\(A=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}=\sqrt{3+1+2\sqrt{3.1}}-\sqrt{3+1-2\sqrt{3.1}}\)

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

\(B=\sqrt{4+5-2\sqrt{4.5}}+\sqrt{4+5+2\sqrt{4.5}}=\sqrt{(\sqrt{4}-\sqrt{5})^2}+\sqrt{(\sqrt{4}+\sqrt{5})^2}\)

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

\(C\sqrt{2}=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7+1-2\sqrt{7.1}}-\sqrt{7+1+2\sqrt{7.1}}\)

\(=\sqrt{(\sqrt{7}-1)^2}-\sqrt{(\sqrt{7}+1)^2}\)

\(=|\sqrt{7}-1|-|\sqrt{7}+1|=-2\Rightarrow C=-\sqrt{2}\)

----------------------------

\(7+4\sqrt{3}=(2+\sqrt{3})^2\Rightarrow 10\sqrt{7+4\sqrt{3}}=10(2+\sqrt{3})\)

\(\Rightarrow \sqrt{48-10\sqrt{7+4\sqrt{3}}}=\sqrt{28-10\sqrt{3}}=\sqrt{(5-\sqrt{3})^2}=5-\sqrt{3}\)

\(\Rightarrow 3+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}=3+5(5-\sqrt{3})=28-5\sqrt{3}\)

\(\Rightarrow D=\sqrt{5\sqrt{28-5\sqrt{3}}}\)

Cách 1:

\(E=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{8-2\sqrt{15}}\)

\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)

\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})(\sqrt{5}-\sqrt{3})=(4+\sqrt{15})(8-2\sqrt{15})\)

\(=2(4+\sqrt{15})(4-\sqrt{15})=2(16-15)=2\)

Cách 2:

\(E^2=(4+\sqrt{15})^2(\sqrt{10}-\sqrt{6})^2(4-\sqrt{15})=(4+\sqrt{15})(4-\sqrt{15})(4+\sqrt{15}).(16-4\sqrt{15})\)

\(=(16-15)(4+\sqrt{15})(4-\sqrt{15}).4=(16-15)(16-15).4=4\)

Vì $E>0$ nên $E=2$

bằng 0 nha bạn

Cách làm nha các bạn

\(1,\left(2+\sqrt{3}\right)\left(7-4\sqrt{3}\right)\\ =14-8\sqrt{3}+7\sqrt{3}-12\\ =2-\sqrt{3}\\ 2,\left(\sqrt{5-2\sqrt{6}}+\sqrt{2}\right)\sqrt{3}\\ =\left(\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{2}\right)\sqrt{3}\\ =\left(\left|\sqrt{3}-\sqrt{2}\right|+\sqrt{2}\right)\sqrt{3}\\ =\left(\sqrt{3}-\sqrt{2}+\sqrt{2}\right)\sqrt{3}\\ =\sqrt{3}.\sqrt{3}\\ =3\\ 3,\sqrt{4+2\sqrt{3}}-\sqrt{5-2\sqrt{6}}+\sqrt{2}\\ =\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{2}\\ =\left|\sqrt{3}+1\right|-\left|\sqrt{3}-\sqrt{2}\right|+\sqrt{2}\\ =\sqrt{3}+1-\sqrt{3}-\sqrt{2}+\sqrt{2}\\ =1\\ 4,\sqrt{3+2\sqrt{2}}+\sqrt{6-4\sqrt{2}}\\ =\sqrt{\left(1+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{4}-\sqrt{2}\right)^2}\\ =\left|1+\sqrt{2}\right|+\left|\sqrt{4}-\sqrt{2}\right|\\ =1+\sqrt{2}+\sqrt{4}-\sqrt{2}\\ =1+\sqrt{4}\\ 5,2+\sqrt{17-4\sqrt{9+4\sqrt{5}}}\\ =2+\sqrt{17-8-4\sqrt{5}}\\ =2+\sqrt{\left(\sqrt{5}-2\right)^2}\\ =2+\left|\sqrt{5}-2\right|\\ =2+\sqrt{5}-2\\ =\sqrt{5}\)

a: =2-căn 3-2-căn 3

=-2căn 3

b: \(=\dfrac{1}{\sqrt{2}}\left(\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}\right)\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{7}+1-\sqrt{7}+1\right)=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)

c: \(A=\sqrt{4-\sqrt{10-2\sqrt{5}}}-\sqrt{4+\sqrt{10-2\sqrt{5}}}\)

=>\(A^2=4-\sqrt{10-2\sqrt{5}}+4+\sqrt{10-2\sqrt{5}}+2\cdot\sqrt{16-10+2\sqrt{5}}\)

\(\Leftrightarrow A^2=8+2\left(\sqrt{5}+1\right)=10+2\sqrt{5}\)

=>\(A=\sqrt{10+2\sqrt{5}}\)

Ta có: 

