Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(=\left(-\sqrt{5}-\sqrt{7}\right)\cdot\left(\sqrt{7}-\sqrt{5}\right)\)
\(=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)
=-2
b: \(=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
\(=\dfrac{\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}-1+\sqrt{3}+1}{\sqrt{2}}=\sqrt{6}\)
c: \(=\dfrac{\sqrt{10}\left(\sqrt{2}-\sqrt{5}\right)}{\sqrt{2}-\sqrt{5}}-2-\sqrt{10}+3\sqrt{7}+2\)
\(=\sqrt{10}-\sqrt{10}+3\sqrt{7}=3\sqrt{7}\)
2]\(\sqrt{3}\)+1+\(\sqrt{4-4\sqrt{3}+3}\)=\(\sqrt{3}+1+\sqrt{\left(2-\sqrt{3}\right)^2}=\sqrt{3}+1+2-\sqrt{3}=3\)
4\(\left(\dfrac{\sqrt{3}.\left(2+\sqrt{3}\right)+2.\left(2-\sqrt{3}\right)}{\left(2-\sqrt{3}\right).\left(2+\sqrt{3}\right)}\right)=\dfrac{\sqrt{3}.\left(2+\sqrt{3}\right)+2.\left(2-\sqrt{3}\right)}{1}\)
1: \(=2\sqrt{7}-12\sqrt{7}+15\sqrt{7}+27\sqrt{7}=32\sqrt{7}\)
3: \(=\sqrt{5}-2-\sqrt{14+6\sqrt{5}}\)
\(=\sqrt{5}-2-3-\sqrt{5}=-5\)
4: \(=2\sqrt{3}+3+4-2\sqrt{3}=7\)
5: \(=3-\sqrt{2}+3+\sqrt{2}+4-3=7\)
6: \(=\sqrt{\dfrac{6+2\sqrt{5}}{4}}+\sqrt{\dfrac{14-6\sqrt{5}}{4}}\)
\(=\dfrac{\sqrt{5}+1+3-\sqrt{5}}{2}=\dfrac{4}{2}=2\)
8: \(=\sqrt{5}-1+\sqrt{\dfrac{\left(3-\sqrt{5}\right)^2}{4}}-\sqrt{\dfrac{\left(3+\sqrt{5}\right)^2}{4}}\)
\(=\sqrt{5}-1+\dfrac{3-\sqrt{5}}{2}-\dfrac{3+\sqrt{5}}{2}\)
\(=\dfrac{2\sqrt{5}-2+3-\sqrt{5}-3-\sqrt{5}}{2}=\dfrac{-2}{2}=-1\)
\(A=-\sqrt{2}-\sqrt{1}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}+....-\sqrt{7}-\sqrt{8}+\sqrt{8}+\sqrt{9}\)
\(A=\sqrt{9}-\sqrt{1}=3-1=2\)
bạn nên tự nghiên cứu rồi giải đi chứ bạn đưa 1 loạt thế thì ai rảnh mà giải, với lại cứ bài gì không biết chưa chịu suy nghĩ đã hỏi rồi thì tiến bộ sao được, đúng không
1)
a. \(\sqrt{\dfrac{25}{7}}.\sqrt{\dfrac{7}{9}}=\sqrt{\dfrac{25.7}{7.9}}=\sqrt{\dfrac{25}{9}}=\dfrac{5}{3}\)
b. \(\left(\sqrt{\dfrac{9}{2}}+\sqrt{\dfrac{1}{2}}-\sqrt{2}\right).\sqrt{2}=3+1-2=2\)
c. \(\left(\sqrt{\dfrac{8}{3}}-\sqrt{24}+\sqrt{\dfrac{50}{3}}\right).\sqrt{6}=4-12+10=2\)
d. \(\left(\sqrt{\dfrac{2}{3}}-\sqrt{\dfrac{3}{2}}\right)^2=\dfrac{2}{3}+\dfrac{3}{2}-2\sqrt{\dfrac{2}{3}.\dfrac{3}{2}}=\dfrac{1}{6}\)
2)
a. \(\sqrt{4+2\sqrt{3}}=\sqrt{3+2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
b. \(\sqrt{8-2\sqrt{7}}=\sqrt{7-2\sqrt{7}+1}=\sqrt{\left(\sqrt{7}-1\right)^2}=\sqrt{7}-1\)
c. \(1+\sqrt{6-2\sqrt{5}}=1+\sqrt{5-2\sqrt{5}+1}=1-\sqrt{\left(\sqrt{5}-1\right)^2}=1-\sqrt{5}+1=2-\sqrt{5}\)
d. \(\sqrt{7-2\sqrt{10}}+\sqrt{2}=\sqrt{5-2.\sqrt{5}.\sqrt{2}+2}+\sqrt{2}=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}+\sqrt{2}=\sqrt{5}-\sqrt{2}+\sqrt{2}=\sqrt{5}\)
3. \(a.A=x^2+2x+16=\left(\sqrt{2}-1\right)^2+2.\left(\sqrt{2}-1\right)+16=2-2\sqrt{2}+1+2\sqrt{2}-2+16=17\)
\(b.B=x^2+12x-14=\left(5\sqrt{2}-6\right)^2+12.\left(5\sqrt{2}-6\right)-14=50+36-60\sqrt{2}+60\sqrt{2}-72-14=0\)
Help me nha 
@Phùng Khánh Linh@Nhã Doanh@Liana@Yukru Cảm ơn trước nhé 
a: \(=\dfrac{2\sqrt{7}-10-6+2\sqrt{7}}{4}+4+2\sqrt{7}-\dfrac{20}{9}+\dfrac{5}{9}\sqrt{7}\)
