Xét \(\sqrt{2}.A=\sqrt{\dfrac{4+2\sqrt{3}}{2}}-\sqrt{\dfrac{4-2\sqrt{3}}{2}}\)

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

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

<=> A = 1

\(\frac{\sqrt{2-\sqrt{3}}}{2}:\left(\frac{\sqrt{2+\sqrt{3}}}{2}-\frac{2}{\sqrt{6}}+\frac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}\right).\)

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

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

\(=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{4}:\left[\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{4}-\frac{2}{\sqrt{6}}+\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{4\sqrt{3}}\right]\)

\(=\frac{\sqrt{3}-1}{4}:\left[\frac{\sqrt{6}\left(\sqrt{3}+1\right)}{4\sqrt{6}}-\frac{2.4}{4\sqrt{6}}+\frac{\sqrt{2}\left(\sqrt{3}+1\right)}{4\sqrt{6}}\right]\)

\(=\frac{\sqrt{3}-1}{4}:\frac{\sqrt{18}+\sqrt{6}-8+\sqrt{6}+\sqrt{2}}{4\sqrt{6}}\)

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

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

HÌNH NHƯ BẰNG 1,414213562

A=\(\sqrt{2}\), cái kết quả này bấm máy tính là ra được, quan trọng là phải làm thế nào để ra

Đặt \(x=2+\sqrt{3};y=2-\sqrt{3}\), ta có

\(A=\frac{x}{\sqrt{2}+\sqrt{x}}+\frac{y}{\sqrt{2}-\sqrt{y}}=\frac{x\left(\sqrt{2}-\sqrt{y}\right)+y\left(\sqrt{2}+\sqrt{x}\right)}{\left(\sqrt{2}+\sqrt{x}\right)\left(\sqrt{2}-\sqrt{y}\right)}\)

\(=\frac{\sqrt{2}\left(x+y\right)-\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{2+\sqrt{2}\left(\sqrt{x}-\sqrt{y}\right)-\sqrt{xy}}\)

Có  # \(x+y=2+\sqrt{3}+2-\sqrt{3}=4\)

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

    ### \(\sqrt{x}-\sqrt{y}=\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}=\frac{\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)

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

Vậy kết luận \(A=\frac{\sqrt{2}.4-\sqrt{2}}{2+\sqrt{2}.\sqrt{2}-1}=\frac{3\sqrt{2}}{3}=\sqrt{2}\)

                                                  Ký tên bài giải: ĐẶNG ĐỨC TRƯỜNG 

a) \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\)

\(=\left[-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right]\cdot\left(\sqrt{2}-\sqrt{5}\right)\)

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

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

\(=-\left(2-5\right)\)

\(=-\left(-3\right)\)

\(=3\)

b) Ta có:

\(x^2-x\sqrt{3}+1\) 

\(=x^2-2\cdot\dfrac{\sqrt{3}}{2}\cdot x+\left(\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)

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

Mà: \(\left(x-\dfrac{\sqrt{3}}{2}\right)^2\ge0\forall x\) nên

\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\forall x\)

Dấu "=" xảy ra:

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

\(\Leftrightarrow x=\dfrac{\sqrt{3}}{2}\)

Vậy: GTNN của biểu thức là \(\dfrac{1}{4}\) tại \(x=\dfrac{\sqrt{3}}{2}\)

a)

\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\\ =\left(-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =\left(-\sqrt{2}-\sqrt{5}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}^2-\sqrt{5}^2\right)\\ =-\left(2-5\right)\\ =-\left(-3\right)\\ =3\)

bạn đặt A=biểu thức rồi tính  \(\frac{1}{\sqrt{2}}A\)  là ra

\(M=\frac{2+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{2-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)

\(M.\frac{1}{\sqrt{2}}=\frac{2+\sqrt{5}}{2+\sqrt{6+2\sqrt{5}}}+\frac{2-\sqrt{5}}{2-\sqrt{6-2\sqrt{5}}}\)

\(M.\frac{1}{\sqrt{2}}=\frac{2+\sqrt{5}}{2+\sqrt{5}+1}+\frac{2-\sqrt{5}}{2-\sqrt{5}-1}\)

\(M.\frac{1}{\sqrt{2}}=\frac{2+\sqrt{5}}{3+\sqrt{5}}+\frac{2-\sqrt{5}}{1-\sqrt{5}}\)

P/s làm tiếp nha , hình như bạn ghi đề sai dấu

bài này dễ bn, bn nhân vs biểu thức liên hợp ở mẫu là ra nka, mik ko bt viết mấy kí tự trên này nên ko hướng dẫn ra cụ thể đc

Gọi biểu thức là A

=>A*\(\sqrt{2}\)=\(\frac{\sqrt{6}}{2+\sqrt{4+2\sqrt{3}}}\)+\(\frac{\sqrt{6}}{2-\sqrt{4-2\sqrt{3}}}\)=\(\frac{\sqrt{6}}{2+\sqrt{\left(1+\sqrt{3}\right)^2}}\)+\(\frac{\sqrt{6}}{2+\sqrt{\left(\sqrt{3}-1\right)^2}}\)=\(\frac{\sqrt{6}}{2+1+\sqrt{3}}\)+\(\frac{\sqrt{6}}{2-\sqrt{3}+1}\)

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

=\(\frac{6\sqrt{6}}{-2\sqrt{3}}\)=-3\(\sqrt{2}\)

=>A=-3

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

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

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

\(=\frac{2}{\sqrt{3}}\left(\frac{4-2\sqrt{3}+4+2\sqrt{3}}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\right)=\frac{2}{\sqrt{3}}.\frac{8}{2}=\frac{8}{\sqrt{3}}=\frac{8\sqrt{3}}{3}\)

1.052631148

có hiểu rút gọn là j ko thế

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

\(=\frac{\sqrt{30}+\sqrt{5}}{\left(\sqrt{3}+1\right)\left(\sqrt{\sqrt{3}}+1\right)+1}=\frac{\left(\sqrt{30}+\sqrt{5}\right)\left(\sqrt{\sqrt{3}}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{\sqrt{3}}+1\right)\left(\sqrt{\sqrt{3}}-1\right)+\sqrt{\sqrt{3}}-1}\)

\(=\frac{\left(\sqrt{30}+\sqrt{5}\right)\left(\sqrt{\sqrt{3}}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)+\sqrt{\sqrt{3}}-1}\)

\(=\frac{\left(\sqrt{30}+\sqrt{5}\right)\left(\sqrt{\sqrt{3}}-1\right)\left(\sqrt{\sqrt{3}}-1\right)}{\left(\sqrt{\sqrt{3}}+1\right)\left(\sqrt{\sqrt{3}}-1\right)}\)

\(=\frac{\left(\sqrt{30}+\sqrt{5}\right)\left(\sqrt{\sqrt{3}}-1\right)^2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

\(=\frac{\left(\sqrt{30}+\sqrt{5}\right)\left(\sqrt{\sqrt{3}}-1\right)^2\left(\sqrt{3}+1\right)}{2}\)\(=2\sqrt{30}+2\sqrt{5}+\sqrt{90}+\sqrt{15}-\sqrt{90\sqrt{3}}-\sqrt{30\sqrt{3}}-\sqrt{15\sqrt{3}}-\sqrt{5\sqrt{3}}\)

mởi tay ùi,có gì thiếu tự giải tiếp ^^