\(2x^2y^3-\frac{x}{4}-4y^6\)

đây là pt bậc 2 của y^3 , ta đặt y^3=z ta được

\(-\left(4z^2+\frac{2.2xz}{2}+\frac{x^2}{4}\right)+\left(\frac{x^2}{4}-\frac{x}{4}\right)\)

\(-\left(2z+\frac{x}{2}\right)^2+\left(\frac{x^2}{4}-\frac{x}{4}\right)\)

\(-\left\{\left(2x+\frac{x}{2}\right)^2-\left(\frac{x^2}{4}-\frac{x}{4}\right)\right\}\)

\(-\left\{\left(2x+\frac{x}{2}+\sqrt{\frac{x^2}{4}-\frac{x}{4}}\right)\left(2x+\frac{x}{2}-\sqrt{\frac{x^2}{4}-\frac{x}{4}}\right)\right\}\)

Bài 2:

1)  \(x^2-4x+4=\left(x-2\right)^2\)

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

3) \(1-8x^3=\left(1-2x\right)\left(1+2x+4x^2\right)\)

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

5) \(\dfrac{1}{25}x^2-64y^2=\left(\dfrac{1}{5}x-8y\right)\left(\dfrac{1}{5}x+8y\right)\)

6) \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)

mình cảm ơn nha

Bài 1:

\(1,Sửa:x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\\ 2,=6\left(x^2+2xy+y^2\right)=6\left(x+y\right)^2\\ 3,=2y\left(y^2+4y+4\right)=2y\left(y+2\right)^2\\ 4,=5\left(x^2-2xy+y^2\right)=5\left(x-y\right)^2\)

Bài 2:

\(1,=x\left(x^2-64\right)=x\left(x-8\right)\left(x+8\right)\\ 2,=2y\left(4x^2-9\right)=2y\left(2x-3\right)\left(2x+3\right)\\ 3,=3\left(x^3-1\right)=3\left(x-1\right)\left(x^2+x+1\right)\)

Bài 3:

\(a,=5\left(x^2+2x+1-y^2\right)=5\left[\left(x+1\right)^2-y^2\right]=5\left(x-y+1\right)\left(x+y+1\right)\\ b,=3x\left(x^2-2x+1-4y^2\right)=3x\left[\left(x-1\right)^2-4y^2\right]\\ =3x\left(x-2y-1\right)\left(x+2y-1\right)\\ c,=ab\left(a-b\right)\left(a+b\right)+\left(a+b\right)^2\\ =\left(a+b\right)\left(a^2b-ab^2+a+b\right)\\ d,=2x\left(x^2-y^2-4x+4\right)=2x\left[\left(x-2\right)^2-y^2\right]\\ =2x\left(x-y-2\right)\left(x+y-2\right)\)

x³+64=x³+4³=(x+4)(x²-4x+4²)=(x+4)(x²-4x+16)

a) Ta có: \(a^3y^3+125\)

\(=\left(ay+5\right)\left(a^2y^2-5ay+25\right)\)

b) Ta có: \(8x^3-y^3-6xy\cdot\left(2x-y\right)\)

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

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

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

a) \(=\left(x-2\right)^2\)

b) \(=\left(2x+1\right)^2\)

c) \(=\left(4x-3y\right)\left(4x+3y\right)\)

d) \(=\left(4-x-3\right)\left(4+x+3\right)=\left(1-x\right)\left(x+7\right)\)

e) \(=\left(2x-3x+1\right)\left(2x+3x-1\right)=\left(1-x\right)\left(5x-1\right)\)

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

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

h) \(=\left(x+2\right)^3\)

i) \(=\left(1-x\right)^3\)

a/ $=(x-2)^2$

b/ $=(2x+1)^2$

c/ $=(4x-3y)(4x+3y)$

d/ $=(1-x)(x+7)$

e/ $=(-x+1)(5x-1)$

f/ $=(x-y)(x^2+xy+y^2)$

g/ $=(3+x)(9-3x+x^2)$

h/ $=(x+2)^3$

i/ $=(1-x)^3$

a: \(x^2-4x+4=\left(x-2\right)^2\)

b: \(4x^2+4x+1=\left(2x+1\right)^2\)

g: \(x^3+27=\left(x+3\right)\left(x^2-3x+9\right)\)

b)  \(64x^3+1=\left(4x+1\right)\left(16x^2-4x+1\right)\)\

c) \(x^3y^6z^9-125=\left(xy^2z^3-5\right)\left(x^2y^4z^6+5xy^2z+25\right)\)

d)  \(27x^6-8x^3=x^3\left(27x^3-8\right)=x^3\left(3x-2\right)\left(9x^2+6x+4\right)\)

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