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

\(=\)\(\dfrac{\sqrt{10}+3\sqrt{2}}{5+\sqrt{5}}+\dfrac{\sqrt{10}-3\sqrt{2}}{5-\sqrt{5}}\)

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

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

Làm câu S tương tự như này rồi đối chiếu kết quả nha

a) Ta có: \(\sqrt{11-2\sqrt{10}}\)

\(=\sqrt{10-2\cdot\sqrt{10}\cdot1+1}\)

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

\(=\left|\sqrt{10}-1\right|=\sqrt{10}-1\)

b) Ta có: \(\sqrt{9-2\sqrt{14}}\)

\(=\sqrt{7-2\cdot\sqrt{7}\cdot\sqrt{2}+2}\)

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

\(=\left|\sqrt{7}-\sqrt{2}\right|\)

\(=\sqrt{7}-\sqrt{2}\)

c) Ta có: \(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\)

\(=\sqrt{3+2\cdot\sqrt{3}\cdot1+1}+\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\)

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

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

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

\(=2\sqrt{3}\)

d) Ta có: \(\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}\)

\(=\sqrt{5-2\cdot\sqrt{5}\cdot2+4}-\sqrt{5+2\cdot\sqrt{5}\cdot2+4}\)

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

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

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

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

\(=-4\)

e) Ta có: \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)

\(=\frac{\sqrt{2}\cdot\left(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\right)}{\sqrt{2}}\)

\(=\frac{\sqrt{2}\cdot\left(\sqrt{4-\sqrt{7}}\right)-\sqrt{2}\cdot\left(\sqrt{4+\sqrt{7}}\right)}{\sqrt{2}}\)

\(=\frac{\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}}{\sqrt{2}}\)

\(=\frac{\sqrt{7-2\cdot\sqrt{7}\cdot1+1}-\sqrt{7+2\cdot\sqrt{7}\cdot1+1}}{\sqrt{2}}\)

\(=\frac{\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}\)

\(=\frac{\left|\sqrt{7}-1\right|-\left|\sqrt{7}+1\right|}{\sqrt{2}}\)

\(=\frac{\sqrt{7}-1-\left(\sqrt{7}+1\right)}{\sqrt{2}}\)

\(=\frac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}\)

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

g) Ta có: \(\sqrt{3}+\sqrt{11+6\sqrt{2}}+\sqrt{5+2\sqrt{6}}\)

\(=\sqrt{3}+\sqrt{9+2\cdot3\cdot\sqrt{2}+2}+\sqrt{2+2\cdot\sqrt{2}\cdot\sqrt{3}+3}\)

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

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

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

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

h) Ta có: \(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)

\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\cdot\sqrt{3+2\cdot\sqrt{3}\cdot2+4}}}\)

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

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

\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{48-10\sqrt{3}-20}}\)

\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{28-10\sqrt{3}}}\)

\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{25-2\cdot5\cdot\sqrt{3}+3}}\)

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

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

\(=\sqrt{5\sqrt{3}+25-5\sqrt{3}}\)

\(=\sqrt{25}=5\)

k) Ta có: \(\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}\)

\(=\sqrt{49-2\cdot7\cdot\sqrt{45}+45}-\sqrt{49+2\cdot7\cdot\sqrt{45}+45}\)

\(=\sqrt{\left(7-\sqrt{45}\right)^2}-\sqrt{\left(7+\sqrt{45}\right)^2}\)

\(=\left|7-\sqrt{45}\right|-\left|7+\sqrt{45}\right|\)

\(=7-\sqrt{45}-\left(7+\sqrt{45}\right)\)

\(=7-\sqrt{45}-7-\sqrt{45}\)

\(=-2\sqrt{45}=-6\sqrt{5}\)

i) Đặt \(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)

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

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

\(=8+2\cdot\sqrt{16-\left(10+2\sqrt{5}\right)}\)

\(=8+2\cdot\sqrt{6-2\sqrt{5}}\)

\(=8+2\cdot\sqrt{\left(\sqrt{5}-1\right)^2}\)

\(=8+2\cdot\left(\sqrt{5}-1\right)\)

\(=8+2\sqrt{5}-2\)

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

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

\(\Leftrightarrow A=\sqrt{5}+1\)

1: \(\sqrt{3+\sqrt{5}}\cdot\sqrt{2}=\sqrt{6+2\sqrt{5}}=\sqrt{5}+1\)

3) \(\left(\sqrt{\dfrac{3}{4}}-\sqrt{3}+5\cdot\sqrt{\dfrac{4}{3}}\right)\cdot\sqrt{12}\)

\(=\left(\dfrac{\sqrt{3}}{2}-\dfrac{2\sqrt{3}}{2}+5\cdot\dfrac{2}{\sqrt{3}}\right)\cdot\sqrt{12}\)

\(=\dfrac{17\sqrt{3}}{6}\cdot2\sqrt{3}\)

\(=\dfrac{34\cdot3}{6}=\dfrac{102}{6}=17\)