Trả lời

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

Phân số bạn làm tiếp nha !

Bài làm nguồn:CHTT , hihi. đg ném đá nha, có ý tốt thoi !

\(A=\frac{\sqrt{1}+\sqrt{2}}{1-2}-\frac{\sqrt{2}+\sqrt{3}}{2-3}+\frac{\sqrt{3}+\sqrt{4}}{3-4}-...-\frac{\sqrt{24}+\sqrt{25}}{24-25}\)

\(=-\sqrt{1}-\sqrt{2}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}+...+\sqrt{24}+\sqrt{25}\)

\(=-\sqrt{1}+\sqrt{25}\)

\(=-1+5\)

\(=4.\)

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

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

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

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

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

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

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

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

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

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

b) \(\frac{1}{2\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}=\frac{2\sqrt{3}}{12}+\frac{2\sqrt{3}}{6}-\frac{6-2\sqrt{3}}{6}\)

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

Có: \(\frac{P}{\sqrt{2}}=\frac{1}{\sqrt{2}}\left(\frac{3+\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}-\frac{3-\sqrt{5}}{\sqrt{10}+\sqrt{3-\sqrt{5}}}\right)\)

\(=\frac{3+\sqrt{5}}{\sqrt{20}+\sqrt{6+2\sqrt{5}}}-\frac{3-\sqrt{5}}{\sqrt{20}+\sqrt{6-2\sqrt{5}}}\)

\(=\frac{3+\sqrt{5}}{\sqrt{20}+\sqrt{\left(\sqrt{5}+1\right)^2}}-\frac{3-\sqrt{5}}{\sqrt{20}+\sqrt{\left(\sqrt{5}-1\right)^2}}\)

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

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

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

\(=\frac{9\sqrt{5}-3+15-\sqrt{5}-9\sqrt{5}-3+15+\sqrt{5}}{9\cdot5-1}\)

\(=\frac{24}{44}=\frac{6}{11}\)

=>P=\(\frac{6}{11}\cdot\sqrt{2}=\frac{6\sqrt{2}}{11}\)

Chính xác 100% mink thử bằng máy tính r

mink làm hơi tắt phần nào k hiểu hói mink nhé