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![](https://rs.olm.vn/images/avt/0.png?1311)
Biến đổi vế trái ta có:
\(\sqrt{3+\sqrt{5}-\sqrt{13+4\sqrt{3}}}=\sqrt{3+\sqrt{5}-\sqrt{\left(2\sqrt{3}+1\right)^2}}=\sqrt{3+\sqrt{5}-2\sqrt{3}-1}=\sqrt{2+\sqrt{5}-2\sqrt{3}}\)
Đề sai
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(\frac{3.\sqrt{6}}{2}+\frac{2.\sqrt{2}}{\sqrt{3}}-\frac{4.\sqrt{3}}{\sqrt{2}}=\frac{3\sqrt{6}}{2}+\frac{2\sqrt{2}.\sqrt{3}}{\sqrt{3}.\sqrt{3}}-\frac{4.\sqrt{3}.\sqrt{2}}{\sqrt{2}.\sqrt{2}}=\frac{3\sqrt{6}}{2}+\frac{2\sqrt{6}}{3}-\frac{4\sqrt{6}}{2}=\frac{2\sqrt{6}}{3}-\frac{\sqrt{6}}{2}=\frac{4\sqrt{6}-3\sqrt{6}}{6}=\frac{\sqrt{6}}{6}\)
--> dpcm
b) \(\left(\frac{-\sqrt{7}.\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\frac{-\sqrt{5}.\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right).\frac{\sqrt{7}-\sqrt{5}}{1}\)
=\(\left(-\sqrt{7}-\sqrt{5}\right).\left(\sqrt{7}-\sqrt{5}\right)\)
=\(-1.\left(\sqrt{7}+\sqrt{5}\right).\left(\sqrt{7}-\sqrt{5}\right)\)
=\(-1.\left(7-5\right)\)
=-1.2
=-2
![](https://rs.olm.vn/images/avt/0.png?1311)
Vế trái : \(\frac{1}{\sqrt{6}+\sqrt{7}}-\sqrt{7}=\frac{\sqrt{7}-\sqrt{6}}{\left(\sqrt{7}-\sqrt{6}\right)\left(\sqrt{7}+\sqrt{6}\right)}-\sqrt{7}=\sqrt{7}-\sqrt{6}-\sqrt{7}=-\sqrt{6}\)
Vế phải : \(\sqrt{\left(\sqrt{5}-\sqrt{6}\right)^2}-2\sqrt{6}+\sqrt{5}=\left(\sqrt{6}-\sqrt{5}\right)-2\sqrt{6}+\sqrt{5}=-\sqrt{6}\)
Vậy ..................................
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
A= \(\dfrac{2\cdot\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)
Đặt B=\(2\cdot\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
Ta có B=\(2\cdot\sqrt{3+\sqrt{5-\sqrt{12+2\cdot\sqrt{12}+1}}}\)
\(2\cdot\sqrt{3+\sqrt{5-\sqrt{12}-1}}\\ =2\sqrt{3+\sqrt{4-\sqrt{12}}}\\ =2\cdot\sqrt{3+\sqrt{3-2\cdot\sqrt{3}+1}}\\ =2\cdot\sqrt{3+\sqrt{3}-1}\\ =2\cdot\sqrt{2+\sqrt{3}}\)
Thay B vào A, ta cũng có:
A=\(\dfrac{2\cdot\sqrt{2+\sqrt{3}}}{\sqrt{6}+\sqrt{2}}\\ =\dfrac{2\cdot\sqrt{2+\sqrt{3}}}{\sqrt{2\cdot\left(\sqrt{3}+1\right)}}\\ =\dfrac{\sqrt{2}\cdot\sqrt{2+\sqrt{3}}}{\sqrt{3}+1}\\ =\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}+1}=\dfrac{\sqrt{3+2\sqrt{3}+1}}{\sqrt{3}+1}=1\)
Vậy A thuộc Z
![](https://rs.olm.vn/images/avt/0.png?1311)
giả sử 2 vế bằng nhau, nhân tích chéo, rồi được 2 vế = nhau là kết luận thỏa mãn
\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}=\frac{1}{1+\sqrt{2}}=\frac{\sqrt{2}-1}{2-1}=\sqrt{2}-1=vp\)
Xét tử \(2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}=2\sqrt{3+\sqrt{5-\sqrt{13+4\sqrt{3}}}}\)
\(=2\sqrt{3+\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}}=2\sqrt{3+\sqrt{5-\left(2\sqrt{3}+1\right)}}\)
\(=2\sqrt{3+\sqrt{4-2\sqrt{3}}}=2\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}=2\sqrt{3+\sqrt{3}-1}\)
\(=2\sqrt{2+\sqrt{3}}=\frac{2\sqrt{4+2\sqrt{3}}}{\sqrt{2}}=\frac{2\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}}=\frac{2\left(\sqrt{3}+1\right)}{\sqrt{2}}=\sqrt{6}+\sqrt{2}\)
Suy ra VT = VP = 1
Đáp án là : VT = VP = 1
vhuv bn hoc gioi tk mk nha