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

n) \(\dfrac{4}{49}-4x^2=\left(\dfrac{2}{7}-2x\right)\left(\dfrac{2}{7}+2x\right)\)

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

Mik cần gấp 

\(1,=-\left(y^2+12y+36\right)=-y^2-12y-36\)

\(2,=-\left(16-8y+y^2\right)=-16+8y-y^2\)

\(3,=-\left(\dfrac{4}{9}+\dfrac{4}{3}x+x^2\right)=-\dfrac{4}{9}-\dfrac{4}{3}x-x^2\)

\(4,=-\left(x^2-3x+\dfrac{9}{4}\right)=-x^2+3x-\dfrac{9}{4}\)

\(5,-\left(2+3y\right)^2=-\left(4+12y+9y^2\right)=-4-12y-9y^2\)

.... mấy ý còn lại bn tự lm nhé, tương tự thhooi

1) \(-\left(y+6\right)^2=-y^2-12y-36\)

2) \(-\left(4-y\right)^2=-y^2+8y-16\)

3) \(-\left(x+\dfrac{2}{3}\right)^2=-x^2-\dfrac{4}{3}x-\dfrac{4}{9}\)

4) \(-\left(x-\dfrac{3}{2}\right)^2=-x^2+3x-\dfrac{9}{4}\)

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

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

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

8) \(-\left(2x-\dfrac{3}{2}\right)^2=-4x^2+6x-\dfrac{9}{4}\)

Chọn B

B

a) (1-5a)^2 = 1- 10a + 25a^2

b) x^2 - 4  = (x-2)(x+2)

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

d) 4a^2 - 9= (2a - 3)(2a + 3)

Các bạn làm cả công thức hộ mình nha

`(1-5a)^2`

`= 1^2 - 2 . 1 . 5a + (5a)^2`

`= 1 - 10a + 25a^2`

`x^2-4`

`=x^2 -2^2`

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

`1-x^2`

`=1^2 - x^2`

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

`4a^2 - 9`

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

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

\(F=\left(3x-2\right)^2+\left(3x+2\right)^2+2\left(9x^2-4\right)\\=\left[\left(3x+2\right)^2+2.\left(3x+2\right)\left(3x-2\right)+\left(3x-2\right)^2\right]\\ =\left[\left(3x+2\right)+\left(3x-2\right)\right]^2\\ =\left(6x\right)^2=36x^2\\ Thay.x=-\dfrac{1}{3}.vào.F.thu.gọn:\\ F=36x^2=36.\left(-\dfrac{1}{3}\right)^2=36.\left(\dfrac{1}{9}\right)=4\)

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

\(=\left(2x^3\right)^3-3\cdot\left(2x^3\right)^2\cdot y^2+3\cdot2x^3\cdot\left(y^2\right)^{^2}-\left(y^2\right)^3\)

\(=8x^9-3\cdot4x^6y^2+3\cdot2x^3y^4-y^6\)

\(=8x^9-12x^6y^2+6x^3y^4-y^6\)

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

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

\(=x^3-27y^3\)

c) \(\left(x+2y+z\right)\left(x+2y-z\right)\)

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

\(=x^2+4xy+4y^2-z^2\)

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

\(=\left(2x^3y-\dfrac{1}{2}x^2\right)^3\)

\(=8x^9y^3-6x^8y^2+\dfrac{3}{2}x^7y-\dfrac{1}{8}x^6\)

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

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

\(=4x^4+9x^2-12x^2-27\)

\(=4x^4-3x^2-27\)

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

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

\(=8x^3-1\)

\(a,\left(2x^3-y^2\right)^3=8x^9-12x^6y^2+6x^3y^4-y^6\)\(b,\left(x-3y\right)\left(x^2+3xy+9y^2\right)=x^3-27y^3\)

\(c,\left(x+2y+z\right)\left(x+2y-z\right)=\left(x+2y\right)^2-z^2=x^2+4xy+4y^2-z^2\)\(d,\left(2x^3y-0,5x^2\right)^3=8x^9y^3-6x^4y^2x^2+3x^3yx^4-0,125x^6=8x^9y^3-6x^6y^2+3x^7y-0,125x^6\)

(2x-1)(2x+1)(4x² +1)

= (4x² - 1)(4x²+1)

=16x⁴ - 1 

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

= (4x^2 - 1).(4x^2+1)

=  (4x^2)^2  -  1

=  16x^4  - 1

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

`=[(2x)^2-1^2](4x^2+1)`

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

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

`=16x^4-1`

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

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

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

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

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

\(1,=x\left(2x+3\right)-\left(2x+3\right)=2x^2+3x-2x-3=2x^2+x-3\\ 2,=\left(2x\right)^2-2.2.x.1+1^2=4x^2-4x+1\)

1. =2x²-2x+3x-3

2. (2x)²-2.2x.1+1²

=4x²-4x+1