Ta có:
1,A=√1+20082+2008220092+20082009=√(1+2.2008+2008)2−2.2008+(20082009)2+20082009=√(1+2008)2−2.2008+(20082009)2+20082009=√20092−2.2009.20082009+(20082009)2+20082009=√(2009−20082009)2+20082009=2009−20082009+20082009=20092,B=√x−2−2√x−3+√x+1−4√x−3=√(x−3)−2√x−3+1+√(x−3)−4√x−3+4=√√x−32−2.√x−3.1+12+√√x−32−2.√x−3.2+22=√(√x−3−1)2+√(√x−3−2)2=∣∣√x−3−1∣∣+∣∣√x−3−2∣∣3≤x≤4⇔0≤x−3≤1⇔0≤√x−3≤1⇔{√x−3−1≤0√x−3−2≤0⇒B=−(√x−3−1)−(√x−3−2)=3−2√x−3
Đáp án:
Giải thích các bước giải: