a, \(y^2+6xy+9x^2=\left(3x\right)^2+2.3x.y+y^2=\left(3x+y\right)^2\)

b, \(2x^3+x^2+x^4=x^2\left(2x+1+x^2\right)=x^2\left(x+1\right)^2\)

\(\left(5y^4-3y^3\right)\div2y^3=\frac{1}{2}\)

\(\frac{5}{2}y-\frac{3}{2}=\frac{1}{2}\)

\(\frac{5}{2}y=\frac{1}{2}+\frac{3}{2}\)

\(\frac{5}{2}y=2\)

\(y=2\div\frac{5}{2}\)

\(y=\frac{4}{5}\)

a) x² - 6x + 10 = ( x² - 6x + 9 ) + 1 = ( x - 3 )² + 1 ≥ 1 ∀ x

Dấu "=" xảy ra khi x = 3

=> GTNN của biểu thức = 1 <=> x = 3

b) 2x² - 6x = 2( x² - 3x + 9/4 ) - 9/2 = 2( x - 3/2 )² - 9/2 ≥ -9/2 ∀ x

Dấu "=" xảy ra khi x = 3/2

=> GTNN của biểu thức = -9/2 <=> x = 3/2

c) 4x - x² + 3 = -( x² - 4x + 4 ) + 7 = -( x - 2 )² + 7 ≤ 7 ∀ x

Dấu "=" xảy ra khi x = 2

=> GTLN của biểu thức = 7 <=> x = 2

a. x² - 6x + 10 = x² - 6x + 9 + 1 = ( x - 3 )² + 1\(\ge\)1\(\forall\)x

Dấu "=" xảy ra <=> ( x - 3 )² = 0 <=> x = 3

Vậy GTNN của bt trên = 1 <=> x = 3

b. 2x² - 6x = 2x² - 6x + 9/2 - 9/2 = 2 ( x² - 3x + 9/4 ) - 9/2

= 2 ( x - 3/2 )² - 9/2\(\ge\)- 9/2\(\forall\)x

Dấu "=" xảy ra <=> 2 ( x - 3/2 )² = 0 <=> x - 3/2 = 0 <=> x = 3/2

Vậy GTNN của bt trên = - 9/2 <=> x = 3/2

c. 4x - x² + 3 = - x² + 4x - 4 + 7 = - ( x² - 4x + 4 ) + 7 = - ( x - 2 )² + 7\(\le\)7\(\forall\)x

Dấu "=" xảy ra <=> - ( x - 2 )² = 0 <=> x - 2 = 0 <=> x = 2

Vậy GTLN của bt trên = 7 <=> x = 2 

a, 3x(x-1)=3x^2+9

\(\Leftrightarrow\)3x(x-1)-3x^2=9

\(\Leftrightarrow\)3x^2-3x-3x^2=9

\(\Leftrightarrow\)-3x=9

\(\Leftrightarrow\)x=9:(-3) \(\Leftrightarrow\)x=-3

3x ( x - 1 ) = 3x² + 9

<=> 3x² - 3x - 3x² - 9 = 0

<=> - 3x - 9 = 0

<=> x = - 3

2x ( 1 - x ) - 6 ( x - 1 ) = 0

<=> 2x ( 1 - x ) + 6 ( 1 - x ) = 0

<=> ( 2x + 6 ) ( 1 - x ) = 0

<=> 2x + 6 = 0 hoặc 1 - x = 0

<=> x = - 3 hoặc x = 1

( 2x - 3 )² - ( x + 1 ) = 0

<=> 4x² - 12x + 9 - x - 1 = 0

<=> 4x² - 13x + 8 = 0

<=> ( 4x² - 13x + 169/16 ) - 41/16 = 0

<=> 4 ( x² - 13x/4 + 169/64 ) = 41/16

<=> ( x - 13/8 )² = 41/64

<=>\(\orbr{\begin{cases}x-\frac{13}{8}=\frac{\sqrt{41}}{8}\\x-\frac{13}{8}=-\frac{\sqrt{41}}{8}\end{cases}}\) <=> \(\orbr{\begin{cases}x=\frac{13+\sqrt{41}}{8}\\x=\frac{13-\sqrt{41}}{8}\end{cases}}\)

20a²b³ + 25a²b² - 15ab² = 5ab²( 4ab + 5a - 3 )

a) 5x( x - 1 ) + ( x - 4 )( x + 4 )

= 5x² - 5x + x² - 16

= 6x² - 5x - 16

b) ( x - 2y )² + ( x³ + 3x²y + 3xy² + y³ ) : ( x + y )

= x² - 4xy + 4y² + ( x + y )³ : ( x + y )

= x² - 4xy + 4y² + ( x + y )²

= x² - 4xy + 4y² + x² + 2xy + y²

= 2x² - 2xy + 5y²

\(\frac{x+1}{2014}+\frac{x+2}{2013}+\frac{x+3}{2012}+\frac{x+4}{2011}+\frac{x+2023}{2}=0\)

\(\left(\frac{x+1}{2014}+1\right)+\left(\frac{x+2}{2013}+1\right)+\left(\frac{x+3}{2012}+1\right)+\left(\frac{x+2011}{4}+1\right)+\left(\frac{x+2023}{2}-4\right)=0\)

\(\frac{x+2015}{2014}+\frac{x+2015}{2013}+\frac{x+2015}{2012}+\frac{x+2015}{2011}+\frac{x+2015}{2}=0\)

\(\left(x+2015\right)\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2}\right)=0\)

Vì \(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2}\ne0\) 

nên \(x+2015=0\)

\(\Rightarrow x=-2015\)

Vậy \(x=-2015\).

x⁵ + 1 - x³ - x²

= ( x⁵ - x³ ) - ( x² - 1 )

= x³ ( x² - 1 ) - ( x² - 1 )

= ( x³ - 1 ) ( x² - 1 )

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

= ( x - 1 )² ( x² + x + 1 ) ( x + 1 )

x^5+1-x^3-x^2

=(x^5-x^2)-(x^3-1)

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

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

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

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

kĩ nhất rồi đấy =')