Câu 1 :
a)` ( 2x - 3 )( 3 - 4x ) = 0`
=>` \(\left[ \begin{array}{l}2x-3=0\\3-4x=0\end{array} \right.\) `
=> `\(\left[ \begin{array}{l}2x=3\\4x=3\end{array} \right.\) `
=>` \(\left[ \begin{array}{l}x=3/2\\x=3/4\end{array} \right.\) `
b) `( 2x-5 )²`= ` 9 `
`( 2x-5 )² = ±3²`
=> `2x - 5 = ± 3`
=> \(\left[ \begin{array}{l}2x-5=3\\2x-5=-3\end{array} \right.\) `
=> \(\left[ \begin{array}{l}2x=8\\2x=2\end{array} \right.\) `
=> \(\left[ \begin{array}{l}x=4\\x=1\end{array} \right.\) `
Câu 2 :
`A= ( x+1 )( x²- x + 1 ) - ( x - 1 )(x² + x + 1 ) + x`
`A = x³ - x² + x + x² - x + 1 - x³ + x² + x - x² - x - 1 + x`
`A = x `
Tại `x = 0,001`
=> `A = 0,001`
Vậy `A = 0,001` tại `x = 0,001`
