a) \(x^2-\frac{1}{49}=0\)

<=> \(\left(x-\frac{1}{7}\right)\left(x+\frac{1}{7}\right)=0\)

<=> \(\orbr{\begin{cases}x-\frac{1}{7}=0\\x+\frac{1}{7}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{7}\\x=-\frac{1}{7}\end{cases}}\)

Vậy x = \(\pm\frac{1}{7}\)

b) \(64-\frac{1}{4}x^2=0\)

<=> \(\left(8-\frac{1}{2}x\right)\left(8+\frac{1}{2}x\right)=0\)

<=> \(\orbr{\begin{cases}8-\frac{1}{2}x=0\\8+\frac{1}{2}x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=16\\x=-16\end{cases}}\)

Vậy \(x=\pm16\)

c) 9x² + 12x + 4 = 0

<=> (3x + 2)² = 0

<=> 3x + 2 = 0 

<=> x = -2/3

Vậy x = -2/3

e) \(x^2+\frac{1}{4}=x\) 

<=> \(x^2-x+\frac{1}{4}=0\)

<=> \(\left(x-\frac{1}{2}\right)^2=0\)

<=> \(x=\frac{1}{2}\)

Vậy \(x=\frac{1}{2}\)

d, sửa đề : \(x^2+4=4x\Leftrightarrow x^2-4x+4=0\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x=2\)

i, \(4-\frac{12}{x}+\frac{9}{x^2}=0\)ĐK : \(x\ne0\)

Vì \(x\ne0\)Nhân 2 vế với \(x^2\)phương trình có dạng 

\(4x^2-12x+9=0\Leftrightarrow\left(2x-3\right)^2=0\Leftrightarrow x=\frac{3}{2}\)

`P=27-27x+9x^2-x^3=3^3-3.3^2 .x+3.3.x^2-x^3=(3-x)^3`

Thay `x=-17`: `(3+17)^3=20^3=8000`

`Q=x^3+3x^2+3x=x(x^2+3x+3)`

Thay `x=99`: `Q=99 . (99^2 +3.99+3)=99. 10101=999 999`

`N=(3x-1)^2-2(9x^2-1)+(3x+1)^2

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

`=(3x-1-3x-1)^2=(-2)^2=4`

`x^3-6x^2y+12xy^2-8y^3`

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

`=(x-2y)^3`

\(=9x^3-3x^2-9x^2+6x-1\)1

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

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

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

a.\(27x^3+27x^2+9x+1=\left(3x+1\right)^3\)

b.\(x^3-6x^2+12x-8=\left(x-2\right)^3\)

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

a)\(69^2-31^2=\left(69-31\right)\left(69+31\right)=38.100=3800\)

\(1023^2-23^2=\left(1023-23\right)\left(1023+23\right)=1000.1046=1046000\)

a)\(6x-9-x^2\)

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

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

b)\(x^2+4y^2+4xy\)

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

c)\(x^2+8x+16\)

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

d)\(9x^2-12xy+4y^2\)

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

e)\(-25x^2y^2+10xy-1\)

\(=-\left(25x^2y^2-10xy+1\right)\)

\(=-\left(5xy-1\right)^2\)

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

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

j)\(x^2+6x+9\)

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

h)\(9x^2-6x+1\)

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

#H

a, 6x - 9 - x² = - x² + 6x - 9 = - (x² - 6x + 9) = - (x - 3)²

b, x² + 4y² + 4xy = x² + 2. x . 2y + (2y)² = (x + 2y)²

c, x² + 8x + 16 = x² + 2 . x . 4 + 4² = (x + 4)²

d, 9x² - 12xy + 4y² = (3x)² - 2 . 3x . 2y + (2y)² = (3x - 2y)²

e, - 25x²y² + 10xy - 1 = - (25x²y² - 10xy + 1) = - [(5xy)² - 2 . 5xy + 1] = - (5xy - 1)²

f, 4x² - 4x + 1 = (2x)² - 2 . 2x + 1 = (2x - 1)²

j, x² + 6x + 9 = x² + 2 . x . 3 + 3² = (x + 3)²

h, 9x² - 6x + 1 = (3x)² - 2 . 3x + 1 = (3x - 1)²