24.11 C

24.12 A

24.13 A

a) \(45,23+56,78-12\cdot10\)

a = 89,91

b =24,78913

c = không biết và minh nha kết bạn nữa

1.

$\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}=\sqrt{3+1+2\sqrt{3}}-\sqrt{3+1-2\sqrt{3}}$

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

$=|\sqrt{3}+1|-|\sqrt{3}-1|=2$

2.

\(\sqrt{12+6\sqrt{3}+\sqrt{12-6\sqrt{3}}}=\sqrt{12+6\sqrt{3}+\sqrt{9+3-2\sqrt{9.3}}}=\sqrt{12+6\sqrt{3}+\sqrt{(3-\sqrt{3})^2}}\)

\(=\sqrt{12+6\sqrt{3}+3-\sqrt{3}}=\sqrt{15+5\sqrt{3}}\)

3.

\(\sqrt{9-4\sqrt{2}+\sqrt{9+4\sqrt{2}}}=\sqrt{9-4\sqrt{2}+\sqrt{8+1+2\sqrt{8.1}}}\)

\(=\sqrt{9-4\sqrt{2}+\sqrt{2\sqrt{2}+1)^2}}=\sqrt{9-4\sqrt{2}+2\sqrt{2}+1}=\sqrt{10-2\sqrt{2}}\)

4.

\(\sqrt{\sqrt{2}+2+\sqrt{4+\sqrt{9-\sqrt{32}}}}=\sqrt{\sqrt{2}+2+\sqrt{4+\sqrt{8+1-2\sqrt{8.1}}}}\)

\(=\sqrt{\sqrt{2}+2+\sqrt{4+\sqrt{(\sqrt{8}-1)^2}}}\) \(=\sqrt{\sqrt{2}+2+\sqrt{4+\sqrt{8}-1}}=\sqrt{\sqrt{2}+2+\sqrt{3+2\sqrt{2}}}\)

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

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

5.

\(\sqrt{6+2\sqrt{5}-\sqrt{29+12\sqrt{5}}}=\sqrt{6+2\sqrt{5}-\sqrt{20+9+2\sqrt{20.9}}}\)

\(=\sqrt{6+2\sqrt{5}-\sqrt{(\sqrt{20}+3)^2}}=\sqrt{6+2\sqrt{5}-(\sqrt{20}+3)}=\sqrt{3}\)

6.

\(\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}-\sqrt{\sqrt{49}+\sqrt{40}}\)

\(=\sqrt{8+2\sqrt{2}+2\sqrt{5}+2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)

\(=\sqrt{(2+5+2\sqrt{2.5})+2(\sqrt{2}+\sqrt{5})+1}-\sqrt{2+5+2\sqrt{2.5}}\)

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

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

a: \(=\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{4-2-\sqrt{2+\sqrt{3}}}\)

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

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

b: \(=\sqrt{2+\sqrt{2}}\cdot\sqrt{3+\sqrt{7+\sqrt{2}}}\cdot\sqrt{9-6-\sqrt{7+\sqrt{2}}}\)

\(=\sqrt{2+\sqrt{2}}\cdot\sqrt{9-7-\sqrt{2}}\)

\(=\sqrt{2}\)

a: \(2\sqrt{8\sqrt{3}}-\sqrt{2\sqrt{3}}-\sqrt{9\sqrt{12}}\)

\(=2\sqrt{4\cdot2\sqrt{3}}-\sqrt{2\sqrt{3}}-\sqrt{9\cdot2\sqrt{3}}\)

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

=0

b: \(\sqrt{3}+\sqrt{7-4\sqrt{3}}\)

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

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

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

=2

c: \(\sqrt{\left(\sqrt{7}-4\right)^2}-\sqrt{28}+\sqrt{63}\)

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

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

=4

d: \(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}\)

\(=\dfrac{\sqrt{10}\left(15\sqrt{5}+5\sqrt{20}-3\sqrt{45}\right)}{\sqrt{10}}\)

\(=15\sqrt{5}+5\sqrt{20}-3\sqrt{45}\)

\(=15\sqrt{5}+5\cdot2\sqrt{5}-3\cdot3\sqrt{5}\)

\(=16\sqrt{5}\)

e: \(\sqrt{3}-2\sqrt{48}+3\sqrt{75}-4\sqrt{108}\)

\(=\sqrt{3}-2\cdot4\sqrt{3}+3\cdot5\sqrt{3}-4\cdot6\sqrt{3}\)

\(=\sqrt{3}-8\sqrt{3}+15\sqrt{3}-24\sqrt{3}\)

\(=-16\sqrt{3}\)

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

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

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

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

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

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

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

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

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

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

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

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

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 $E=2$

a) Ta có: \(\sqrt{3-2\sqrt{2}}-\sqrt{11+6\sqrt{2}}\)

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

=-4

b) Ta có: \(\sqrt{4-2\sqrt{3}}-\sqrt{7-4\sqrt{3}}+\sqrt{19+8\sqrt{3}}\)

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

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

c) Ta có: \(\sqrt{6-2\sqrt{5}}+\sqrt{9+4\sqrt{5}}-\sqrt{14-6\sqrt{5}}\)

\(=\sqrt{5}-1+\sqrt{5}-2-3+\sqrt{5}\)

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

d) Ta có: \(\sqrt{11-4\sqrt{7}}+\sqrt{23-8\sqrt{7}}+\sqrt{\left(-2\right)^6}\)

\(=\sqrt{7}-2+4-\sqrt{7}+8\)

=10