giúp với

\(D=\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)

\(=\sqrt{4+2\sqrt{2}}.\sqrt{\left(2+\sqrt{2+\sqrt{2}}\right)\left(2-\sqrt{2+\sqrt{2}}\right)}\)

\(=\sqrt{4+2\sqrt{2}}.\sqrt{4-\left(2+\sqrt{2}\right)}\)

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

\(=\sqrt{2\left(2+\sqrt{2}\right)\left(2-\sqrt{2}\right)}\)

\(=\sqrt{2.2}=2\)

Nếu x>3 thì \(2^x⋮\:8\Rightarrow2^x+7\)chia 8 dư 7 , không là số chính phương.

Vậy x\(\le\)3 \(\Rightarrow x\in\){0;1;2}

x=0 \(\Rightarrow7=y^2\).Phương trình không có nghiệm nguyên.

x=1 \(\Rightarrow9=y^2\Leftrightarrow y=\pm3\)

x=2 \(\Rightarrow11=y^2\).Phương trình không có nghiệm nguyên.

Phương trình có 2 nghiệm nguyên (1;3),(1;-3)

\(C=\left(1+\frac{5+\sqrt{5}}{1+\sqrt{5}}\right).\left(\frac{5-\sqrt{5}}{\sqrt{5}-1}+1\right)\)

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

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

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

\(=1+2\sqrt{5}+5\)

\(=6+2\sqrt{5}\)

=.= hok tốt !!

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

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

\(=\left(1-\sqrt{5}\right).\sqrt{5}\)

\(=\sqrt{5}-5\)

\(B=\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\left(2+\sqrt{3}\right)\)

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

\(=\sqrt{3}+2+\sqrt{2}-2-\sqrt{3}\)

\(=\sqrt{2}\)

=.= hok tốt!!

\(\sqrt{9x}-5\sqrt{x}=6-4\sqrt{x}\)

\(\Leftrightarrow\sqrt{9}.\sqrt{x}-5\sqrt{x}+4\sqrt{x}=6\)

\(\Leftrightarrow3.\sqrt{x}-5\sqrt{x}+4\sqrt{x}=6\)

\(\Leftrightarrow\sqrt{x}\left(3-5+4\right)=6\)

\(\Leftrightarrow\sqrt{x}2=6\)

\(\Leftrightarrow\sqrt{x}=3\)

\(\Leftrightarrow\sqrt{x}=\sqrt{9}\)

=> x = 9

cám ơn bn nhieu 

\(A=\sqrt{11-6\sqrt{2}}+3+\sqrt{2}\)

     \(=\sqrt{\left(\sqrt{9}\right)^2-2.\sqrt{9}.\sqrt{2}+\left(\sqrt{2}\right)^2}+3+\sqrt{2}\)

      \(=\sqrt{\left(\sqrt{9}-\sqrt{2}\right)^2}+3+\sqrt{2}\)

        \(=\left|\sqrt{9}-\sqrt{2}\right|+3+\sqrt{2}=\sqrt{9}-\sqrt{2}+3+\sqrt{2}\)

         \(=3-\sqrt{2}+3+\sqrt{2}=6\)

a)  \(A=\sqrt{10+\sqrt{99}}=\sqrt{10+3\sqrt{11}}=\frac{1}{\sqrt{2}}.\sqrt{20+6\sqrt{11}}\)

\(=\frac{1}{\sqrt{2}}.\sqrt{\left(3+\sqrt{11}\right)^2}=\frac{3+\sqrt{11}}{2}\)

b)  \(B=\sqrt{21+6\sqrt{6}}-\sqrt{21-6\sqrt{6}}=\sqrt{\left(3\sqrt{2}+\sqrt{3}\right)^2}-\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}\)

\(=3\sqrt{2}+\sqrt{3}-3\sqrt{2}+\sqrt{3}=2\sqrt{3}\)

c) bn ktra lại đề

d) ĐK:  \(x\ge0\)

 \(\sqrt{x+1+2\sqrt{x}}=\sqrt{\left(\sqrt{x}+1\right)^2}=\sqrt{x}+1\)

e) đk:  \(x\ge-1\)

 \(\sqrt{2x+3+2\sqrt{x^2+3x+2}}=\sqrt{x+1+2\sqrt{\left(x+1\right)\left(x+2\right)}+x+2}\)

\(=\sqrt{\left(\sqrt{x+1}+\sqrt{x+2}\right)^2}=\sqrt{x+1}+\sqrt{x+2}\)