a)
Vì ⎧⎨⎩22=4(√3)2=3{22=4(3)2=3
mà 4>34>3 nên √4>√3⇔2>√3⇔2−√3>04>3⇔2>3⇔2−3>0.
⇔∣∣2−√3∣∣=2−√3⇔|2−3|=2−3.
Do đó: √(2−√3)2=∣∣2−√3∣∣=2−√3(2−3)2=|2−3|=2−3
b)
Vì ⎧⎨⎩32=9(√11)2=11{32=9(11)2=11
mà 9<</span>119<</mo>11 nên √9<</span>√11⇔3<</span>√11⇔3−√11<</span>09<</mo>11⇔3<</mo>11⇔3−11<</mo>0
⇔∣∣3−√11∣∣=−(3−√11)=−3+√11⇔|3−11|=−(3−11)=−3+11
=√11−3=11−3.
Do đó: √(3−√11)2=∣∣3−√11∣∣=√11−3(3−11)2=|3−11|=11−3.
c) Ta có: 2√a2=2|a|=2a2a2=2|a|=2a (vì a≥0a≥0 )
d) Vì a<</span>2a<</mo>2 nên a−2<</span>0a−2<</mo>0.
⇔|a−2|=−(a−2)=−a+2=2−a⇔|a−2|=−(a−2)=−a+2=2−a
Do đó: 3√(a−2)2=3|a−2|=3(2−a)3(a−2)2=3|a−2|=3(2−a)
=6−3a=6−3a.