a) \(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)

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

\(=3\sqrt{5}\)

b) \(\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}\)

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

=2

c) \(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}\)

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

\(=2\sqrt{2}\)

Lời giải:
a.

\(=\sqrt{5+2.2\sqrt{5}+2^2}-\sqrt{5-2.2\sqrt{5}+2^2}\)

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

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

b.

$=\sqrt{3-2.3\sqrt{3}+3^2}+\sqrt{3+2.3.\sqrt{3}+3^2}$

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

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

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

c.

$=\sqrt{2+2.3\sqrt{2}+3^2}-\sqrt{2-2.3\sqrt{2}+3^2}$

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

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

Mình dùng máy casio nhé bạn.

KQ; 0,6151214812.

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Áp dụng: \(\left(a+b\right)^3=a^3+3a^2b+3ab^2+b^3=a^3+b^3+3ab\left(a+b\right)\)

Ta đặt: \(x=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)

\(\Rightarrow x^3=9+4\sqrt{5}+9-4\sqrt{5}+3\sqrt[3]{\left(9+4\sqrt{5}\right)\left(9-4\sqrt{5}\right)}.x\)

           \(=18+3\sqrt[3]{81-80}.x\)

             \(=18+3x\)

\(\Rightarrow x^3-18-3x=0\)

\(\Rightarrow x^3-3x^2+3x^2-9x+6x-18=0\)

\(\Leftrightarrow x^2\left(x-3\right)+3x\left(x-3\right)+6\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2+3x+6\right)=0\)

Vì \(x^2+3x+6=x^2+2.x.\frac{3}{2}+\frac{9}{4}+\frac{15}{4}=\left(x+\frac{3}{2}\right)^2+\frac{15}{4}>0\)

Suy ra: x - 3 = 0

=> x = 3

Vâỵ \(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}=3\)

\(A=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)

\(\Leftrightarrow A^3=9+4\sqrt{5}+9-4\sqrt{5}\)

\(+3\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)\sqrt[3]{\left(9+4\sqrt{5}\right)\left(9-4\sqrt{5}\right)}\)

\(\Leftrightarrow A^3=18+3A\Leftrightarrow A^3-3A-18=0\)

\(\Leftrightarrow\left(A-3\right)\left(A^2+3A+6\right)=0\)

Dễ thấy : \(A^2+3A+6=\left(A+\frac{3}{2}\right)^2+\frac{15}{4}\ge0\forall A\)

\(\Leftrightarrow A=3\)

Chúc bạn học tốt !!!

\(A=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)

\(\Leftrightarrow A^3=9+4\sqrt{5}+9-4\sqrt{5}\)

\(+3\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)\sqrt[3]{\left(9+4\sqrt{5}\right)\left(9-4\sqrt{5}\right)}\)

\(\Leftrightarrow A^3+18+3A\Leftrightarrow A^3-3A-18=0\)

\(\Leftrightarrow\left(A-3\right)\left(A^2+3A+6\right)=0\)

Dễ thấy : \(A^2+3A+6=\left(A+\frac{3}{2}\right)^2+\frac{15}{4}\ge0\forall A\)

\(\Leftrightarrow A=3\)

Chúc bạn học tốt !!!

\(c,\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\\ =\sqrt{\sqrt{3^2}+2\sqrt{3}.1+1}+\sqrt{\sqrt{3^2}-2\sqrt{3}.1+1}\\ =\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\\ =\left|\sqrt{3}+1\right|+\left|\sqrt{3}-1\right|\\ =\sqrt{3}+1+\sqrt{3}-1\\ =2\sqrt{3}\)

\(d,\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\\ =\sqrt{\sqrt{5^2}+2.2\sqrt{5}+2^2}-\sqrt{\sqrt{5^2}-2.2\sqrt{5} +2^2}\\ =\sqrt{\left(\sqrt{5}+2\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}\\ =\left|\sqrt{5}+2\right|-\left|\sqrt{5}-2\right|\\ =\sqrt{5}+2-\sqrt{5}+2\\ =4\)

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

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

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

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

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

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

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

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

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

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

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

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

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 $E=2$