x³ + y³ - 3xy

= ( x + y )( x² - xy + y² ) - 3xy

= -1( x² - xy + y² ) - 3xy

= -x² + xy - y² - 3xy

= -x² - 2xy - y²

= -( x² + 2xy + y² )

= -( x + y )²

= -(-1)² = -1

4x² - 9 - 2(2x + 3) = 0

=> 4x² - 4x - 15 = 0

=> 4x² - 4x + 1 - 16 = 0

=> (2x - 1)² - 4² = 0

=> (2x - 1 - 4)(2x - 1 + 4) = 0

=> (2x - 5)(2x + 3) = 0

=> \(\orbr{\begin{cases}2x-5=0\\2x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{3}{2}\end{cases}}\)

Vậy \(x\in\left\{\frac{5}{2};-\frac{3}{2}\right\}\)là giá trị cần tìm

4x²-9-2(2x+3) =0

(2x)² -3² -2(2x+3) = 0

(2x + 3)(2x - 3) -2(2x+3) = 0

(2x + 3)(2x - 3 - 2) =0

\(\orbr{\begin{cases}2x+3=0\\2x-3-2=0\end{cases}}\)\(\orbr{\begin{cases}2x=-3\\2x=5\end{cases}}\)\(\orbr{\begin{cases}x=\frac{-3}{2}\\x=\frac{5}{2}\end{cases}}\)

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

\(\frac{2x}{x^2+x+1}\)( ĐKXĐ : x ∈ R )

\(\frac{x}{x-1}\)( ĐKXĐ : x ≠ 1 )

MTC : ( x - 1 )( x² + x + 1 )

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

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

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

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

( ĐKXĐ : x ≠ 1 ; x ≠ 2 )

\(\frac{1}{x^2+5x-6}=\frac{1}{x^2-x+6x-6}=\frac{1}{x\left(x-1\right)+6\left(x-1\right)}=\frac{1}{\left(x-1\right)\left(x+6\right)}\)

( ĐKXĐ : x ≠ 1 ; x ≠ -6 )

\(\frac{1}{-x^2+4x-3}=\frac{-1}{x^2-4x+3}=\frac{-1}{x^2-x-3x+3}=\frac{-1}{x\left(x-1\right)-3\left(x-1\right)}=\frac{-1}{\left(x-1\right)\left(x-3\right)}\)

( ĐKXĐ : x ≠ 1 ; x ≠ 3 )

MTC : ( x - 1 )( x - 2 )( x - 3 )( x + 6 )

=> \(\frac{-1}{\left(x-1\right)\left(x-2\right)}=\frac{-1\left(x+6\right)\left(x-3\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x+6\right)}=\frac{-x^2-3x+18}{\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x+6\right)}\)

=>\(\frac{1}{\left(x-1\right)\left(x+6\right)}=\frac{1\left(x-2\right)\left(x-3\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x+6\right)}=\frac{x^2-5x+6}{\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x+6\right)}\)

=> \(\frac{-1}{\left(x-1\right)\left(x-3\right)}=\frac{-1\left(x-2\right)\left(x+6\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x+6\right)}=\frac{-x^2-4x+12}{\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x+6\right)}\)

\(\frac{1}{x^2-2x+1}=\frac{1}{\left(x-1\right)^2}\)( ĐKXĐ : x ≠ 1 )

\(\frac{2}{x^2+2x}=\frac{2}{x\left(x+2\right)}\)( ĐKXĐ : x ≠ 0 ; x ≠ -2 )

MTC : x( x + 2 )( x - 1 )²

Thực hiện phép tính

\(5x^2\left(4x^2-2x+5\right)\)

\(=20x^4-10x^3+25x^2\)

b) \(\left(6x^2-5\right)\left(2x+3\right)\)

\(=6x^2\left(2x+3\right)-5\left(2x+3\right)\)

\(=12x^3+18x^2-10x-15\)

a) x² + 7x + 12

= x² + 3x + 4x + 12

= ( x² + 3x ) + ( 4x + 12 )

= x( x + 3 ) + 4( x + 3 )

= ( x + 3 )( x + 4 )

b) x⁴ + 8x² + 16

= ( x² )² + 2.x².4 + 4²

= ( x² + 4 )²

c) x² + 26x + 25

= x² + x + 25x + 25

= ( x² + x ) + ( 25x + 25 )

= x( x + 1 ) + 25( x + 1 )

= ( x + 1 )( x + 25 )

a) \(x^2+7x+12\)

\(x^2+4x+3x+12\)

\(x\left(x+4\right)+3\left(x+4\right)\)

\(\left(x+4\right)\left(x+3\right)\)

b) \(x^4+8x^2+16\)

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

c) \(x^2+26x+25\)

\(x^2+x+25x+25\)

\(x\left(x+1\right)+25\left(x+1\right)\)

\(\left(x+25\right)\left(x+1\right)\)