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Võ Thị Ngọc Tú
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hoàng thiên
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Akai Haruma
7 tháng 8 2019 lúc 23:50

Lời giải:
Biểu thức \(=\frac{(5+2\sqrt{6})(25+24-2\sqrt{25.24})\sqrt{3+2-2\sqrt{3.2}}}{9\sqrt{3}-11\sqrt{2}}\)

\(=\frac{(5+2\sqrt{6})(\sqrt{25}-\sqrt{24})^2.\sqrt{(\sqrt{3}-\sqrt{2})^2}}{9\sqrt{3}-11\sqrt{2}}\)

\(=\frac{(5+2\sqrt{6})(5-2\sqrt{6})^2(\sqrt{3}-\sqrt{2})}{9\sqrt{3}-11\sqrt{2}}\)

\(=\frac{(5+2\sqrt{6})(5-2\sqrt{6})(5-2\sqrt{6})(\sqrt{3}-\sqrt{2})}{9\sqrt{3}-11\sqrt{2}}\)

\(=\frac{(5-2\sqrt{6})(\sqrt{3}-\sqrt{2})}{9\sqrt{3}-11\sqrt{2}}=\frac{9\sqrt{3}-11\sqrt{2}}{9\sqrt{3}-11\sqrt{2}}=1\)

Quyết Tâm Chiến Thắng
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alibaba nguyễn
18 tháng 9 2019 lúc 16:44

d/ \(x=\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)

\(\Leftrightarrow x^3=3+\sqrt{9+\frac{125}{27}}+3-\sqrt{9+\frac{125}{27}}-3\left(\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\right)\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}.\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)

\(\Leftrightarrow x^3=6-3x\sqrt[3]{9-9-\frac{125}{27}}\)

\(\Leftrightarrow x^3=6-5x\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+x+6\right)=0\)

\(\Leftrightarrow x=1\)

alibaba nguyễn
19 tháng 9 2019 lúc 13:39

c/

\(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)

\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)

\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{12}+4}}}\)

\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)

\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)

\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)

\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)

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

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

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

\(=3-1=2\)

alibaba nguyễn
19 tháng 9 2019 lúc 14:05

b/ \(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)

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

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

\(=\sqrt{4+\sqrt{25}}\)

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

khanhhuyen6a5
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Akai Haruma
3 tháng 7 2021 lúc 17:18

Bạn tham khảo 

https://hoc24.vn/cau-hoi/rut-gonfracleft52sqrt6rightleft49-20sqrt6rightsqrt5-2sqrt69sqrt3-11sqrt2.227145517764

KHANH QUYNH MAI PHAM
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Phú Nguyễn Duy
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Tiểu Ma Bạc Hà
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Thiên An
30 tháng 6 2017 lúc 19:42

\(\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)

\(=\frac{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)^2\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)

\(=\frac{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)\sqrt{\left(5-2\sqrt{6}\right)^2.\left(5-2\sqrt{6}\right)}}{9\sqrt{3}-11\sqrt{2}}\)

\(=\frac{\left[25-\left(2\sqrt{6}\right)^2\right]\sqrt{\left(5-2\sqrt{6}\right)^3}}{9\sqrt{3}-11\sqrt{2}}\)

\(=\frac{\sqrt{125-150\sqrt{6}+360-48\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)

\(=\frac{\sqrt{485-198\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)

\(=\frac{\sqrt{243-2.9\sqrt{3}.11\sqrt{2}+242}}{9\sqrt{3}-11\sqrt{2}}\)

\(=\frac{\sqrt{\left(9\sqrt{3}-11\sqrt{2}\right)^2}}{9\sqrt{3}-11\sqrt{2}}=1\)

Trần Quang Minh
7 tháng 3 2019 lúc 19:56

Bang 1 nha ban

Kiyotaka Ayanokoji
28 tháng 6 2020 lúc 17:43

\(\frac{\left(5+2\sqrt{6}\right).\left(49-20\sqrt{6}\right).\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)

\(=\frac{\left(5+2\sqrt{6}\right).\left(25-20\sqrt{6}+24\right).\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)

\(=\frac{\left(5+2\sqrt{6}\right).\left(5-2\sqrt{6}\right)^2.\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)

\(=\frac{\left(25-\left(2\sqrt{6}\right)^2\right).\sqrt{\left(5-2\sqrt{6}\right)^2.\left(5-2\sqrt{6}\right)}}{9\sqrt{3}-11\sqrt{2}}\)

\(=\frac{1.\sqrt{\left(5-2\sqrt{6}\right)^3}}{9\sqrt{3}-11\sqrt{2}}\)

\(=\frac{\sqrt{125-150\sqrt{6}+360-48\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)

\(=\frac{\sqrt{485-198\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)

\(=\frac{\sqrt{243-2.9\sqrt{3}.11\sqrt{2}+242}}{9\sqrt{3}-11\sqrt{2}}\)

\(=\frac{\sqrt{\left(9\sqrt{3}-11\sqrt{2}\right)^2}}{9\sqrt{3}-11\sqrt{2}}\)

\(=\frac{9\sqrt{3}-11\sqrt{2}}{9\sqrt{3}-11\sqrt{2}}\)

\(=1\)

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