a) Ta có: \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2\)

\(=\left(4x^2-3x-18-4x^2-3x\right)\left(4x^2-3x-18+4x^2+3x\right)\)

\(=\left(-6x-18\right)\left(8x^2-18\right)\)

\(=-6\left(x+3\right)\cdot2\left(4x^2-9\right)\)

\(=-12\left(x+3\right)\left(2x-3\right)\left(2x+3\right)\)

b) Ta có: \(9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)

\(=\left(3x+3y-3\right)^2-\left(4x+6y+2\right)^2\)

\(=\left(3x+3y-3-4x-6y-2\right)\left(3x+3y-3+4x+6y+2\right)\)

\(=-\left(x+3y+5\right)\left(7x+9y-1\right)\)

c) Ta có: \(-4x^2+12xy-9y^2+25\)

\(=-\left(4x^2-12xy+9y^2-25\right)\)

\(=-\left[\left(2x-3y\right)^2-25\right]\)

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

d) Ta có: \(x^2-2xy+y^2-4m^2+4mn-n^2\)

\(=\left(x^2-2xy+y^2\right)-\left(4m^2-4mn+n^2\right)\)

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

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

a) (4x²-3x-18)²-(4x²+3x)²

=(4x²-3x-18-4x²-3x)(4x²-3x-18+4x²+3x)

=(-6x-18)(8x²-18)

=-48x³+108x-144x²+324

phân tích đa thức trên thành nhân tử

`#040911`

`a)`

`196 - a^2 + 2ab - b^2`

`= 196 - (a^2 - 2ab + b^2)`

`= 196 - (a - b)^2`

`= 14^2 - (a - b)^2`

`= (14 - a + b)(14 + a - b)`

`b)`

`a^2 + 6a - 4b^2 + 9`

`= (a^2 + 6a + 9) - 4b^2`

`= [ (a)^2 + 2*a*3 + 3^2] - (2b)^2`

`= (a + 3)^2 - (2b)^2`

`= (a + 3 - 2b)(a + 3 + 2b)`

`c)`

`4x - 4 + 9y^2 - x^2`

`=  9y^2 - (x^2 - 4x + 4)`

`= (3y)^2 - [ (x)^2 - 2*x*2 + 2^2]`

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

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

`d)`

`5x^2 - 10x + 5 - 45t^2`

`= 5*(x^2 - 2x + 1 - 9t^2)`

`= 5*[ (x^2 - 2x + 1) - 9t^2]`

`= 5*{ [(x)^2 - 2*x*1 + 1^2] - (3t)^2}`

`= 5*[ (x - 1)^2 - (3t)^2]`

`= 5*(x - 1 - 3t)(x - 1 + 3t)`

`e)`

`x^2 - 36y^2t^2 - 10x +25`

`= (x^2 - 10x + 25) - 36y^2t^2`

`= [ (x)^2 - 2*x*5 + 5^2] - (6yt)^2`

`= (x - 5)^2 - (6yt)^2`

`= (x - 5 - 6yt)(x - 5 + 6yt)`

a: =196-(a^2-2ab+b^2)

=196-(a-b)^2

=(14-a+b)(14+a-b)

b: \(=\left(a^2+6a+9\right)-4b^2\)

\(=\left(a+3\right)^2-4b^2\)

\(=\left(a+3-2b\right)\left(a+3+2b\right)\)

c: \(=9y^2-\left(x^2-4x+4\right)\)

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

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

d: \(=5\left(x^2-2x+1-9t^2\right)\)

\(=5\left[\left(x-1\right)^2-\left(3t\right)^2\right]\)

\(=5\left(x-1-3t\right)\cdot\left(x-1+3t\right)\)

e: \(=x^2-10x+25-36y^2t^2\)

\(=\left(x-5\right)^2-\left(6yt\right)^2\)

\(=\left(x-5-6yt\right)\left(x-5+6yt\right)\)

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

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

c. \(4x^2-4x+1=\left(2x\right)^2-2.2x.1+1^2=\left(2x-1\right)^2\)

d. \(x^3-3x^2+3x-1\)

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

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

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

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

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

g. \(4x^2+12xy+9y^2=\left(2x\right)^2+2.2x.3y+\left(3y\right)^2=\left(2x+3y\right)^2\)

a) \(5x^2-12xy+9y^2-4x+4=\left(4x^2-12xy+9y^2\right)+x^2-4x+4=\left(2x-3y\right)^2+\left(x-2\right)^2\ge0\)
b) \(-x^2-2y^2+12x-4y+7=-\left(x^2-12x+36\right)-2\left(y^2+2y+1\right)+45=-\left(x-6\right)^2-2\left(y+1\right)^2+45\le45\)

c)\(4y^2+10x^2+12xy+6x+7=\left(4y^2+12xy+9x^2\right)+x^2+6x+9-2=\left(2y+3x\right)^2+\left(x+3\right)^2-2\ge-2\)

d) \(3-10x^2-4xy-4y^2=3-\left(4y^2+4xy+x^2\right)-9x^2=-\left(2y+x\right)^2-9x^2+3\le3\)

e)\(x^2-5x+y^2-xy-4y+16=\left(\frac{1}{2}x^2-xy+\frac{1}{2}y^2\right)+\frac{1}{2}\left(x^2-10x+25\right)+\frac{1}{2}\left(y^2-8y+16\right)-\frac{9}{2}=\frac{1}{2}\left(x-y\right)^2+\frac{1}{2}\left(x-5\right)^2+\frac{1}{2}\left(y-4\right)^2-\frac{9}{2}\ge-\frac{9}{2}\)Phần e) mới nghĩ đk v, tui biết đáp án sao do k xảy ra dấu bằng

a) \(x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2\)

b) \(x^2+12xy+36y^2=\left(x+6y\right)^2\)

c) \(4x^2-12xy+9y^2=\left(2x-3y\right)^2\)

d) Không phải hằng đẳng thức \(\left(x^2-2x+4=\left(x-1\right)^2+3\right)\)

e) \(25x^2+4y^2-20xy=\left(5x-2y\right)^2\)

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

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

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

b: \(=\left(6x+9\right)^2-\left(2x+2\right)^2\)

\(=\left(6x+9+2x+2\right)\left(6x+9-2x-2\right)\)

\(=\left(8x+11\right)\left(4x+7\right)\)

c: \(=\left(4x^2-3x-18-4x^2-3x\right)\left(4x^2-3x-18+4x^2+3x\right)\)

\(=\left(-6x-18\right)\left(8x^2-18\right)\)

\(=-6\cdot2\cdot\left(x+3\right)\left(4x^2-9\right)\)

\(=-12\left(x+3\right)\left(2x-3\right)\left(2x+3\right)\)

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

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

i)\(10x-x^2-25=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)

k)\(4x^2-12xy+9y^2=\left(2x\right)^2-2.2x.3y+\left(3y\right)^2=\left(2x-3y\right)^2\)

mấy bài này cơ bản mà, mở sgk toán 8 ra có các dạng đấy, đăng cũng đăng ít chứ, đăng nhiều quá

a)\(6x^3-9y^2=3\left(2x^3-3y^2\right)\)

b)\(4x^2y-8xy^2+18x^2y^2=2xy\left(2x-4y+9xy\right)\)

c)\(18x^2y-12x^3=6x^2\left(3y-2x\right)\)

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

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

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

\(=4x^2-4x+4-x^2-2x-1\)\(=3x^2-6x+3\)\(=3\left(x^2-2x+1\right)\)

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

Vâng ạ ^^ Cảm ơn ạ.