1: \(A=\left(3\sqrt{2}+2\sqrt{3}\right)^2+\dfrac{12}{\sqrt{6}}-3\sqrt{24}\)
\(=18+2\cdot3\sqrt{2}\cdot2\sqrt{3}+12+2\sqrt{6}-3\cdot2\sqrt{6}\)
\(=30+12\sqrt{6}-4\sqrt{6}=30+8\sqrt{6}\)
2:
a: \(B=\left(\dfrac{1}{x+2\sqrt{x}}-\dfrac{1}{\sqrt{x}+2}\right):\dfrac{1-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x+4\sqrt{x}+4\right)}\)
\(=\left(\dfrac{1}{\sqrt{x}\left(\sqrt{x}+2\right)}-\dfrac{1}{\sqrt{x}+2}\right)\cdot\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)^2}{1-\sqrt{x}}\)
\(=\dfrac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\left(\sqrt{x}+1\right)\cdot\left(\sqrt{x}+2\right)^2}{1-\sqrt{x}}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}}\)
b: Để B=6 thì \(\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)=6\cdot\sqrt{x}\)
=>\(x-3\sqrt{x}+2=0\)
=>\(\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\)
=>\(\left[{}\begin{matrix}x=1\left(loại\right)\\x=4\left(nhận\right)\end{matrix}\right.\)
c: \(B=\dfrac{x+3\sqrt{x}+2}{\sqrt{x}}=\sqrt{x}+3+\dfrac{2}{\sqrt{x}}\)
=>\(B>=2\cdot\sqrt{\sqrt{x}\cdot\dfrac{2}{\sqrt{x}}}+3=2\sqrt{2}+3\forall x\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi x=2
nhanh nh
