a) Ta có: VT=(√3−1)2=(√3)2−2.√3.1+12
=3−2√3+1
=(3+1)−2√3
=4−2√3 = VP
Vậy (√3−1)2=4−2√3 (đpcm)
b) Ta có:
VT=√4−2√3−√3=√(3+1)−2√3−√3
=√3−2√3+1−√3
=√(√3)2−2.√3.1+12−√3
=√(√3−1)2−√3
=|√3−1|−√3.
Lại có:
{(√3)2=3(√1)2=1
Mà 3>1⇔√3>√1⇔√3>1⇔√3−1>0.
⇒|√3−1|=√3−1.
Do đó |√3−1|−√3=√3−1−√3
=(√3−√3)−1=−1 = VP.
Vậy √4−2√3−√3=−1 (đpcm)