a) Ta có: Vì x≥0x≥0 và y≥0y≥0 nên x+y≥0⇔|x+y|=x+yx+y≥0⇔|x+y|=x+y.
2x2−y2√3(x+y)22=2x2−y2√32.(x+y)22x2−y23(x+y)22=2x2−y232.(x+y)2
=2x2−y2.√32.√(x+y)2=2x2−y2.32.(x+y)2
=2x2−y2.√32.|x+y|=2x2−y2.32.|x+y|
=2(x+y)(x−y).√32.(x+y)=2(x+y)(x−y).32.(x+y)
=2x−y.√32=2x−y.32
=1x−y.2.√32=1x−y.2.32
=1x−y.√22.32=1x−y.22.32
=1x−y.√6=1x−y.6 =√6x−y=6x−y
b) Ta có:
22a−1√5a2(1−4a+4a2)22a−15a2(1−4a+4a2)
=22a−1√5a2(1−2.2a+22a2)=22a−15a2(1−2.2a+22a2)
=22a−1√5a2[12−2.1.2a+(2a)2]=22a−15a2[12−2.1.2a+(2a)2]
=22a−1√5a2(1−2a)2=22a−15a2(1−2a)2
=22a−1√5.√a2.√(1−2a)2=22a−15.a2.(1−2a)2
=22a−1√5.|a|.|1−2a|=22a−15.|a|.|1−2a|
Vì a>0,5a>0,5 nên a>0⇔|a|=aa>0⇔|a|=a.
Vì a>0,5⇔2a>2.0,5⇔2a>1a>0,5⇔2a>2.0,5⇔2a>1 hay 1<</span>2a1<</mo>2a
⇔1−2a<</span>0⇔|1−2a|=−(1−2a)⇔1−2a<</mo>0⇔|1−2a|=−(1−2a)
=−1+2a=2a−1=−1+2a=2a−1
Thay vào trên, ta được:
22a−1√5.|a2|.|1−2a|=22a−1√5.a2.(2a−1)22a−15.|a2|.|1−2a|=22a−15.a2.(2a−1)=2√5a=25a.
Vậy 22a−1√5a2(1−4a+4a2)=2√5a22a−15a2(1−4a+4a2)=25a.
chúc bạn học tôts