cái này dễ mà

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

= (2x-1)^3

Lời giải:

$x^3-4x^2-12x+27$

$=(x^3+3x^2)-(7x^2+21x)+(9x+27)$

$=x^2(x+3)-7x(x+3)+9(x+3)$

$=(x+3)(x^2-7x+9)$

\(x^3-4x^2+12x-27\)

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

\(=x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)\)

\(=\left(x-3\right)\left(x^2-x+9\right)\)

\(16y^2-4x^2-12x-9=16y^2-\left(4x^2+12x+9\right)=\left(4y\right)^2-\left(2x+3\right)^2\)\(=\left[4y-\left(2x+3\right)\right]\left(4y+2x+3\right)=\left(4y-2x-3\right)\left(4y+2x+3\right)\)

\(a,x^2+6x+9\)

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

\(b,10x-25-x^2\)

\(=-\left(x^2-10x+25\right)\)

\(=-\left(x-5\right)^2\)

\(c,8x^3-\frac{1}{8}\)

\(=8x^3-\left(\frac{1}{2}\right)^3\)

\(=\left(8x-\frac{1}{2}\right)\left(64x^2+4x+\frac{1}{4}\right)\)

\(d,8x^3+12x^2+6xy^2+y^3\)

\(=2\left(4x^3+6x^2+3xy^2+\frac{1}{2}y^3\right)\)

hok tốt!

ai tra loi dung minh cho

Điệp viên 007 sai c

c, \(8x^3-\frac{1}{8}=\left(2x\right)^3-\left(\frac{1}{2}\right)^3=\left(2x-\frac{1}{2}\right)\left(4x^2+x+\frac{1}{4}\right)\)

bạn đăng ít  thôi dc ko

đăng ít thôi thế này ai làm đc

Bạn học căn thức chưa ?

phan tich cac da thuc sau thanh nhan tu theo mau:

a)\(2x^3-x\)

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

\(=x\left(\left(\sqrt{2}x\right)^2-1^2\right)\)\

\(=x\left(\sqrt{2}x-1\right)\left(\sqrt{2}x+1\right)\)

b)\(5x^2\left(x-1\right)-15x\left(x-1\right)\)

\(=\left(5x^2-15x\right)\left(x-1\right)\)

\(=5x\left(x-3\right)\left(x-1\right)\)

d)\(3x\left(x-2y\right)+6y\left(2y-x\right)\)

\(=3x\left(x-2y\right)-6y\left(x-2y\right)\)

\(=\left(3x-6y\right)\left(x-2y\right)\)

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

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