a) x⁶ + 1 

= (x²)³ + 1³

=(x² +1)(x⁴ - x² + 1)

b) x⁶ - y⁶

= (x²)³ - (y²)³

=(x² -y²)(x⁴ + x²y² + y⁴)

= (x-y)(x+y)[ x⁴ + 2.x²y² + y⁴ - x²y² ]

=(x-y)(x+y) [(x² +y²)² - (xy)² ]

=(x-y)(x+y)(x² - xy + y²)(x² + xy + y²)

c) x⁹ +1

= (x³)³ + 1³

=(x³ +1)(x⁶ - x³ -1)

= (x+1)(x² - x +1)(x⁶ - x³-1) 

D = (2x-1)³  -2x(2x-3)(2x+3) + 13x(x-1)

D= (2x)³ - 3. (2x)² .1 + 3.2x .1² -1³ - 2x(2x-3)(2x+3) + 13x(x-1)

D= 8x³ - 12x² + 6x -1 - 2x(4x² -9) + 13x² -13x

D= 8x³ -12x² + 6x-1 - 8x³ + 18x + 13x² -13x

D= (8x³ - 8x³) -(12x² -13x²) + (6x + 18x -13x) - 1

D= x² + 11x -1 

D = x² + 2x . 11/2 +(11/2)² -125/4

D= (x+ 11/2)² - 125/4

Với mọi x thì (x+11/2)² >= 0

=> (x+11/2)² - 125/4 >= -125/4

Dấu bằng xảy ra khi: (x+11/2)² =0

=> x + 11/2 =0 

=> x= -11/2

Vậy giá trị nhỏ nhất của D là -125/4 khi x= -11/2

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

a, \(\frac{1}{25}x^2-\frac{1}{9}y^2=\left(\frac{1}{5}x\right)^2-\left(\frac{1}{3}y\right)^2=\left(\frac{1}{5}x-\frac{1}{3}y\right)\left(\frac{1}{5}x+\frac{1}{3}y\right)\)

b, \(-\left(4x^2-20x+25\right)=-\left[\left(2x\right)^2-2.2x.5+5^2\right]=-\left(2x-5\right)^2\)

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

e, \(x^6-y^6=\left(x^3-y^3\right)\left(x^3+y^3\right)=\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)\)

3x³ + 10x² + 2 = 3x³ + x² + 9x² + 3x - 3x - 1 + 3

= x²( 3x + 1 ) + 3x( 3x + 1 ) - ( 3x + 1 ) + 3

= ( 3x + 1 )( x² + 3x - 1 ) + 3

Vì ( 3x + 1 )( x² + 3x - 1 ) ⋮ ( 3x + 1 )

=> 3 ⋮ ( 3x + 1 ) <=> ( 3x + 1 ) ∈ Ư(3) ( đến đây bạn tự xét giá trị nhé )

Trả lời:

a, \(x^2=6x\)

\(\Leftrightarrow x^2-6x=0\)

\(\Leftrightarrow x\left(x-6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}}\)

Vậy x = 0; x = 6 là nghiệm của pt.

b, \(x^2-4x+3=0\)

\(\Leftrightarrow x^2-x-3x+3=0\)

\(\Leftrightarrow x\left(x-1\right)-3\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=3\end{cases}}}\)

Vậy x = 1; x = 3 là nghiệm của pt.

Bài 3:

a) \(x^2=6x\)

\(\Leftrightarrow x^2-6x=0\)

\(\Leftrightarrow x\left(x-6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-6=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}\)

Vậy \(S=\left\{0;6\right\}\)

b) \(x^2-4x+3=0\)

\(\Leftrightarrow x^2-4x+4-1=0\)

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

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

\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)

Vậy \(S=\left\{3;1\right\}\)

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 )

Trả lời:

a, 3x²y - 6xy = 3xy ( x - 2 )

b, x² - y² - 9x + 9y 

= ( x² - y² ) - ( 9x - 9y )

= ( x - y )( x + y ) - 9 ( x - y )

= ( x - y )( x + y - 9 )

c, x³ - 6x² - y²x + 9x 

= x ( x² - 6x - y² + 9 )

= x [ ( x² - 6x + 9 ) - y² ]

= x [ ( x - 3 )² - y² ]

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

3x²y - 6xy = 3xy( x - 2 )

x² - y² - 9x + 9y = ( x - y )( x + y ) - 9( x - y ) = ( x - y )( x + y - 9 )

x³ - 6x² - y²x + 9x = x( x² - 6x - y² + 9 ) = x[ ( x - 3 )² - y² ] = x( x - y - 3 )( x + y - 3 )