\(a,4x^2-1\)

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

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

\(b,25x^2-0,09\)

\(=\left(5x\right)^2-\left(0,3\right)^2\)

\(=\left(5x-0,3\right)\left(5x+0,3\right)\)

\(d,\left(x-y\right)^2-4\)

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

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

\(e,9-\left(x-y\right)^2\)

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

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

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

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

\(f,\left(x^2+4\right)^2-16x^2\)

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

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

\(=\left(x^2-2\cdot x\cdot2+2^2\right)\left(x^2+2\cdot x\cdot2+2^2\right)\)

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

#\(Toru\)

`#3107`

a)

`4x^2 - 1`

`= (2x)^2 - 1^2`

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

b)

`25x^2 - 0,09`

`= (5x)^2 - (0,3)^2`

`= (5x - 0,3)(5x + 0,3)`

d)

`(x - y)^2 - 4`

`= (x - y)^2 - 2^2`

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

e)

`9 - (x - y)^2`

`= 3^2 - (x - y)^2`

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

f)

`(x^2 + 4)^2 - 16x^2`

`= (x^2 + 4)^2 - (4x)^2`

`= (x^2 - 4x + 4)(x^2 + 4x + 4)`

`= (x - 2)^2 * (x + 2)^2`

_____

Tất cả các câu trên bạn sử dụng hđt:

`A^2 - B^2 = (A - B)(A + B)`

\(#MaiChangLaAnhDau..\)

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

a: =64x^4+16x^2y^2+y^4-16x^2y^2

=(8x^2+y^2)^2-(4xy)^2

=(8x^2+y^2-4xy)(8x^2+y^2+4xy)

b: =x^8+2x^4+1-x^4

=(x^4+1)^2-x^4

=(x^4-x^2+1)(x^4+x^2+1)

=(x^4-x^2+1)(x^4+2x^2+1-x^2)

=(x^4-x^2+1)(x^2+1-x)(x^2+x+1)

c: =(x+1)(x^2-x+1)+2x(x+1)

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

=(x+1)(x^2+x+1)

d: =(x^2-1)(x^2+1)-2x(x^2-1)

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

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

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

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

\(b,x^3-2x^2-4xy^2+x\)

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

\(=x\left[\left(x^2-2x+1\right)-4y^2\right]\)

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

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

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

\(---\)

\(c,\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8\)

\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-8\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\) (1)

Đặt \(y=x^2+7x+10\), thay vào (1) ta được:

\(y\left(y+2\right)-8\)

\(=y^2+2y+1-9\)

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

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

\(=\left(y-2\right)\left(y+4\right)\)

\(=\left(x^2+7x+10-2\right)\left(x^2+7x+10+4\right)\)

\(=\left(x^2+7x+8\right)\left(x^2+7x+14\right)\)

#Ayumu

\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)

(x+1)(x+2)(x+3)(x+4)-8

=[(x+1).(x+4)].[(x+2).(x+3)]-8

=(x²+5x+4).(x²+5x+6)-8

Đặt (x²+5x+4)=t =>(x²+5x+6)=t+2

Thay vào biểu thức ta có:

(x²+5x+4).(x²+5x+6)-8

t.(t+2)-8

=t²+2t+1-9

=(t+1)²-3²

=(x²+5x+4+1)-3²

=(x²+5x+5+3).(x²+5x+5-3)

=(x²+5x+8).(x²+5x+2)

=

ta làm như sau : 

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

\(\Rightarrow\left(x^2+5X+4\right)\left(x^2+5x+6\right)-8\)

Đặt \(x^2+5x+4=t\)

\(\Leftrightarrow t\left(t+2\right)-8\)

\(\Leftrightarrow t^2+2t-8\Leftrightarrow t^2+2t+1-9\)

\(\Leftrightarrow\left(t+1\right)^2-3^2\)

\(\Leftrightarrow\left(t-2\right)\left(t+4\right)\)

\(\Leftrightarrow\left(x^2+5x+2\right)\left(x^2+5x+8\right)\)