Question

giải BPT :

\(\dfrac{2x-1}{3}-\dfrac{x+3}{2}\le1\)

Answer 1

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

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

=>4x -2 +3x+9\(\le\)6

<=>7x+7\(\le\)6

<=>7x\(\le\)6-7

<=>7x\(\le\)-1

<=>x\(\le\)\(\dfrac{-1}{7}\)

vậy bất phương trình có nghiệm là x\(\le\)\(\dfrac{-1}{7}\)

Answer 2

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Answer 3

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

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

=>4x-2-3x-9\(\le\)6

<=>x-11\(\le\)6

<=>x\(\le\)6+11

<=>x\(\le\)17

Vậy bất phương trình có nghiệm là x\(\le\)17

Answer 4

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

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

\(\Leftrightarrow\) 2.(2x-1)- 3.(x+3)\(\le\) 6

\(\Leftrightarrow\) 4x-2- 3x- 3\(\le\) 6

\(\Leftrightarrow\) 4x- 3x\(\le\) 6+ 2+ 3

\(\Leftrightarrow\) x\(\le\) 11

vậy bpt có n_o là x\(\le\) 11

Answer 5

Phương trình có nghiệm khi x < hoặc = 17 chứ bạn.