a) Ta có: \(x^2-2xy+y^2-2x+2y\)

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

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

b) Ta có: \(x^2-4x+4-x^2y+2xy\)

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

\(=\left(x-2\right)\left(x-2-xy\right)\)

c) Ta có: \(ax^2-3axy-x^2+6xy-9y^2\)

\(=ax\left(x-3y\right)-\left(x^2-6xy+9y^2\right)\)

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

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

d) Ta có: \(2a^2x-5a^2y-4x^2+30xy-25y^2\)

\(=a^2\left(2x-5y\right)-\left(4x^2-30xy+25y^2\right)\)

\(=a^2\left(2x-5y\right)-\left(2x-5y\right)^2\)

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

a) \(4x^2y^2-12xy+9y^2\)

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

b) \(25x^2-36y^2\)

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

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

a) \(x^2+2x+1=\left(x+1\right)^2\)

b) \(x^2+8x+16=\left(x+4\right)^2\)

c) \(x^2+6x+9=\left(x+3\right)^2\)

d) \(4x^2+4x+1=\left(2x+1\right)^2\)

e) \(36+x^2-12x=x^2-12x+36=\left(x-6\right)^2\)

f) \(4x^2+12x+9=\left(2x+3\right)^2\)

g) \(x^4+81+18x^2=x^4+18x^2+81=\left(x^2+9\right)^2\)

h) \(9x^2+30xy+25y^2=\left(3x+5y\right)^2\)

a, \(x^2\) + 2\(x\) + 1 = (\(x\) + 1)²

b, \(x^2\) + 8\(x\) + 16 = (\(x\) + 4)²

c, \(x^2\) + 6\(x\) + 9 = (\(x\) + 3)²

d, 4\(x^2\) + 4\(x\) + 1 = (2\(x\) + 1)²

a) \(39x-39y=39\left(x-y\right)\)

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

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

c) \(16x^2+24xy+9y^2=\left(4x\right)^2+4x.3y.2+\left(3y\right)^2=\left(4x+3y\right)^2\)

d) \(25x^2-\frac{1}{25y^2}=\left(5x\right)^2-\left(\frac{1}{5y}\right)^2=\left(5x-\frac{1}{5y}\right)\left(5x+\frac{1}{5y}\right)\)

e) \(7x^2-7xy+5x-5y=7x\left(x-y\right)+5\left(x-y\right)=\left(x-y\right)\left(7x+5\right)\)

f) \(5x^2-45y^2-30y-5=5\left(x^2-9y^2-6y-1\right)=5\left[x^2-\left(9y^2+6y+1\right)\right]\)

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

g) \(x^2+2x+1-y^2-4y-1=\left(x^2+2x+1\right)-\left(y^2+2y+1\right)\) ( Chắc đề vậy :v ) 

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

h) \(4x^2+8x-5=4x^2-2x+10x-5=2x\left(2x-1\right)+5\left(2x-1\right)\)

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

Bài 1: Viết các biểu thức sau dưới dạng bình phương của 1 tổng hoặc 1 hiệu

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

b) \(25x^2-20xy+4y^2=\left(5x\right)^2-2.5x.2y+\left(2y\right)^2=\left(5x-2y\right)^2\)

c) \(9x^2+y^2-6xy=\left(3x\right)^2-2.3xy+y^2=\left(3x-y\right)^2\)

d) \(x^2+6xy+9y^2=x^2+2x.3y+\left(3y\right)^2=\left(x+3y\right)^2\)

e) \(x^2-10xy+25y^2=x^2-2x.5y+\left(5y\right)^2=\left(x-5y\right)^2\)

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

Câu cuối mình sửa lại đề nhé bạn! Nếu để như trên đề thì không thể viết đáp án dưới dạng bình phương của 1 tổng hoặc 1 hiệu được.

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

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

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

\(x^2+6xy+9y^2=\left(x+3y\right)^2\)

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

\(\left(3x+2y\right)+2\left(3x+2y\right)+1=3\left(3x+2y\right)+1=9x+6y+1\)

a) \(x^2-9\)

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

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

b) \(4x^2-1\)

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

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

c) \(9x^2-y^2\)

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

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

d) \(25x^2-9y^2\)

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

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

e) \(36x^2-25y^2\)

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

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

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

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

tương tự

\(8x^3-64y^3=\left(2x\right)^3-\left(4y\right)^3=\left(2x-4y\right)\left(4x^2+8xy+16y^2\right)\)

\(9x^2-30xy+25y^2=\left(3x\right)^2-2\cdot3x\cdot5y+\left(5y\right)^2=\left(3x-5y\right)^2\)

\(4x^2+16x+7=\left(2x^2\right)+2\cdot2x\cdot4+4^2-9=\left(2x+4\right)^2-3^2=\left(2x+1\right)\left(2x+7\right)\)

\(-5+18y-9y^2=-\left[\left(3y\right)^2-2\cdot3y\cdot3+3^2-4\right]=-\left[\left(3y-3\right)^2-2^2\right]=-\left(3y-5\right)\left(3y-1\right)\)

đến đó giải tiếp hộ mik nx ddi đc k

Giúp mình với ạ,cảm ơn mọi người

b: Ta có: \(B=x^2+4x+9y^2-6y-1\)

\(=x^2+4x+4+9y^2-6y+1-6\)

\(=\left(x+2\right)^2+\left(3y-1\right)^2-6\ge-6\forall x,y\)

Dấu '=' xảy ra khi x=-2 và \(y=\dfrac{1}{3}\)

`#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: \(=x^2-10x+25-36y^2t^2\)

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

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