Bài 3: 

a: \(=\left(2x-5\right)\left(4x^2+10x+25\right)\)

ghi rõ cách làm cho mình với ạ

câu b sai r 

\(\dfrac{1}{3}xy+x^2z+xz=3x\left(\dfrac{1}{9}y+\dfrac{1}{3}xz+\dfrac{1}{3}z\right)\)

Lời giải:

a.

$=\frac{1}{2}(x^2-4y^2)=\frac{1}{2}[x^2-(2y)^2]=\frac{1}{2}(x-2y)(x+2y)$

b.

$=\frac{1}{3}x(y+3xz+3z)$

c.

$=\frac{2}{25}x(225x^2-4)=\frac{2}{25}(15x-2)(15x+2)$

d.

$=\frac{1}{5}x^2(2+25x+5y)$

\(a,=\dfrac{1}{2}\left[\left(x^2+y^2\right)^2-4x^2y^2\right]\\ =\dfrac{1}{2}\left(x^2-2xy+y^2\right)\left(x^2+2xy+y^2\right)\\ =\dfrac{1}{2}\left(x-y\right)^2\left(x+y\right)^2\\ b,=\left(3x-\dfrac{1}{2}y\right)\left(9x^2+\dfrac{3}{2}xy+\dfrac{1}{4}y^2\right)\\ c,=\dfrac{1}{2}\left(x^2+\dfrac{1}{2}x+\dfrac{1}{16}\right)=\dfrac{1}{2}\left(x+\dfrac{1}{4}\right)^2\)

a) 8x³ - 125=(2x)³-5³=(2x-5)(2x+5)

b) 1 - 27x³y⁶=1-3³x³(y²)³=1-(3xy²)³=(1-3xy²)(1-3xy²+9x²y⁴)

c)27x³ + y³/64=(3x)³ + y³/4³=(3x)³ + (y/4)³=(3x+y/4)(9x²-3xy/4+y²/16)

d) 27x³ + 125y³ = (3x)³+(5y)³=(3x+5y)(9x²-15xy+25y²)

Đúng hết mà?

A

A

x đầu ở đa thức A là x^3 chăng?

a/ \(A=x^3-5x^2+8x-4\)

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

\(=x^2\left(x-1\right)-4\left(x-1\right)\left(x+1\right)+8\)

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

b/ \(B=\dfrac{x^5}{30}-\dfrac{x^3}{6}+\dfrac{2x}{15}\)

\(=\dfrac{x^5}{30}-\dfrac{5x^3}{30}+\dfrac{4x}{30}\)

\(=\dfrac{x\left(x^4-5x^2+4\right)}{30}\)

\(=\dfrac{x\left(x^4-x^2-4x^2+4\right)}{30}\)

\(=\dfrac{x\left(x+2\right)\left(x-1\right)\left(x+1\right)\left(x-2\right)}{30}\)

\(a,=2\left(\dfrac{1}{4}x^2-y^2\right)=2\left(\dfrac{1}{2}x-y\right)\left(\dfrac{1}{2}x+y\right)\\ b,=\dfrac{1}{3}x\left(y+3xz+3z\right)\\ c,=2x\left(9x^2-\dfrac{4}{25}\right)=2x\left(3x-\dfrac{2}{5}\right)\left(3x+\dfrac{2}{5}\right)\)

\(d,=x^2\left(\dfrac{2}{5}+5x+y\right)\\ e,=\dfrac{1}{2}\left[\left(x^2+y^2\right)^2-4x^2y^2\right]\\ =\dfrac{1}{2}\left(x^2-2xy+y^2\right)\left(x^2+2xy+y^2\right)\\ =\dfrac{1}{2}\left(x-y\right)^2\left(x+y\right)^2\\ f,=\left(3x-\dfrac{1}{2}y\right)\left(9x^2+\dfrac{3}{2}xy+\dfrac{1}{4}y^2\right)\\ g,=\dfrac{1}{2}\left(x^2+\dfrac{1}{2}x+\dfrac{1}{16}\right)=\dfrac{1}{2}\left(x+\dfrac{1}{4}\right)^2\)