Đặt bth đã cho là A, ta có:

A²=3−√5+3+√5+2√3−√5.√3+√53−5+3+5+23−5.3+5

A²=6+2√(3−√5)(3+√5)6+2(3−5)(3+5)

A^{2=6+2√9−56+29−5}

A²=6+4=10

 ( Tôi giúp ng ae rồi đấy, ok thì kb nhoa) =33

A=√10

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

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

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

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

=\(\frac{2\sqrt{5}}{\sqrt{2}}\)

=\(\sqrt{2}\sqrt{5}\)=\(\sqrt{10}\)

Trả lời:

\(\frac{a-\sqrt{a}}{1-a}\)

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

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

\(=\frac{\sqrt{a}}{1-\sqrt{a}}\)

Trả lời:

Bài 1:

a, \(3\sqrt{2}-4\sqrt{18}+2\sqrt{32}\)

\(=3\sqrt{2}-4\sqrt{3^2.2}+2\sqrt{4^2.2}\)

\(=3\sqrt{2}-12\sqrt{2}+8\sqrt{2}\)

\(=-\sqrt{2}\)

b, \(\sqrt{\left(2+\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}\)

\(=\left|2+\sqrt{5}\right|+\sqrt{5-6\sqrt{5}+9}\)

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

\(=2+\sqrt{5}+\sqrt{\left(\sqrt{5}-3\right)^2}\)

\(=2+\sqrt{5}+\left|\sqrt{5}-3\right|\)

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

\(=2+\sqrt{5}+3-\sqrt{5}=5\)

c, \(\frac{2}{\sqrt{3}-1}+\frac{2}{\sqrt{3}+1}\)

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

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

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

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

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

Bài 2:

a, \(\sqrt{2x+1}-3=1\left(ĐK:x\ge-\frac{1}{2}\right)\)

\(\Leftrightarrow\sqrt{2x+1}=4\)

\(\Leftrightarrow2x+1=16\)

\(\Leftrightarrow2x=15\)

\(\Leftrightarrow x=\frac{15}{2}\left(tm\right)\)

Vậy x = 15/2 

Trả lời:

a, \(5\sqrt{x}-2=13\left(ĐK:x\ge0\right)\)

\(\Leftrightarrow5\sqrt{x}=15\)

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

\(\Leftrightarrow x=9\left(tm\right)\)

Vậy x = 9 

b, \(2\sqrt{8x}+7\sqrt{18x}=9-\sqrt{50x}\left(ĐK:x\ge0\right)\)

\(\Leftrightarrow2\sqrt{2^2.2x}+7\sqrt{3^2.2x}=9-\sqrt{5^2.2x}\)

\(\Leftrightarrow4\sqrt{2x}+21\sqrt{2x}=9-5\sqrt{2x}\)

\(\Leftrightarrow4\sqrt{2x}+21\sqrt{2x}+5\sqrt{2x}=9\)

\(\Leftrightarrow30\sqrt{2x}=9\)

\(\Leftrightarrow\sqrt{2x}=\frac{3}{10}\)

\(\Leftrightarrow2x=\frac{9}{100}\)

\(\Leftrightarrow x=\frac{9}{200}\left(tm\right)\)

Vậy x = 9/200 

TL

HT:

TL:

\(\sqrt{x+1}\) \(+\sqrt{x-1}\)

~HT~