1: \(A=\left(\sqrt{28}+3\sqrt{63}-\sqrt{\dfrac{4}{7}}\right)\cdot\sqrt{7}\)
\(=\left(2\sqrt{7}+3\cdot3\sqrt{7}-\dfrac{2}{\sqrt{7}}\right)\cdot\sqrt{7}\)
\(=2\cdot7+9\cdot7-2\)
=77-2
=75
2:
\(b=\sqrt[3]{27}=3=\sqrt{9}\)
\(a=\sqrt{12}\)
mà 9<12
nên b<a
3:
a: \(M=\left(\dfrac{a-\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}+1}{a+\sqrt{a}}\right):\dfrac{\sqrt{a}+1}{a}\)
\(=\left(\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-\dfrac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right)\cdot\dfrac{a}{\sqrt{a}+1}\)
\(=\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\cdot\dfrac{a}{\sqrt{a}+1}\)
\(=\dfrac{a-1}{\sqrt{a}}\cdot\dfrac{a}{\sqrt{a}+1}\)
\(=\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)a}{\left(\sqrt{a}+1\right)\cdot\sqrt{a}}=\sqrt{a}\left(\sqrt{a}-1\right)\)
b: \(M=\sqrt{a}\left(\sqrt{a}-1\right)=a-\sqrt{a}\)
=>\(M=a-\sqrt{a}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}>=-\dfrac{1}{4}\forall a\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi \(\sqrt{a}-\dfrac{1}{2}=0\)
=>a=1/4
