Question

[33. Nguyễn Minh Ngọc]

Tìm x biết: 

a) x³ - 8 = (x - 2)(x - 12)

b) x² (x² + 4) - x² = 4

Answer 1

[l҉o҉n҉g҉ d҉z҉]

a) x³ - 8 = ( x - 2 )( x - 12 )

<=> ( x - 2 )( x² + 2x + 4 ) - ( x - 2 )( x - 12 ) = 0

<=> ( x - 2 )( x² + 2x + 4 - x + 12 ) = 0

<=> ( x - 2 )( x² + x + 16 ) = 0

<=> \(\orbr{\begin{cases}x-2=0\\x^2+x+16=0\end{cases}}\Leftrightarrow x=2\)( vì x² + x + 16 = ( x² + x + 1/4 ) + 63/4 = ( x + 1/2 )² + 63/4 ≥ 63/4 > 0 ∀ x )

b) x²( x² + 4 ) - x² = 4

<=> x²( x² + 4 ) - x² - 4 = 0

<=> x²( x² + 4 ) - ( x² + 4 ) = 0

<=> ( x² + 4 )( x² - 1 ) = 0

<=> \(\orbr{\begin{cases}x^2+4=0\\x^2-1=0\end{cases}}\Leftrightarrow x=\pm1\)( vì x² + 4 ≥ 4 > 0 ∀ x )

Answer 2

[ミ★Ƙαї★彡]

Ko khó nè :3, đừng tách ra nhé ! 

a, \(x^3-8=\left(x-2\right)\left(x-12\right)\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)=\left(x-2\right)\left(x-12\right)\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4-x+12\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+x+16\right)=0\Leftrightarrow x=2\)

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

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

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

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

\(\Leftrightarrow\left(-x^2+1\right)\left(2-x\right)\left(2+x\right)=0\Leftrightarrow x=\pm1;\pm2\)

Check hộ dáp án nhá :), ko chắc lắm nếu khai triển sẽ dễ nhìn hơn đấy.