mn giải chi tiết giúp mk vs ạ! mk đang cần gấp
Bài 2:
a: \(A=\sqrt{a}\left(\sqrt{a}+1\right)-2\sqrt{a}-1+1=a-\sqrt{a}\)
c: Để A=2 thì \(a-\sqrt{a}-2=0\)
=>căn a-2=0
hay a=4
d: \(A=a-\sqrt{a}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}>=-\dfrac{1}{4}\)
Dấu '=' xảy ra khi a=1/4
Bài 3
B=(\(\dfrac{3}{\sqrt{1+a}}+\sqrt{1-a}\))\(\left(\dfrac{\sqrt{1-a}\sqrt{1+a}}{3+\sqrt{1-a^{^2}}}\right)\)
=\(\left(\dfrac{3+\sqrt{1-a^{^2}}}{\sqrt{1-a}}\right)\left(\dfrac{\sqrt{1-a}\sqrt{1+a}}{3+\sqrt{1-a^{^2}}}\right)=\sqrt{1+a}\)
b) a=\(\dfrac{\sqrt{3}}{2+\sqrt{3}}=>B=\sqrt{1+\dfrac{\sqrt{3}}{2+\sqrt{3}}}=>\)B=1,21
