\(a,A=\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}\)

\(=\sqrt{\left(\sqrt{5}^2+2\sqrt{5}+2\sqrt{2}\cdot\sqrt{5}\right)+\sqrt{2}^2+2\sqrt{2}\cdot1+1^2}\)

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

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

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

\(b,B=\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)

\(=\left(\frac{3\cdot\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}{\sqrt{6}+1}+\frac{2\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}{\sqrt{6}-2}-\frac{4\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)

\(=\left[3\cdot\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)\right]\left(\sqrt{6}+11\right)\)

\(=\left(\sqrt{6}+11\right)\left(\sqrt{6}-11\right)=-115\)

Bạn có thể tham khảo tại đây:

Câu hỏi của Nguyễn Ngọc Gia Hân - Toán lớp 9 | Học trực tuyến

\(\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+1\right)=-5-10\sqrt{6}\)        

Mình không chắc lắm

Câu 1,2,3 Ez quá rồi :3

Câu 4:

Tổng quát:

\(\frac{1}{\sqrt{a}+\sqrt{a+1}}=\frac{\sqrt{a}-\sqrt{a+1}}{a-a-1}=\sqrt{a+1}-\sqrt{a}.\) Game là dễ :v

Câu 5 ko khác câu 4 lắm :v

Câu 5: 

Tổng quát:

\(\frac{1}{\sqrt{a}-\sqrt{a+1}}=\frac{\sqrt{a}+\sqrt{a+1}}{a-a-1}=-\sqrt{a}-\sqrt{a+1}.\) Game là dễ :v

Sao làm hổng ai bảo đú.n/g vậy :(((

Lời giải:

a)

\(\sqrt{6+\sqrt{24}+\sqrt{12}+\sqrt{8}}-\sqrt{4-2\sqrt{3}}\)

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

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

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

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

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

b)

\(\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right)(\sqrt{6}+11)\)

\(=\left(\frac{15(\sqrt{6}-1)}{(\sqrt{6}+1)(\sqrt{6}-1)}+\frac{4(\sqrt{6}+2)}{(\sqrt{6}-2)(\sqrt{6}+2)}-\frac{12(3+\sqrt{6})}{(3-\sqrt{6})(3+\sqrt{6})}\right)(\sqrt{6}+11)\)

\(=\left(\frac{15(\sqrt{6}-1)}{5}+\frac{4(\sqrt{6}+2)}{2}-\frac{12(3+\sqrt{6})}{3}\right)(\sqrt{6}+11)\)

\(=[3(\sqrt{6}-1)+2(\sqrt{6}+2)-4(3+\sqrt{6})](\sqrt{6}+11)\)

\(=(\sqrt{6}-11)(\sqrt{6}+11)=6-11^2=-115\)

Lời giải:

a)

\(\sqrt{6+\sqrt{24}+\sqrt{12}+\sqrt{8}}-\sqrt{4-2\sqrt{3}}\)

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

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

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

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

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

b)

\(\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right)(\sqrt{6}+11)\)

\(=\left(\frac{15(\sqrt{6}-1)}{(\sqrt{6}+1)(\sqrt{6}-1)}+\frac{4(\sqrt{6}+2)}{(\sqrt{6}-2)(\sqrt{6}+2)}-\frac{12(3+\sqrt{6})}{(3-\sqrt{6})(3+\sqrt{6})}\right)(\sqrt{6}+11)\)

\(=\left(\frac{15(\sqrt{6}-1)}{5}+\frac{4(\sqrt{6}+2)}{2}-\frac{12(3+\sqrt{6})}{3}\right)(\sqrt{6}+11)\)

\(=[3(\sqrt{6}-1)+2(\sqrt{6}+2)-4(3+\sqrt{6})](\sqrt{6}+11)\)

\(=(\sqrt{6}-11)(\sqrt{6}+11)=6-11^2=-115\)

B=\(\frac{6-6\sqrt{3}}{1-\sqrt{3}}+\frac{3\sqrt{3}+3}{\sqrt{3}+1}=\frac{6\left(1-\sqrt{3}\right)}{1-\sqrt{3}}+\frac{3\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=6+3=9\)

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

D=\(\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}=\frac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)

E=\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}=\frac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}=\sqrt{3}+\frac{1}{2-\sqrt{3}}=\frac{2\sqrt{3}-1}{2-\sqrt{3}}\)