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

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

\(A=\sqrt{5}-1-\sqrt{5}-1\)

\(A=-2\)

\(B=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)

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

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

\(B=4\)

Sửa đề :

\(C=\sqrt{14-6\sqrt{5}}-\sqrt{14+6\sqrt{5}}\)

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

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

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

Ta có : \(\sqrt{54-14\sqrt{5}}-\sqrt{14+6\sqrt{5}}\)

\(=\sqrt{7^2-2.7.\sqrt{5}+\left(\sqrt{5}\right)^2}-\sqrt{3^2+2.3.\sqrt{5}+\left(\sqrt{5}\right)^2}\)

\(=\sqrt{\left(7-\sqrt{5}\right)^2}-\sqrt{\left(3+\sqrt{5}\right)^2}\)

\(=7-\sqrt{5}-\left(3+\sqrt{5}\right)=7-\sqrt{5}-3-\sqrt{5}=4\)

xin chào em mới học dưới lớp tám thôi khó quá không biết làm 

1)
\(=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(\sqrt{11}-\sqrt{3}\right)^2}=\sqrt{11}-\sqrt{3}\)
2)
\(=\sqrt{\left(\sqrt{7}\right)^2-2.\sqrt{7}\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{7}-\sqrt{5}\right)^2}=\sqrt{7}-\sqrt{5}\)
3)
\(=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{11}-\sqrt{5}\right)}=\sqrt{11}-\sqrt{5}\)
4)
\(=\sqrt{3^2-2.3.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(3-\sqrt{5}\right)^2}=3-\sqrt{5}\)
5)
\(=\sqrt{3^2-2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}=\sqrt{\left(3-2\sqrt{2}\right)^2}=3-2\sqrt{2}\)

 

\(\dfrac{15}{\sqrt{6}-1}+\dfrac{8}{\sqrt{6}+2}+\dfrac{6}{3-\sqrt{6}}-9\sqrt{6}\)

\(=\dfrac{15\left(\sqrt{6}+1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{8\left(\sqrt{6}-2\right)}{\left(\sqrt{6}+2\right)\left(\sqrt{6}-2\right)}+\dfrac{6\left(3+\sqrt{6}\right)}{\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}-9\sqrt{6}\)

\(=\dfrac{15\left(\sqrt{6}+1\right)}{6-1}+\dfrac{8\left(\sqrt{6}-2\right)}{6-4}+\dfrac{6\left(3+\sqrt{6}\right)}{9-6}-9\sqrt{6}\)

\(=3\left(\sqrt{6}+1\right)+4\left(\sqrt{6}-2\right)+2\left(3+\sqrt{6}\right)-9\sqrt{6}\)

\(=3\sqrt{6}+3+4\sqrt{6}-8+6+2\sqrt{6}-9\sqrt{6}\)

\(=9\sqrt{6}+1-9\sqrt{6}\)

\(=1\)

\(\sqrt{\left(\sqrt{5}-1\right)\sqrt{14-6\sqrt{5}}}\)

\(=\sqrt{\left(\sqrt{5}-1\right)\sqrt{9-6\sqrt{5}+5}}\)

\(=\sqrt{\left(\sqrt{5}-1\right)\sqrt{3^2-2\cdot3\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}}\)

\(=\sqrt{\left(\sqrt{5}-1\right)\sqrt{\left(3-\sqrt{5}\right)^2}}\)

\(=\sqrt{\left(\sqrt{5}-1\right)\left|3-\sqrt{5}\right|}\)

\(=\sqrt{\left(\sqrt{5}-1\right)\left(3-\sqrt{5}\right)}\)

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

\(=\sqrt{4\sqrt{5}-8}\)

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

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

\(=6.\)

chỉ mk cách làm

bấm máy ra.

\(P=\sqrt{14+6\sqrt{5}}+\sqrt{14-6\sqrt{5}}\)

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

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

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

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

\(=6\)

\(\frac{14-6\sqrt{5}+8}{\sqrt{14-6\sqrt{5}}+1}=\frac{22-6\sqrt{5}}{\sqrt{\left(3-\sqrt{5}\right)^2}+1}=\frac{22-6\sqrt{5}}{4-\sqrt{5}}=\frac{\left(22-6\sqrt{5}\right)\left(4+\sqrt{5}\right)}{16-5}=\frac{58-2\sqrt{5}}{11}\)