`4x^2-1=0`

`<=>(2x-1)(2x+1)=0`

`<=>[(2x-1=0),(2x+1=0):}`

`<=>[(2x=1),(2x=-1):}`

`<=>[(x=1/2),(x=-1/2):}`

Vậy `x=1/2` hoặc `x=-1/2`

Trong đề đâu có x đâu bạn

Ta có: \(4x^2-1=0\)

\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

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

\(\frac{1}{4}x^4-9\)

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

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

nó dễ ợt mà -_- 

x²y²+4xy+4=(xy+2)² xong :)) 

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

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

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

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

\(=2x^4+12x^2+2\)

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 )

\(8x^3+12x^2+6x+1=\left(2x+1\right)^3\)

\(=\left(2\cdot24.5+1\right)^3=50^3=125000\)

\(S=1^3+2^3+3^3+...+n^3=\left(1+2+3+...+n\right)^2\)

\(=\left[\dfrac{n\left(n+1\right)}{2}\right]^2=\dfrac{n^2\cdot\left(n+1\right)^2}{4}\)