Question

Tìm x biết,

a) \(\left(x^2-1\right)\left(x-101\right)+101x\left(x+1\right)=101\)

b) \(x^4-3x^2\left(2x-3\right)=0\)

Answer 1

a) ( x² - 1 )( x - 101 ) + 101x( x + 1 ) = 101

<=> x³ - 101x² - x + 101 + 101x² + 101x - 101 = 0

<=> x³ + 100x = 0

<=> x( x² + 100 ) = 0

<=> \(\orbr{\begin{cases}x=0\\x^2+100=0\end{cases}}\Leftrightarrow x=0\)( vì x² + 100 ≥ 100 > 0 ∀ x )

b) x⁴ - 3x²( 2x - 3 ) = 0

<=> x⁴ - 6x³ + 9x² = 0

<=> x²( x² - 6x + 9 ) = 0

<=> x²( x - 3 )² = 0

<=> \(\orbr{\begin{cases}x^2=0\\\left(x-3\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

Answer 2

a,\(\left(x^2-1\right)\left(x-101\right)+101x\left(x+1\right)=101\)

\(\Leftrightarrow x^3-101x^2-x+101+101x^2+101x=101\)

\(\Leftrightarrow x^3+100x=101-101\)

\(\Leftrightarrow x^3+101x=0\)

\(\Leftrightarrow x\left(x^2+101\right)=0\)

\(\Rightarrow\hept{\begin{cases}x=0\\x^2+101\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x^2=-101\end{cases}\Rightarrow}x=0}\)