a) A=3x+2+|5x|A=3x+2+|5x|
- Khi x≥0x≥0 ta có 5x≥05x≥0 nên |5x|=5x|5x|=5x.
Do đó A=3x+2+5x=8x+2A=3x+2+5x=8x+2 khi x≥0x≥0.
- Khi x<</span>0x<</mo>0 ta có 5x<</span>05x<</mo>0 nên |5x|=−5x|5x|=−5x.
Do đó A=3x+2−5x=−2x+2A=3x+2−5x=−2x+2 khi x<</span>0x<</mo>0.
Vậy A=8x+2A=8x+2 khi x≥0x≥0;
A=−2x+2A=−2x+2 khi x<</span>0x<</mo>0.
b) B=|−4x|−2x+12B=|−4x|−2x+12
- Khi x≤0x≤0 ta có −4x≥0−4x≥0 nên |−4x|=−4x|−4x|=−4x.
Do đó B=−4x−2x+12=−6x+12B=−4x−2x+12=−6x+12 khi x≤0x≤0.
- Khi x>0x>0 ta có −4x<</span>0−4x<</mo>0 nên |−4x|=−(−4x)=4x|−4x|=−(−4x)=4x.
Do đó B=4x−2x+12=2x+12B=4x−2x+12=2x+12 khi x<</span>0x<</mo>0.
Vậy B=−6x+12B=−6x+12 khi x≤0x≤0;
B=2x+12B=2x+12 khi x<</span>0x<</mo>0.
c) C=|x−4|−2x+12C=|x−4|−2x+12
Với x>5x>5 ta có x−4>1x−4>1 hay x−4>0x−4>0 nên |x−4|=x−4|x−4|=x−4.
Do đó: C=x−4−2x+12=−x+8C=x−4−2x+12=−x+8.
Vậy với x>5x>5 thì C=−x+8C=−x+8.
d) D=3x+2+|x+5|D=3x+2+|x+5|
- Khi x+5≥0x+5≥0 hay x≥−5x≥−5 ta có |x+5|=x+5|x+5|=x+5.
Do đó: D=3x+2+x+5=4x+7D=3x+2+x+5=4x+7 khi x≥−5x≥−5
- Khi x+5<</span>0x+5<</mo>0 hay x<</span>−5x<</mo>−5 ta có |x+5|=−(x+5)|x+5|=−(x+5).
Do đó: D=3x+2−(x+5)D=3x+2−(x+5) =3x+2−x−5=2x−7=3x+2−x−5=2x−7 khi x<</span>−5x<</mo>−5
Vậy D=4x+7D=4x+7 khi x≥−5x≥−5
D=2x−3D=2x−3 khi x<</span>−5