c: Ta có: \(C=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)

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

\(=\dfrac{\sqrt{5}-1+\sqrt{5}+1}{\sqrt{2}}=\sqrt{10}\)

\(\sqrt{11-2\sqrt{18}}=3-\sqrt{2}\)

=> a=3; b=-1

Chờ từ trưa không idol nào đụng thì thôi em xin vậy :))

BT1:

Ta có: \(A\cdot B=\sqrt{4+\sqrt{10+2\sqrt{5}}}\cdot\sqrt{4-\sqrt{10+2\sqrt{5}}}\)

\(=\sqrt{16-10-2\sqrt{5}}\)

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

Từ đó thay vào: \(\left(A-B\right)^2\)

\(=A^2-2AB+B^2\)

\(=4+\sqrt{10+2\sqrt{5}}-2\left(\sqrt{5}-1\right)+4-\sqrt{10+2\sqrt{5}}\)

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

\(\Rightarrow A-B=\sqrt{10-2\sqrt{5}}\)

BT2:

Đặt \(B=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)

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

\(=8-2\sqrt{16-7}=8-2\cdot3=2\)

\(\Rightarrow B=\sqrt{2}\)

\(\Rightarrow A=B-\sqrt{2}=\sqrt{2}-\sqrt{2}=0\)

BT3:

đk: \(\orbr{\begin{cases}x\ge2\\x< -2\end{cases}}\)

\(C=\frac{x+2+\sqrt{x^2-4}}{x+2-\sqrt{x^2-4}}+\frac{x+2-\sqrt{x^2-4}}{x+2+\sqrt{x^2-4}}\)

\(C=\frac{\left(x+2+\sqrt{x^2-4}\right)^2}{\left(x+2\right)^2-\left(x^2-4\right)}+\frac{\left(x+2-\sqrt{x^2-4}\right)^2}{\left(x+2\right)^2-\left(x^2-4\right)}\)

\(C=\frac{\left(x+2\right)^2+2\left(x+2\right)\sqrt{x^2-4}+x^2-4+\left(x+2\right)^2-2\left(x+2\right)\sqrt{x^2-4}+x^2-4}{x^2+4x+4-x^2+4}\)

\(C=\frac{2x^2+8x+8+2x^2-8}{4x+8}\)

\(C=\frac{4x^2+8x}{4x+8}=x\)

Vậy C = x

a: Ta có: \(A=\sqrt{12-3\sqrt7}-\sqrt{12+3\sqrt7}\)

=>\(A^2=12-3\sqrt7+12+3\sqrt7-2\cdot\sqrt{\left(12-3\sqrt7\right)\left(12+3\sqrt7\right)}\)

=>\(A^2=24-2\cdot\sqrt{144-63}=24-2\cdot\sqrt{81}=24-2\cdot9=24-18=6\)

=>\(A=\sqrt6\)

b: \(B=\sqrt{4+\sqrt{10+2\sqrt5}}+\sqrt{4-\sqrt{10+2\sqrt5}}\)

=>\(B^2=4+\sqrt{10+2\sqrt5}+4-\sqrt{10+2\sqrt5}+2\cdot\sqrt{4^2-\left(10+2\sqrt5\right)}\)

=>\(B^2=8+2\cdot\sqrt{16-10-2\sqrt5}=8+2\cdot\sqrt{6-2\sqrt5}\)

=>\(B^2=8+2\cdot\sqrt{\left(\sqrt5-1\right)^2}=8+2\left(\sqrt5-1\right)=6+2\sqrt5=\left(\sqrt5+1\right)^2\)

=>\(B=\sqrt5+1\)

c: \(C=\sqrt{3-\sqrt5}+\sqrt{3+\sqrt5}\)

=>\(C^2=3-\sqrt5+3+\sqrt5+2\cdot\sqrt{\left(3-\sqrt5\right)\left(3+\sqrt5\right)}\)

=>\(C^2=6+2\cdot\sqrt{9-5}=6+2\cdot\sqrt4=6+2\cdot2=10\)

=>\(C=\sqrt{10}\)

1.

a, \(\sqrt{7-2\sqrt{10}}+\sqrt{7+2\sqrt{10}}=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}\)

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

b, \(\sqrt{8-2\sqrt{15}}+\sqrt{8-2\sqrt{15}}=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\)

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

c, \(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}=\frac{\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)

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

\(A=\sqrt{7-2\sqrt{10}}+\sqrt{7+2\sqrt{10}}\)

\(A^2=\left(7+2\sqrt{10}+7-2\sqrt{10}\right)+2\sqrt{\left(7-2\sqrt{10}\right)\left(7+2\sqrt{10}\right)}\)

\(=14+2\sqrt{49-40}=14+6=20\)

Khi đó:\(A=\sqrt{20}\)

Các câu còn lại bạn làm nốt nhé

a) \(A=\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\Rightarrow A^2=12-3\sqrt{7}+12+3\sqrt{7}-2\sqrt{\left(12-3\sqrt{7}\right)\left(12+3\sqrt{7}\right)}\Rightarrow A^2=24-2\sqrt{144-63}\Rightarrow A^2=24-18\Rightarrow A^2=6\Rightarrow A=\pm\sqrt{6}\)Ta có \(12-3\sqrt{7}< 12+3\sqrt{7}\Rightarrow\sqrt{12-3\sqrt{7}}< \sqrt{12+3\sqrt{7}}\Rightarrow\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}< 0\Rightarrow A< 0\)Vậy A=-6

b) \(B=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\Rightarrow B^2=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}\Rightarrow B^2=8+2\sqrt{16-10-2\sqrt{5}}\Rightarrow B^2=8+2\sqrt{5-2\sqrt{5}+1}\Rightarrow B^2=8+2\sqrt{\left(\sqrt{5}-1\right)^2}\Rightarrow B^2=8+2\sqrt{5}-2\Rightarrow B=\pm\sqrt{5+2\sqrt{5}+1}\Rightarrow B=\pm\left(\sqrt{5}+1\right)\)Ta có B>0⇒B=\(\sqrt{5}+1\)

c) \(C=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\Rightarrow C^2=3-\sqrt{5}+3+\sqrt{5}+2\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\Rightarrow C^2=6+2\sqrt{9-5}\Rightarrow C^2=6+4=10\Rightarrow C=\pm\sqrt{10}\)Ta có C>0⇒C=\(\sqrt{10}\)

a, A= \(\frac{\sqrt{48-12\sqrt{7}}}{2}-\frac{\sqrt{48+12\sqrt{7}}}{2}\)

       = \(\frac{\sqrt{\left(\sqrt{42}-\sqrt{6}\right)^2}}{2}-\frac{\sqrt{\left(\sqrt{42}+\sqrt{6}\right)^2}}{2}\)

       = \(\frac{-2\sqrt{6}}{2}\)

       = \(-\sqrt{6}\)

a: \(A^2=12-2\sqrt{7}+12+2\sqrt{7}-2\cdot\sqrt{116}\)

\(\Leftrightarrow A^2=24-4\sqrt{29}\)

hay \(A=\sqrt{24-4\sqrt{29}}\)

c: \(C=\dfrac{\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{5}-1+\sqrt{5}+1}{\sqrt{2}}=\dfrac{2\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)