a) \(=\sqrt{\left(3\sqrt{5}-2\right)^2}+\sqrt{\left(3\sqrt{5}+2\right)^2}=3\sqrt{5}-2+3\sqrt{5}+2=6\sqrt{5}\)

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

a: Sửa đề: \(A=\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)

\(=4-\sqrt{15}+\sqrt{15}=4\)

b: \(A=2-\sqrt{3}+\sqrt{3}-1=1\)

c: \(C=3\sqrt{5}-2-3\sqrt{5}-2=-4\)

d: Sửa đề: \(D=\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)

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

=6

a) \(A=\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)

\(A=\left|4-\sqrt{15}\right|+\sqrt{15}\)

\(A=4-\sqrt{15}+\sqrt{15}\)

\(A=4\)

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

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

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

\(B=1\)

c) \(C=\sqrt{49-12\sqrt{5}}-\sqrt{49+12\sqrt{5}}\)

\(C=\sqrt{\left(3\sqrt{5}\right)^2-2\cdot3\sqrt{15}\cdot2+2^2}-\sqrt{\left(3\sqrt{5}\right)^2+2\cdot3\sqrt{5}\cdot2+2^2}\)

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

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

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

\(C=-4\)

d) \(D=\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)

\(D=\sqrt{\left(2\sqrt{5}\right)^2+2\cdot2\sqrt{5}\cdot3+3^2}-\sqrt{\left(2\sqrt{5}\right)^2-2\cdot2\sqrt{5}\cdot3+3^3}\)

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

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

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

\(D=6\)

\(3\sqrt{144}-5\sqrt{49}+\dfrac{1}{2}\sqrt{36}\)

\(=3.12-5.7+\dfrac{1}{2}.6\)

\(=36-35+3=4\)

Bài 1 : \(\sqrt{49-12\sqrt{5}}+\sqrt{49+12\sqrt{5}}\)

\(=\sqrt{45-4\sqrt{45}+4}+\sqrt{45+4\sqrt{45}+4}\)

\(=\sqrt{\left(\sqrt{45}-2\right)^2}+\sqrt{\left(\sqrt{45}+2\right)^2}\)

\(=\sqrt{45}-2+\sqrt{45}+2=2\sqrt{45}\)

Bài 2 : \(\sqrt{29+12\sqrt{5}}+\sqrt{29-12\sqrt{5}}\)

\(=\sqrt{20+6\sqrt{20}+9}+\sqrt{20-6\sqrt{20}+9}\)

\(=\sqrt{\left(\sqrt{20}+3\right)^2}+\sqrt{\left(\sqrt{20}-3\right)^2}\)

\(=\sqrt{20}+3+\sqrt{20}-3=2\sqrt{20}\)

Bài 3 : \(\sqrt{31-12\sqrt{3}}+\sqrt{31+12\sqrt{3}}\)

\(=\sqrt{27-4\sqrt{27}+4}+\sqrt{27+4\sqrt{27}+4}\)

\(=\sqrt{\left(\sqrt{27}-2\right)^2}+\sqrt{\left(\sqrt{27}+2\right)^2}\)

\(=\sqrt{27}-2+\sqrt{27}+2=2\sqrt{27}\)

Chúc bạn học tốt

4 , Ta có :

\(\sqrt{39-12\sqrt{3}}-\sqrt{39+12\sqrt{3}}\)

\(=\sqrt{3-2.6.\sqrt{3}+6^2}-\sqrt{3+2.6.\sqrt{3}+6^2}\)

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

\(=\left|\sqrt{3}-6\right|-\left|\sqrt{3}+6\right|\)

\(=6-\sqrt{3}-\sqrt{3}-6\)

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

1,\(\sqrt{4-2.2.3\sqrt{5}+45}+\sqrt{4+2.2.3\sqrt{5}+45}\)

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

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

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

\(=6\sqrt{5}\)

câu đầu có \(3-12\sqrt{6}< 0\) nên không căn được nên đề bạn sai

\(\sqrt{31-8\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)

\(=\sqrt{4^2-2.4.\sqrt{15}+\left(\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}\right)^2-2.\sqrt{15}.3+3^2}\)

\(=\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}-3\right)^2}=\left|4-\sqrt{15}\right|+\left|\sqrt{15}-3\right|\)

\(=4-\sqrt{15}+\sqrt{15}-3=1\)

\(\sqrt{49-5\sqrt{96}}-\sqrt{49+5\sqrt{96}}=\sqrt{49-20\sqrt{6}}-\sqrt{49+20\sqrt{6}}\)

\(=\sqrt{5^2-2.5.2\sqrt{6}+\left(2\sqrt{6}\right)^2}-\sqrt{5^2+2.5.4\sqrt{6}+\left(2\sqrt{6}\right)^2}\)

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

\(=5-2\sqrt{6}-5-2\sqrt{6}=-4\sqrt{6}\)

\(\sqrt{31-8\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)

\(=4-\sqrt{15}+\sqrt{15}-3\)

=1

1) \(\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)

\(=\sqrt{2^2+2\cdot2\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}-\sqrt{2^2-2\cdot2\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}\)

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

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

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

\(=4\)

2) \(\sqrt{12-6\sqrt{3}}+\sqrt{12+6\sqrt{3}}\)

\(=\sqrt{3^2-2\cdot3\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}+\sqrt{3^2+2\cdot3\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}\)

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

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

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

\(=6\)

\(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3\sqrt{3}-2\sqrt{2}\right)^2}\)

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

\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}=\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3+2\sqrt{2}\right)^2}\)

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

\(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}=\sqrt{\left(5-2\sqrt{6}\right)^2}+\sqrt{\left(5+2\sqrt{6}\right)^2}\)

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

\(\sqrt{13-\sqrt{160}}+\sqrt{53+4\sqrt{90}}=\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}+\sqrt{\left(3\sqrt{5}+2\sqrt{2}\right)^2}\)

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

a: \(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}\)

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

b: \(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)

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

=6

c: Ta có: \(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}\)

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

=10

d: Ta có: \(\sqrt{13-\sqrt{160}}+\sqrt{53+4\sqrt{90}}\)

\(=\sqrt{13-4\sqrt{10}}+\sqrt{53+4\sqrt{90}}\)

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

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

a: \(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)

\(=4-\sqrt{15}+\sqrt{15}=4\)

b: \(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)

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

\(=2\sqrt{3}\)

c: \(\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)

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

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