\(a,=x^2+x+\dfrac{1}{4}\\ b,=4x^2+2x+\dfrac{1}{4}\\ c,=x^2-2+\dfrac{1}{x^2}\\ d,=4x^2+\dfrac{8}{3}x+\dfrac{4}{9}x^2\\ e,=a^2-1\\ f,=25x^4-4\)

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

\(b,\left(2x+\dfrac{1}{2}\right)^2=4x^2+2x+\dfrac{1}{4}\)

\(c,\left(x-\dfrac{1}{x}\right)^2=x^2-2+\dfrac{1}{x^2}\)

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

\(e,\left(a-1\right).\left(a+1\right)=a^2-1\)

\(f,\left(5x^2-2\right).\left(5x^2+2\right)=25x^4-4\)

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

b. \(\left(2x+\dfrac{1}{2}\right)^2=4x^2+2x+\dfrac{1}{4}\)

c. \(\left(x-\dfrac{1}{x}\right)^2=x^2-2+\dfrac{1}{x^2}\)

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

e. (a - 1)(a + 1) = a² - 1

f. (5x² - 2)(5x² + 2) = 25x⁴ - 4 

g. (2a - 3)(2a + 3) = 4a² - 9

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

= 2x(4x + 2)

Em áp dụng hđt số 3 trong sgk nhé.

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

                                      = 4x(2x+1)

1,2,3,4 không tính được.

`5)(2x-1/2)^2`

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

`=4x^2-2+1/4`

`6)(x+1/4)^2`

`=x^2+1/2x+1/16`

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

4) \(5x-y^2=\left(\sqrt{5x}-y\right)\left(\sqrt{5x}+y\right)\)

Phân tích thôi hả :v 

( 3x - 7 )² - 25 = ( 3x - 7 )² - 5² = ( 3x - 7 - 5 )( 3x - 7 + 5 ) = ( 3x - 12 )( 3x - 2 ) = 3( x - 4 )( 3x - 2 )

( x - 1/2 )² - 9/4 = ( x - 1/2 )² - ( 3/2 )² = ( x - 1/2 - 3/2 )( x - 1/2 + 3/2 ) = ( x - 2 )( x + 1 )

49 - ( x + 7 )² = 7² - ( x + 7 )² = [ 7 - ( x + 7 ) ][ 7 + ( x + 7 ) ] = ( 7 - x - 7 )( 7 + x + 7 ) = -x( x + 14 )

25 - ( x - 3 )² = 5² - ( x - 3 )² = [ 5 - ( x - 3 ) ][ 5 + ( x - 3 ) ] = ( 5 - x + 3 )( 5 + x - 3 ) = ( 8 - x )( x + 2 )

\(A=x^3+3x^2+3x+6\)

\(=x^3+3x^2+3x+1+5\)

\(=\left(x+1\right)^3+5\)

Thay x = 19 vào biểu thức \(A=\left(x+1\right)^3+5\)ta được:

\(A=\left(19+1\right)^3+5=20^3+5=8000+5=8005\)

Vậy giá trị của biểu thức A tại x = 19 là 8005.

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

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

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

Thay x = 11 vào biểu thức \(B=\left(x-1\right)^3+1\)ta được:

\(B=\left(11-1\right)^3+1=10^3+1=1000+1=1001\)

Vậy giá trị của biểu thức B tại x = 11 là 1001.

5: \(=4b^2-2b+\dfrac{1}{4}-\dfrac{1}{4}+a-a^2\)

\(=\left(2b\right)^2-2\cdot2b\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\left(a^2-a+\dfrac{1}{4}\right)\)

\(=\left(2b-\dfrac{1}{2}\right)^2-\left(a-\dfrac{1}{2}\right)^2\)

\(=\left(2b-\dfrac{1}{2}-a+\dfrac{1}{2}\right)\left(2b-\dfrac{1}{2}+a-\dfrac{1}{2}\right)\)

\(=\left(2b-a\right)\left(2b+a-1\right)\)

6:

\(=b^2-4b+4-9c^2\)

\(=\left(b-2\right)^2-9c^2\)

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

5) \(4b^2-2b+a-a^2\)

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

\(=\left(2b-a\right)\left(2b+a-1\right)\)

6) \(b^2-9c^2+4+4b\)

\(=\left(b+2\right)^2-9c^2\)

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

=(3x + 1)(3x - 1).2 = (9x² - 1).2 = 18x² - 2

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

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

   \(=\left(9x^2-1\right).2\)

\(=18x^2-2\)