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17 tháng 8 2018

Đặt \(\sqrt{2+\sqrt{3}}=a\left(a>0\right)\)

Ta có x=\(\sqrt{2+a}-\sqrt{3\left(2-a\right)}\Rightarrow x^2=2+a+3\left(2-a\right)-2\sqrt{3\left(2+a\right)\left(2-a\right)}\)\(=8-2a-2\sqrt{3\left(4-a^2\right)}=8-2a-2\sqrt{3\left(4-2-\sqrt{3}\right)}=8-2a-\sqrt{6}\sqrt{4-2\sqrt{3}}\)

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

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

\(\Rightarrow x^2-8=-4\sqrt{2}\Rightarrow\left(x^2-8\right)^2=32\Rightarrow x^4-16x^2+64=32\Rightarrow x^4-16x^2+32=0\left(ĐPCM\right)\)

26 tháng 6 2023

Giải

Ta có:

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

Khi đó:

\(x^2=\left(\sqrt{2+\sqrt{2+\sqrt{3}}-\sqrt{6-3\sqrt{2+\sqrt{3}}}}\right)^2\\ =2+\sqrt{2+\sqrt{3}}+6-3\sqrt{2+\sqrt{3}}-2\sqrt{\left(2+\sqrt{2+\sqrt{3}}\right)\left(6-3\sqrt{2+\sqrt{3}}\right)}\\ =8-2\sqrt{2+\sqrt{3}}-2\sqrt{12-3\left(2+\sqrt{3}\right)}\\ =8-\sqrt{2}.\sqrt{4+2\sqrt{3}}-2\sqrt{6-3\sqrt{3}}\\ =8-\sqrt{2}.\sqrt{4+2\sqrt{3}}-\sqrt{2}.\sqrt{12-6\sqrt{3}}\\ =8-\sqrt{2}.\left(\sqrt{4+2\sqrt{3}}+\sqrt{12-6\sqrt{3}}\right)\\ =8-\sqrt{2}.\left(\sqrt{\left(\sqrt{3}\right)^2+2\sqrt{3}+1}+\sqrt{9-2.3\sqrt{3}+\left(\sqrt{3}\right)^2}\right)\\ 8-\sqrt{2}.\left(\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(3-\sqrt{3}\right)^2}\right)\\ =8-\sqrt{2}.\left(\sqrt{3}+1+3-\sqrt{3}\right)\\ =8-4\sqrt{2}\\ \Rightarrow x^4-16x^2=\left(8-4\sqrt{2}\right)^2-16.\left(8-4\sqrt{2}\right)\\ =96-64\sqrt{2}-128+64\sqrt{2}=-32\)

Vậy \(S=-32\)

31 tháng 7 2017

Câu hỏi của Phạm Thị Thu Trang - Toán lớp 9 - Học toán với OnlineMath

29 tháng 7 2017

Xem lại đề đi. Thế x ngược lại đâu có đúng

a) Ta có: \(a^3\)

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

\(=5\sqrt{5}+15\sqrt{3}+9\sqrt{5}+3\sqrt{3}\)

b) Ta có: \(a^4-16a^2+4=0\)

\(\Leftrightarrow\left(\sqrt{5}+\sqrt{3}\right)^4-16\left(\sqrt{5}+\sqrt{3}\right)^2+4=0\)

\(\Leftrightarrow\left(8+2\sqrt{15}\right)^2-16\left(8+2\sqrt{15}\right)+4=0\)

\(\Leftrightarrow64+32\sqrt{15}+60-128-32\sqrt{15}+4=0\)

\(\Leftrightarrow0=0\)(đúng)

31 tháng 7 2018

\(x^2=8-2\sqrt{2+\sqrt{3}}-2\sqrt{3.\left(2-\sqrt{3}\right)}\)

\(\Leftrightarrow8-x^2=2\sqrt{2+\sqrt{3}}+2\sqrt{3.\left(2-\sqrt{3}\right)}\)

\(\Leftrightarrow x^4-16x^2+64=4\left(2+\sqrt{3}+6-3\sqrt{3}+2\sqrt{3}\right)\)

\(\Leftrightarrow x^4-16x^2+64=32\)

\(\Leftrightarrow x^4-16x^2+32=0\)

Vậy có điều phải chứng minh.

16 tháng 5 2017

Ta có:

\(x_0^2=8-2\sqrt{2+\sqrt{3}}-2\sqrt{3\left(2-\sqrt{3}\right)}\)

\(\Leftrightarrow\left(\dfrac{8-x_0^2}{2}\right)^2=\left(\sqrt{2+\sqrt{3}}+\sqrt{3\left(2-\sqrt{3}\right)}\right)^2\)

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

\(\Rightarrow x_0^4-16x_0^2+64=32\)

\(\Rightarrow x_0^4-16x_0^2+32=0\)

Vậy ......

8 tháng 7 2019

làm tắt quá mình ko hiểu

26 tháng 7 2015

\(x^4-16x^2+32=0\Leftrightarrow x^2=8+4\sqrt{2}\text{ hoặc }x^2=8-4\sqrt{2}\)

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

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

\(a^2=\frac{4+\sqrt{6}+\sqrt{2}}{2}+3.\frac{4-\sqrt{6}-\sqrt{2}}{2}-2\sqrt{3}\sqrt{\frac{\left(4+\sqrt{6}+\sqrt{2}\right)\left(4-\sqrt{6}-\sqrt{2}\right)}{2.2}}\)

\(=8-\left(\sqrt{6}+\sqrt{2}\right)-2\sqrt{3}.\frac{1}{2}.\sqrt{4^2-\left(\sqrt{6}+\sqrt{2}\right)^2}\)

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

\(=8-\sqrt{2}-\sqrt{6}-\sqrt{\left(3\sqrt{2}-\sqrt{6}\right)^2}\)

\(=8-\sqrt{2}-\sqrt{6}-\left(3\sqrt{2}-\sqrt{6}\right)\)

\(=8-4\sqrt{2}\)

\(\Rightarrow a\text{ là nghiệm phương trình }x^4-16x^2+32=0\)

26 tháng 7 2015

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

\(x^2=8-2\sqrt{2+\sqrt{3}}-2.\sqrt{3.\left(2+\sqrt{2+\sqrt{3}}\right).\left(2-\sqrt{2+\sqrt{3}}\right)}\)

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

\(x^2=8-\sqrt{2}\sqrt{4+2.\sqrt{3}}-\sqrt{6}.\sqrt{4-2.\sqrt{3}}=8-\sqrt{2}.\sqrt{\left(1+\sqrt{3}\right)^2}-\sqrt{6}.\sqrt{\left(\sqrt{3}-1\right)^2}\)

\(x^2=8-\sqrt{2}.\left(1+\sqrt{3}\right)-\sqrt{6}.\left(\sqrt{3}-1\right)=8-\sqrt{2}-\sqrt{6}-3\sqrt{2}+\sqrt{6}=8-4\sqrt{2}\)

=> \(x^4=\left(x^2\right)^2=\left(8-4\sqrt{2}\right)^2=\left(4\sqrt{2}\right)^2.\left(\sqrt{2}-1\right)^2=32.\left(2-2\sqrt{2}+1\right)=96-64\sqrt{2}\)

=> \(x^4-16x^2+32=96-64\sqrt{2}-16.\left(8-4\sqrt{2}\right)+32=\left(96-96\right)-64\sqrt{2}+64\sqrt{2}=0\)

=> đpcm