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

\(2x^2-7x+3\)

\(=2\left(x^2-\frac{7}{2}x+\frac{3}{2}\right)\)

Vậy thôi đâu cần dùng HĐT

2x²-7x+3

=2x²-x-6x+3

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

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

\(=\left(4-a+b\right)\left(4+a-b\right)\)

\(=\left(4-a-b\right)\left(4+a-b\right)\), đằng trước là dấu trừ thì khi bỏ ngoặc phải đổi dấu chứ nhỉ :0

\(\left(25-m^2\right)=\left(5-m\right)\left(5+m\right)\)

giai chi tiết hộ với em mới học

\(10a-25-a^2=-\left(a^2-10a+25\right)=-\left(a^2-2.a.5+5^2\right)=-\left(a-5\right)^2\)

`10a-25-a^2`

`=-(a^2-10a+25)`

`=-(a^2-2.a.5+5^2)`

`=-(a-5)^2`

Áp dụng hằng đẳng thức : `(a-b)^2=a^2-2ab+b^2`

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

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

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

10x - 25 - x²

= x²- 10x - 25

= - ( x² +10x +25)

= -(x² + 2.x.5+5² )

= - (x+5 )²

đề sai rùi phải là : \(36\left(x-y\right)^2-25\left(2x-1\right)^2\)

\(=>\left[6\left(x-y\right)\right]^2-\left[5\left(2x-1\right)\right]^2=\left[6\left(x-y\right)-5\left(2x-1\right)\right]\left[6\left(x-y\right)+5\left(2x-1\right)\right]\)

\(=>\left(6x-6y-10x+5\right)\left(6x-6y+10x-5\right)=\left(5-4x-6y\right)\left(16x-6y-5\right)\)

Áp dụng HDT : x^2 -y^2 =(x-y) (x+y)

Ủng hộ = 1 cái t i c k nha cảm ơn

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:

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