a, \(x^2-6x+9=\left(x-3\right)^2\)

b, \(x^2-12x+36=\left(x-4\right)^2\)

c, \(9x^2-25=\left(3x-5\right)\left(3x+5\right)\)

d, \(x^2-x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2\)

e, \(x^4-8x^2+16=\left(x^2-4\right)^2=\left[\left(x-2\right)\left(x+2\right)\right]^2\)

f, \(x^4-81=\left(x^2-9\right)\left(x^2+9\right)=\left(x-3\right)\left(x+3\right)\left(x^2+9\right)\)

g, \(\left(4x+5\right)^2-\left(5x+4\right)^2=\left(4x+5-5x-4\right)\left(4x+5+5x+4\right)=9\left(1-x\right)\left(x+1\right)\)

h, \(\left(2x-3\right)^2-2\left(2x-3\right)\left(x+2\right)+\left(-x-2\right)^2\)

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

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

a) x² - 6x + 9 = (x-3)²

b) x² - 12 + 36 = (x-6)²

c) 9x² - 25 = (3x - 25)(3x + 25)

d) x² - x + 1/4 = (x - 1/2)²

a, \(x^2-6x+9=4< =>\left(x-3\right)^2=4< =>\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\)

\(< =>\orbr{\begin{cases}x=5\\x=1\end{cases}}\)

b,\(x^2\left(x-3\right)-4\left(x-3\right)=0< =>\left(x-2\right)\left(x+2\right)\left(x-3\right)=0\)

\(< =>\orbr{\begin{cases}x=2\\x=-2\end{cases}orx=3}\)

c nhường mấy bn khácccc

a) x^2-6x+9=4.

 x=1, x=5

b) x^2(x-3)-(4X-12)=0

x=-2, x=2, x=3

c) (2x+3)^2-4(x+2)^2=12

x=-19/4

(x+2)(x+3)(x+4)(x+5)=24

x=-6,

x=-1;

x = -(căn bậc hai(3)căn bậc hai(5)i+7)/2

;x = (căn bậc hai(3)căn bậc hai(5)i-7)/2;

nha bạn chúc bạn học tốt nha 

(x + 2)(x + 3)(x + 4)(x + 5) = 24

<=> [(x + 2)(x + 5][(x + 3)(x + 4] = 24

<=> (x² + 7x + 10)(x² + 7x + 12) - 24 = 0

<=> (x² + 7x + 11 - 1)(x² + 7x + 11 + 1) - 24 = 0

<=> (x² + 7x + 11)² - 25 = 0 

<=> (x² + 7x + 16)(x² + 7x + 6) = 0

\(\Leftrightarrow\left(x+1\right)\left(x+6\right)\left[\left(x+\frac{7}{2}\right)^2+\frac{15}{4}\right]=0\)

<=> (x + 1)(x + 6) = 0 

<=> \(\orbr{\begin{cases}x+1=0\\x+6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-6\end{cases}}\)

Vậy \(x\in\left\{-1;-6\right\}\)

271​x3−32​x2+4x−8

 x  -  x  +  4x  -  8

=  4x  -  8

=  4( x  -  8)

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

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

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

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

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

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

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

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

192=(20−1)2=202−2.20.1+12=400−40+1=361

a)

19^2=(20−1)^2=20^2−2.20.1+1^2=400−40+1=361

28^2=(30−2)^2=30^2−2.30.2+2^2=900−120+4=784

81^2=(80+1)^2=80^2+2.80.1+1^2=6400+160+1=6561

91^2=(90+1)^2=90^2+2.90.1+1^2=8100+180+1=8281

c) (xy^2+1)^2

d) (1/3-y^4)^2

e) (1/2a-2b^2)^2

f) (5-x)^2

c) 2xy² + x² y⁴ + 1

= (xy²)² + 2xy² .1 + 1²

=(xy² +1)²

d) 1/9 - 2/3 y⁴ + y⁸

= (1/3)²  - 2 . 1/3 y⁴ + (y⁴)²

=(1/3 - y⁴)²

e) 1/4 a² - 2ab²  + 4b⁴

= (1/2 a)² - 2 1/2 a . 2b  + (2b²)²

=(1/2 a  - 2b²)²

f) 25 - 10x + x²

= x² + 2 . 5x + 5²

= (x+5)²

x² + x + 1/4 = ( x + 1/2 )²

Trả lời:

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

...

1) =\(x^7-x+x^2+x\)+1

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

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

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

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

=(x^5-x^4+x^2-x)(x^2+x+1)

Trả lời:

1, x⁷ + x² + 1 

= x⁷ + x² + 1 + x⁶ - x⁶ + x⁵ - x⁵ + x⁴ - x⁴ + x³ - x³ + x² - x² + x - x 

= ( x⁷ + x⁶ + x⁵ ) - ( x⁶ + x⁵ + x⁴ ) + ( x⁴ + x³ + x² ) - ( x³ + x² + x ) + ( x² + x + 1 )

= x⁵ ( x² + x + 1 ) - x⁴ ( x² + x + 1 ) + x² ( x² + x + 1 ) - x ( x² + x + 1 ) + ( x² + x + 1 )

= ( x² + x + 1 )( x⁵ - x⁴ + x² - x + 1 )

b, x⁸ + x⁷ + 1 

= x⁸ + x⁷ + 1 + x⁶ - x⁶ + x⁵ - x⁵ + x⁴ - x⁴ + x³ - x³ + x² - x² + x - x 

= ( x⁸ + x⁷ + x⁶ ) - ( x⁶ + x⁵ + x⁴ ) + ( x⁵ + x⁴ + x³ ) - ( x³ + x² + x ) + ( x² + x + 1 ) 

= x⁶ ( x² + x + 1 ) - x⁴ ( x² + x + 1 ) + x³ ( x² + x + 1 ) - x ( x² + x + 1 ) + ( x² + x + 1 )

= ( x² + x + 1 )( x⁶ - x⁴ + x³ - x + 1 )

=(x+y)²+2(x+y)+1-16

=(x+y+1)²-4²

=(x+y+5)(x+y-3)

Sửa đề: \(x^2+2xy+y^2+2x+2y-15\)

\(=\left(x+y\right)^2+2\left(x+y\right)+1-16\)

Đặt \(x+y=t\)

\(\Rightarrow t^2+2t+1-16\)

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

\(=\left(t+1-4\right)\left(t+1+4\right)\)

\(=\left(t-3\right)\left(t+5\right)\)

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