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

\(\Leftrightarrow x^3-3x^2+3x-1+3x^2+6x+3=x^3+8\)

\(\Leftrightarrow x^3+9x+2=x^3+8\)

\(\Leftrightarrow9x=6\)

\(\Leftrightarrow x=\frac{2}{3}\)

vậy......

( x - 1 )³ + 3( x + 1 )² = ( x² - 2x + 4 )( x + 2 )

⇔ x³ - 3x² + 3x - 1 + 3( x² + 2x + 1 ) = x³ + 8

⇔ x³ - 3x² + 3x - 1 + 3x² + 6x + 3 = x³ + 8

⇔ x³ + 9x + 2 = x³ + 8

⇔ x³ + 9x + 2 - x³ - 8 = 0

⇔ 9x - 6 = 0

⇔ 9x = 6

⇔ x = 6/9 = 2/3

٩(๑~▽~๑)۶(●￣(工)￣●)

5x - 5y + x² - 2xy + y²

= ( 5x - 5y ) + ( x² - 2xy + y² )

= 5( x - y ) + ( x - y )²

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

olm hiển the thiếu đấy nhé.

25a²b² - 4x² + 4x - 1

= (5ab)² - (4x² + 4x - 1)²

= (5ab)² - (1 - 2x)²

= (5ab - 1 + 2x).(5ab + 1 - 2x)

25a²b² - 4x² + 4x - 1

= 25a²b² - ( 4x² - 4x + 1 )

= ( 5ab )² - ( 2x - 1 )²

= ( 5ab - 2x + 1 )( 5ab + 2x - 1 )

b) \(ĐKXĐ:x\ne0\)

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

\(\Leftrightarrow x^3.\left(5x-2\right):2x^3=\frac{1}{2}\)

\(\Leftrightarrow\frac{5x-2}{2}=\frac{1}{2}\)\(\Leftrightarrow5x-2=1\)

\(\Leftrightarrow5x=3\)\(\Leftrightarrow x=\frac{3}{5}\)( thỏa mãn ĐKXĐ )

Vậy \(x=\frac{3}{5}\)

c) \(ĐKXĐ:x\ne2\)

\(\frac{x^4-2x^2-8}{x-2}=0\)\(\Rightarrow x^4-2x^2-8=0\)

\(\Leftrightarrow\left(x^4-4x^2\right)+\left(2x^2-8\right)=0\)

\(\Leftrightarrow x^2.\left(x^2-4\right)+2\left(x^2-4\right)=0\)

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

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

Vì \(x^2\ge0\forall x\)\(\Rightarrow x^2+2\ge2\)

\(\Rightarrow\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)

So sánh với ĐKXĐ ta thấy: \(x=-2\)thỏa mãn 

Vậy \(x=-2\)