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AT
1
VT
17 tháng 8 2018
Đặt \(\sqrt{2+\sqrt{3}}=a\left(a>0\right)\)
Ta có x=\(\sqrt{2+a}-\sqrt{3\left(2-a\right)}\Rightarrow x^2=2+a+3\left(2-a\right)-2\sqrt{3\left(2+a\right)\left(2-a\right)}\)\(=8-2a-2\sqrt{3\left(4-a^2\right)}=8-2a-2\sqrt{3\left(4-2-\sqrt{3}\right)}=8-2a-\sqrt{6}\sqrt{4-2\sqrt{3}}\)
\(=8-2\sqrt{2+\sqrt{3}}-\sqrt{6}\left(\sqrt{3}-1\right)=8-\sqrt{2}\sqrt{4+2\sqrt{3}}-3\sqrt{2}+\sqrt{6}\)
\(=8-\sqrt{2}\left(\sqrt{3}+1\right)-3\sqrt{2}+\sqrt{6}=8-\sqrt{6}-\sqrt{2}-3\sqrt{2}+\sqrt{6}=8-4\sqrt{2}\)
\(\Rightarrow x^2-8=-4\sqrt{2}\Rightarrow\left(x^2-8\right)^2=32\Rightarrow x^4-16x^2+64=32\Rightarrow x^4-16x^2+32=0\left(ĐPCM\right)\)
ML
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
6 tháng 8 2017
\(\frac{A}{\sqrt{2}}=\frac{1+\sqrt{7}}{2+\sqrt{8+2\sqrt{7}}}+\frac{1-\sqrt{7}}{2-\sqrt{8-2\sqrt{7}}}\)
\(=\frac{1+\sqrt{7}}{2+1+\sqrt{7}}+\frac{1-\sqrt{7}}{2-\sqrt{7}+1}\)
\(=\frac{1+\sqrt{7}}{3+\sqrt{7}}+\frac{1-\sqrt{7}}{3-\sqrt{7}}\)
=\(\frac{\left(1+\sqrt{7}\right)\left(3-\sqrt{7}\right)+\left(1-\sqrt{7}\right)\left(3+\sqrt{7}\right)}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}\)
\(=\frac{-8}{2}=-4\)
\(\Rightarrow A=-4\sqrt{2}\)
23 tháng 9 2017
3.
\(•x=3+\sqrt{2}\\ x^2=\left(3+\sqrt{2}\right)^2\\ x^2=9+2.3.\sqrt{2}+2\\ x^2=11+6\sqrt{2}\\• y=\sqrt{11+6\sqrt{2}}\\ y^2=\left(\sqrt{11+6\sqrt{2}}\right)^2\\ y^2=11+6\sqrt{2}\)
\(\Rightarrow x^2=y^2=11+6\sqrt{2}\)
LD
23 tháng 9 2017
1. ta có : \(4\sqrt{7}=\sqrt{112}\)
\(3\sqrt{3}=\sqrt{27}\)
ta thấy : \(\sqrt{112}>\sqrt{27}\) hay \(4\sqrt{7}>3\sqrt{3}\)
2. \(\dfrac{1}{4}\sqrt{82}=\sqrt{\dfrac{41}{8}}\)
\(6\sqrt{\dfrac{1}{7}}=\sqrt{\dfrac{36}{7}}\)
ta thấy :\(\sqrt{\dfrac{41}{8}}< \sqrt{\dfrac{36}{7}}\) hay \(\dfrac{1}{4}\sqrt{82}< 6\sqrt{\dfrac{1}{7}}\)
3. \(x^2=\left(3+\sqrt{2}\right)^2\)
\(y^2=11+6\sqrt{2}\)=\(\left(3+\sqrt{2}\right)^2\)
ta thấy : \(x^2=y^2\Rightarrow x=y\)
NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 8 2020
6.
Đặt \(\left\{{}\begin{matrix}\sqrt{5x^2+6x+5}=a\\4x=b\end{matrix}\right.\)
\(\Rightarrow a\left(a^2+1\right)=b\left(b^2+1\right)\)
\(\Leftrightarrow a^3-b^3+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+b^2+ab+1\right)=0\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow\sqrt{5x^2+6x+5}=4x\left(x\ge0\right)\)
\(\Leftrightarrow5x^2+6x+5=16x^2\)
\(\Leftrightarrow11x^2-6x-5=0\)
\(\Rightarrow x=1\)
NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 8 2020
4. Bạn coi lại đề (chính xác là pt này ko có nghiệm thực)
5.
\(\Leftrightarrow x^2+x+6-\left(2x+1\right)\sqrt{x^2+x+6}+6x-6=0\)
Đặt \(\sqrt{x^2+x+6}=t>0\)
\(t^2-\left(2x+1\right)t+6x-6=0\)
\(\Delta=\left(2x+1\right)^2-4\left(6x-6\right)=\left(2x-5\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}t=\frac{2x+1+2x-5}{2}=2x-2\\t=\frac{2x+1-2x+5}{2}=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+x+6}=2x-2\left(x\ge1\right)\\\sqrt{x^2+x+6}=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+6=4x^2-8x+4\left(x\ge1\right)\\x^2+x+6=9\end{matrix}\right.\)
<=> \(x^2=2+\sqrt{2+\sqrt{3}}+6-3\sqrt{2+\sqrt{3}}-2\sqrt{\left(2+\sqrt{2+\sqrt{3}}\right)\left(6-3\sqrt{2+\sqrt{3}}\right)}\)
<=> \(x^2=8-2\sqrt{2+\sqrt{3}}-2\sqrt{12-6\sqrt{2+\sqrt{3}}+6\sqrt{2+\sqrt{3}}-3\left(2+\sqrt{3}\right)}\)
<=> \(x^2=8-\sqrt{2}.\sqrt{4+2\sqrt{3}}-2\sqrt{12-6-3\sqrt{3}}\)
<=> \(x^2=8-\sqrt{2}.\sqrt{\left(\sqrt{3}+1\right)^2}-2\sqrt{6-3\sqrt{3}}\)
<=> \(x^2=8-\sqrt{2}\left(\sqrt{3}+1\right)-\sqrt{2}.\sqrt{12-6\sqrt{3}}\)
<=> \(x^2=8-\sqrt{6}-\sqrt{2}-\sqrt{2}.\sqrt{\left(3-\sqrt{3}\right)^2}\)
<=> \(x^2=8-\sqrt{6}-\sqrt{2}-\sqrt{2}\left(3-\sqrt{3}\right)\)
<=> \(x^2=8-\sqrt{6}-\sqrt{2}-3\sqrt{2}+\sqrt{6}\)
<=> \(x^2=8-4\sqrt{2}\)
<=> \(8-x^2=4\sqrt{2}\)
<=> \(\left(8-x^2\right)^2=\left(4\sqrt{2}\right)^2\)
<=> \(x^4-16x^2+64=32\)
<=> \(x^4-16x^2=-32\)
VẬY \(x^4-16x^2=-32\)
*** ĐÂY LÀ 1 BÀI TOÁN RẤT CỔ RỒI !!!!!!