\(=\sqrt{7}-4+\dfrac{23}{9}\sqrt{7}+\dfrac{16}{9}\)
\(=\dfrac{32}{9}\sqrt{7}-\dfrac{20}{9}\)
b:\(=\dfrac{2\sqrt{6}+4+2\sqrt{6}-4}{2}+\dfrac{5}{6}\sqrt{6}\)
\(=2\sqrt{6}+\dfrac{5}{6}\sqrt{6}=\dfrac{17}{6}\sqrt{6}\)
c: \(=\dfrac{1}{3}\sqrt{3}+\dfrac{1}{6}\sqrt{2}+\dfrac{1}{\sqrt{3}}\cdot\sqrt{\dfrac{5-2\sqrt{6}}{12}}\)
\(=\dfrac{1}{3}\sqrt{3}+\dfrac{1}{6}\sqrt{2}+\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}}\)
\(=\dfrac{2\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}}{6}=\dfrac{3\sqrt{3}}{6}=\dfrac{\sqrt{3}}{2}\)
a: \(=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{2}\sqrt{2}\right)\cdot5\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+\dfrac{9}{2}\sqrt{2}\right)\cdot5\sqrt{6}\)
\(=60-20\sqrt{18}+\dfrac{45}{2}\sqrt{12}\)
\(=60-60\sqrt{2}+45\sqrt{3}\)
b: \(=\dfrac{\sqrt{5}+\sqrt{2}+\sqrt{5}-\sqrt{2}+3}{3}\cdot\dfrac{1}{3+2\sqrt{2}}\)
\(=\dfrac{2\sqrt{5}+3}{3}\cdot\dfrac{1}{3+2\sqrt{2}}=\dfrac{2\sqrt{5}+3}{9+6\sqrt{2}}\)
\(a.\dfrac{\sqrt{7}-5}{2}-\dfrac{6}{\sqrt{7}-2}+\dfrac{1}{3+\sqrt{7}}+\dfrac{3}{5+2\sqrt{7}}=\dfrac{\sqrt{7}-5}{2}+\dfrac{3-\sqrt{7}}{2}+\dfrac{6\sqrt{7}-15}{3}-\dfrac{6\sqrt{7}+12}{3}=-10\)
\(b.\left(\sqrt{10}+\sqrt{2}\right)\left(6-2\sqrt{5}\right)\sqrt{3+\sqrt{5}}=\left(\sqrt{5}+1\right)\left(6-2\sqrt{5}\right)\sqrt{5+2\sqrt{5}+1}=\left(\sqrt{5}+1\right)^2\left(6-2\sqrt{5}\right)=\left(6+2\sqrt{5}\right)\left(6-2\sqrt{5}\right)=36-20=16\)
\(A=\dfrac{1}{\sqrt{7-\sqrt{24}}+1}-\dfrac{1}{\sqrt{7+\sqrt{24}}+1}\)
\(=\dfrac{\sqrt{7-2\sqrt{6}}-1}{7-2\sqrt{6}-1}-\dfrac{\sqrt{7+2\sqrt{6}}-1}{7+2\sqrt{6}-1}\)
\(=\dfrac{\sqrt{\left(\sqrt{6}-1\right)^2}-1}{6-2\sqrt{6}}-\dfrac{\sqrt{\left(\sqrt{6}+1\right)^2}-1}{6+2\sqrt{6}}\)
\(=\dfrac{\sqrt{6}-2}{\sqrt{6}\left(\sqrt{6}-2\right)}-\dfrac{\sqrt{6}}{\sqrt{6}\left(\sqrt{6}+2\right)}\)
\(=\dfrac{1}{\sqrt{6}}-\dfrac{1}{\sqrt{6}+2}=\dfrac{\sqrt{6}+2-\sqrt{6}}{\sqrt{6}\left(\sqrt{6}+2\right)}\)
\(=\dfrac{2}{\sqrt{12}\left(\sqrt{3}+\sqrt{2}\right)}=\dfrac{2\left(\sqrt{3}-\sqrt{2}\right)}{2\sqrt{3}\left(3-2\right)}=\dfrac{3-\sqrt{6}}{3}\)
\(5-2\sqrt{6}=\left(\sqrt{2}\right)^2-2\times\sqrt{2}\times\sqrt{3}+\left(\sqrt{3}\right)^2=\left(\sqrt{3}-\sqrt{2}\right)^2\)
\(7+2\sqrt{10}=\left(\sqrt{2}\right)^2+2\times\sqrt{2}\times\sqrt{5}+\left(\sqrt{5}\right)^2=\left(\sqrt{2}+\sqrt{5}\right)^2\)
\(8-2\sqrt{15}=\left(\sqrt{5}\right)^3-2\times\sqrt{5}\times\sqrt{3}+\left(\sqrt{3}\right)^2=\left(\sqrt{5}-\sqrt{3}\right)^2\)
\(B=\dfrac{2}{\sqrt{8-2\sqrt{15}}}-\dfrac{1}{\sqrt{5-2\sqrt{6}}}-\dfrac{3}{\sqrt{7+2\sqrt{10}}}\)
\(=\dfrac{2}{\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}}-\dfrac{1}{\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}-\dfrac{3}{\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}}\)
\(=\dfrac{2\left(\sqrt{5}+\sqrt{3}\right)}{5-3}-\dfrac{1\left(\sqrt{3}+\sqrt{2}\right)}{3-2}-\dfrac{3\left(\sqrt{5}-\sqrt{2}\right)}{5-2}\)
\(=\sqrt{5}+\sqrt{3}-\sqrt{3}-\sqrt{2}-\sqrt{5}+\sqrt{2}=0\)