - 2xy + y²) - 4]

= x[(x - y)² - 2²]

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

e) Ta có: \(x^4-2x^3+2x-1\)

\(=\left(x^4-1\right)-2x\left(x^2-1\right)\)

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

\(=\left(x-1\right)\left(x+1\right)\cdot\left(x^2-2x+1\right)\)

\(=\left(x+1\right)\cdot\left(x-1\right)^3\)

h) Ta có: \(3x^2-3y^2-2\left(x-y\right)^2\)

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

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

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

\(=\left(x-y\right)\left(x+5y\right)\)

a) Ta có: \(x^2-y^2-2x-2y\)

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

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

b) Ta có: \(x^2\left(x+2y\right)-x-2y\)

\(=\left(x+2y\right)\left(x^2-1\right)\)

\(=\left(x+2y\right)\left(x-1\right)\left(x+1\right)\)

a: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)

\(=4x^2-4x+1+4-2\left(4x^2-12x+9\right)\)

\(=4x^2-4x+5-8x^2+24x-18\)

\(=-4x^2+20x-13\)

e: \(\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)=8x^3+27y^3\)

a: Ta có: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)

\(=4x^2-4x+1-2\left(4x^2-12x+9\right)+4\)

\(=4x^2-4x+5-8x^2+24x-18\)

\(=-4x^2+20x-13\)

b: \(\left(3x+2\right)^2+2\left(3x+2\right)\left(1-2y\right)+\left(1-2y\right)^2\)

\(=\left(3x+2+1-2y\right)^2\)

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