\(A=\frac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)

\(\sqrt{2}A=\frac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{2}\left(\sqrt{5}+1\right)}=\frac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{\sqrt{2}\left(\sqrt{5}+1\right)}\)

\(\sqrt{2}A=\frac{2\left(\sqrt{5}+1\right)}{\sqrt{2}\left(\sqrt{5}+1\right)}=\frac{2}{\sqrt{2}}\)

\(A=\frac{2}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}}=1\)

\(\frac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}=\frac{2\left[\sqrt{3-\sqrt{5}}.\left(3+\sqrt{5}\right)\right]}{2\left(\sqrt{10}+\sqrt{2}\right)}\)

\(=\frac{\sqrt{2}.\sqrt{3-\sqrt{5}}.\sqrt{2}\left(3+\sqrt{5}\right)}{2\sqrt{10}+2\sqrt{2}}\)

\(=\frac{\sqrt{6-2\sqrt{5}}\left(6+2\sqrt{5}\right)}{2\sqrt{10}+2\sqrt{2}}\)

\(=\frac{\sqrt{5-2\sqrt{5}+1}\left(5+2\sqrt{5}+1\right)}{2\sqrt{10}+2\sqrt{2}}\)

\(=\frac{\sqrt{\left(\sqrt{5}-1\right)^2}.\left(\sqrt{5}+1\right)^2}{2\sqrt{10}+2\sqrt{2}}\)

\(=\frac{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)^2}{2\sqrt{10}+2\sqrt{2}}=\frac{\left(5-1\right)\left(\sqrt{5}+1\right)}{2\sqrt{10}+2\sqrt{2}}\)

\(=\frac{4\sqrt{5}+4}{2\sqrt{10}+2\sqrt{2}}=\frac{2\sqrt{5}+2}{\sqrt{10}+\sqrt{2}}\)

\(=\frac{\sqrt{2}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{2}\left(\sqrt{5}+1\right)}=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}+1}\)

a, ĐKXĐ: \(x>0;x\ne1;x\ne4\)

\(M=\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)

\(=\frac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)

\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{x-1-x+2}\)

\(=\frac{\sqrt{x}-2}{\sqrt{x}}\)

a/ rút gọn A

b/tìm a để giá trị của A đạt GTLN

\(A=\left(1-\frac{2\sqrt{a}-2}{a-1}\right):\left(\frac{1}{1+\sqrt{a}}-\frac{a}{1+a\sqrt{a}}\right)\)

\(=\left(\frac{a-1-\left(2\sqrt{a}-2\right)}{a-1}\right):\)\(\left(\frac{1}{\sqrt{a}+1}-\frac{a}{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}\right)\)

\(=\left(\frac{a-1-2\sqrt{a}-2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}+1\right)}\right):\)\(\left(\frac{a-\sqrt{a}+1-a}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)

\(=\frac{\left(a-2\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}:\frac{-\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)

\(=-\frac{\left(\sqrt{a}-1\right)^2\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}=-\left(\sqrt{a}-1\right)=1-\sqrt{a}\)

mình nghĩ bài này sai đề, 

ĐÚng phải là\(\sqrt[3]{2+\sqrt{3}}\)

(   KHÔNG CHẮC NỮA   :D   )

\(\text{sai đề chú ơi}\)

\(A=\left(1+\frac{5}{\sqrt{x}-2}\right).\left(\sqrt{x}-\frac{x+2\sqrt{x}+4}{\sqrt{x}+3}\right).\)

\(=\frac{\sqrt{x}-2+5}{\sqrt{x}-2}.\frac{x+3\sqrt{x}-x-2\sqrt{x}-4}{\sqrt{x}+3}\)

\(=\frac{\sqrt{x}+3}{\sqrt{x}-2}.\frac{\sqrt{x}-4}{\sqrt{x}+3}=\frac{\sqrt{x}-4}{\sqrt{x}-2}\)

ĐK \(x\ge-2\)

pT<=> \(2\left(x+1\right)\sqrt{x+2}+2\left(x+6\right)\sqrt{x+7}=2x^2+14x+24\)

<=>\(\left(x+1\right)\left(x+2-2\sqrt{x+2}\right)+\left(x+6\right)\left(x+4-2\sqrt{x+7}\right)+x-2=0\)

<=>\(\frac{\left(x+1\right)\left(x^2-4\right)}{x+2+2\sqrt{x+2}}+\frac{\left(x+6\right)\left(x^2+4x-12\right)}{x+4+2\sqrt{x+7}}+x-2=0\forall x>-2\)

=> \(\orbr{\begin{cases}x=2\\\frac{\left(x+1\right)\left(x+2\right)}{x+2+2\sqrt{x+2}}\end{cases}}+\frac{x+6}{x+4+2\sqrt{x+7}}+1=0\left(2\right)\)

Pt (2) + \(x\ge-1\)=> \(VT>0\)=> PT (2) vô nghiệm

+  \(-2< x\le-1\)=> \(\frac{\left(x+1\right)\left(x+2\right)}{x+2+2\sqrt{x+2}}>-1\)=> \(VT>0\)=> PT vô nghiệm

Vậy x=2

a, c.Câu hỏi của Nữ hoàng sến súa là ta - Toán lớp 9 - Học toán với OnlineMath