Question

Dau coroana, exercitiul 14, va rog frumos.​

Answer 1

a) |x+5| + |3-x| $\geq$ 8.

Noi stim ca |x+5| + |3-x| $\geq$ |x+5+3-x|, deci |x+5| + |3-x| $\geq$ 8.

b) |2x+1| + 2*|1-x| $\geq$ 3 <=> |2x+1| + |2-2x| $\geq$ 3

Noi stim ca |2x+1| + |2-2x| $\geq$ |2x+1+2-2x|, adica |2x+1| + |2-2x| $\geq$ 3

c) |4x+11| + 2*|2x-5| $\geq$ 21, adica |4x+11| + |4x-10| $\geq$ 21

Dar |4x+11| + |4x-10| = |4x+11| + |10-4x|, si $\geq$ |4x+11+10-4x| deci $\geq$ 21

d) 3*|4x+7| + 4*|7-3x| $\geq$ 49, deci |12x+21| + |28-12x| $\geq$ 49.

Stim ca |12x+21| + |28-12x| $\geq$|12x+21+28-12x|

Deci |12x+21| + |28-12x| $\geq$ 49

e) 2*|2x+3| + |x-5| + |3x-19| $\geq$ 30, deci |4x+6| + |x-5| + |3x-19| $\geq$ 30

Se poate scrie si ca |4x+6| + |5-x| + |19-3x| $\geq$ 30

Dar |4x+6| + |5-x| + |19-3x| $\geq$ |4x+6+5-x+19-3x|, deci

|4x+6| + |5-x| + |19-3x| $\geq$ 30

Sper ca se intelege!

Answer 2

Ideea e ca pui semnele alea(+ sau -) astfel incat sa scapi de termenii cu x

La punctul e se merge ai se arata ca inegalitatea se extinde si pt 3 termeni