Question

giải BPT \(\dfrac{2x-1}{3}+\dfrac{x-1}{2}\le3\)

Answer 1

\(\dfrac{2x-1}{3}\)+\(\dfrac{x-1}{2}\)\(\le3\)

<=> \(\dfrac{2\left(2x-1\right)}{6}\)+\(\dfrac{3\left(x-1\right)}{6}\)\(\le\dfrac{18}{6}\)

<=> 4x -2+3x-3\(\le\)18

<=>7x-5\(\le\)18

<=>7x\(\le\)23

<=>x\(\le\)\(\dfrac{23}{7}\)

Vậy bất phương trình có nghiệm là x\(\le\)\(\dfrac{23}{7}\)

Answer 2

\(\dfrac{2x-1}{3}\)+ \(\dfrac{x-1}{2}\)\(\le\) 3

\(\Leftrightarrow\) \(\dfrac{2.\left(2x-1\right)+3.\left(x-1\right)}{6}\)\(\le\) \(\dfrac{18}{6}\)

\(\Leftrightarrow\) 2.(2x-1)+ 3.( x-1)\(\le\) 18

\(\Leftrightarrow\) 4x- 2+ 3x- 3\(\le\) 18

\(\Leftrightarrow\) 4x+ 3x\(\le\) 18+ 2+ 3

\(\Leftrightarrow\) 7x\(\le\) 23

\(\Leftrightarrow\) x\(\le\) \(\dfrac{23}{7}\)

vậy bpt có n_o là x\(\le\) \(\dfrac{23}{7}\)