\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(\sqrt{3}+\sqrt{4}\right)^2}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)

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

\(=\sqrt{4+\sqrt{25}}=\sqrt{4+5}=\sqrt{9}=3\)

\(=\sqrt{5-\sqrt{3-\sqrt{\left(2\sqrt{5}\right)^2-2.2\sqrt{5}.3+3^2}}}\)

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

\(=\sqrt{5-\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{5-\left(\sqrt{5}-1\right)}\)\(=\sqrt{6-\sqrt{5}}\)

\(=\sqrt{5-\sqrt{3-\sqrt{20-2.3.2\sqrt{5}+9}}}\)

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

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

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

\(=\sqrt{5-\sqrt{5-2.1.\sqrt{5}+1}}\)

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

Nếu là căn 5 thì = 1 

chẳng bít có đúng không

\(=\sqrt{\frac{\sqrt{a}-1}{\sqrt{b}+1}.\frac{\sqrt{a}+1}{\sqrt{b}-1}}=\sqrt{\frac{a-1}{b-1}}=\sqrt{\frac{7,25-1}{3,25-1}}=\sqrt{\frac{625}{225}}=\frac{5}{3}\)

\(\sqrt{\frac{\sqrt{a}-1}{\sqrt{b}-1}}:\sqrt{\frac{\sqrt{b}-1}{\sqrt{a}+1}}=\frac{\sqrt{\sqrt{a}-1}}{\sqrt{\sqrt{b+1}}}\cdot\frac{\sqrt{\sqrt{a}+1}}{\sqrt{\sqrt{b}-1}}\)

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

Thay a = 7,25 ; b = 3,25 ta có

 \(\frac{\sqrt{7,25-1}}{\sqrt{3,25-1}}=\frac{\sqrt{6,25}}{\sqrt{2,25}}=\frac{2,5}{1,5}\)

Đúng nha

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

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

1)

a)\(\left(\sqrt{x}-\sqrt{y}\right).\left(\sqrt{x}+\sqrt{y}\right)=\left(\sqrt{x}\right)^2-\left(\sqrt{y}\right)^2=x-y\)

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

c)\(\sqrt{\left(2-\sqrt{5}\right)^2.\left(2+\sqrt{5}\right)^2}=\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)=4-\left(\sqrt{5}\right)^2\)

=\(4-5=1\)

d)\(\sqrt{\left(-5\right)^2.3^2}=\left(5\right).3=15\)

e)\(\sqrt{\frac{5}{27}.\frac{8}{20}}=\sqrt{\frac{2}{27}}\)

ĐÂy này nhớ **** vài câu nha

a, \(\left(\sqrt{14}+\sqrt{6}\right)\sqrt{5-\sqrt{21}}=\left(\sqrt{7}+\sqrt{3}\right)\sqrt{2}\sqrt{5-\sqrt{21}}\)

\(=\left(\sqrt{7}+\sqrt{3}\right)\sqrt{10-2\sqrt{21}}=\left(\sqrt{7}+\sqrt{3}\right)\sqrt{7-2.\sqrt{3}.\sqrt{7}+3}\)

\(=\left(\sqrt{7}+\sqrt{3}\right)\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}=\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)=7-3=4\)

Rất muốn tính nhưng mà dài quá

 = \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\) ( vì căn 16 = 4)

=\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)  (vì căn 4 = 2 mà 2 + 2 = 4 nên tách luôn thành căn 4 )

= \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

= \(\frac{\left(1+\sqrt{2}\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)

Đúng nha lần sau mình giải tiếp cho

(\(3\sqrt{2}\)+\(\sqrt{6}\))\(\sqrt{6-3\sqrt{2}}\)

Rút y từ phương trình số 2 rồi thay vào phương trrình 1 => 3x + m^2x - m = 5 => m^2x+3x=m+5 => x(m^2+3)=m+5

a) đk: a>=0\(M=\frac{x^4-x}{x^2+x+1}-\frac{x^4+x}{x^2-x+1}=\frac{x\left(x-1\right)\left(x^2+x+1\right)}{x^2+x+1}-\frac{x\left(x+1\right)\left(x^2-x+1\right)}{x^2-x+1}=x^2-x-x^2-x=-2x\)

\(\Leftrightarrow M=-2\sqrt{a}\)

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