Question

(2x-3/2) . ( 2x + 1) > 0

ai giả hội mk đi mk đang cần rất gấp T_T

Answer 1

\(\left(2x-\frac{3}{2}\right).\left(2x+1\right)>0\Leftrightarrow4x+2x-3x-\frac{3}{2}>0\Leftrightarrow3x>\frac{3}{2}\Leftrightarrow x>\frac{1}{2}\)

Answer 2

ta có (2x - \(\frac{3}{2}\)) . (2x + 1) > 0

mà 2x + 1 là số lẻ

=> 2x - \(\frac{3}{2}\) = 0

=> 2x = 0 + \(\frac{3}{2}\)

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

=> x = \(\frac{3}{2}\) : 2

=> x = \(\frac{3}{2}\) . \(\frac{1}{2}\)

=> x = \(\frac{3}{4}\)(T/M)

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