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<=> (x-4)(x-3) = \(\sqrt{3}\)(y+1)
Nếu y là số nguyên khác -1 thì y+1 là số nguyên; \(\sqrt{3}\)là số vô tỉ nên \(\sqrt{3}\left(y+1\right)\)là số vô tỉ
mà x-4 và x-3 đều là số nguyên nên (x-3)(x-4) là số nguyên => vô lý
vậy y = -1 => (x-4)(x-3)=0 <=> x=4 hoặc x= 3
vậy có 2 nghiêm thỏa mãn (x;y) = (4;-1); (x;y) = (3;-1)
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 \(S=-32\)
Đặt \(\sqrt{2+\sqrt{3}}=a\left(a>0\right)\)
Ta có x=\(\sqrt{2+a}-\sqrt{3\left(2-a\right)}\Rightarrow x^2=2+a+3\left(2-a\right)-2\sqrt{3\left(2+a\right)\left(2-a\right)}\)\(=8-2a-2\sqrt{3\left(4-a^2\right)}=8-2a-2\sqrt{3\left(4-2-\sqrt{3}\right)}=8-2a-\sqrt{6}\sqrt{4-2\sqrt{3}}\)
\(=8-2\sqrt{2+\sqrt{3}}-\sqrt{6}\left(\sqrt{3}-1\right)=8-\sqrt{2}\sqrt{4+2\sqrt{3}}-3\sqrt{2}+\sqrt{6}\)
\(=8-\sqrt{2}\left(\sqrt{3}+1\right)-3\sqrt{2}+\sqrt{6}=8-\sqrt{6}-\sqrt{2}-3\sqrt{2}+\sqrt{6}=8-4\sqrt{2}\)
\(\Rightarrow x^2-8=-4\sqrt{2}\Rightarrow\left(x^2-8\right)^2=32\Rightarrow x^4-16x^2+64=32\Rightarrow x^4-16x^2+32=0\left(ĐPCM\right)\)
a, \(\sqrt{2}x-\sqrt{6}=0\Leftrightarrow\sqrt{2}x=\sqrt{6}\Leftrightarrow x=\sqrt{3}\)
b, \(\frac{x^2}{\sqrt{3}}-\sqrt{12}=0\Leftrightarrow\frac{x^2}{\sqrt{3}}=\sqrt{12}\Leftrightarrow x^2=\sqrt{12}.\sqrt{3}\Leftrightarrow x^2=\sqrt{36}\Leftrightarrow x=36\)
c, \(\sqrt{3}x+\sqrt{3}=\sqrt{12}+\sqrt{27}\Leftrightarrow\sqrt{3}x=\sqrt{12}+\sqrt{27}-\sqrt{3}\)
\(\Leftrightarrow\sqrt{3}x=2\sqrt{3}+3\sqrt{3}-\sqrt{3}\Leftrightarrow\sqrt{3}x=4\sqrt{3}\Leftrightarrow x=4\)
CMR: \(x^4-16.x^2+32=0\)
có 1 nghiệm là:
\(x=\sqrt{6-3.\sqrt{2+\sqrt{3}}}+\sqrt{2+\sqrt{2+\sqrt{3}}}\)
Câu hỏi của Phạm Thị Thu Trang - Toán lớp 9 - Học toán với OnlineMath
\(x^4-16x^2+32=0\Leftrightarrow x^2=8+4\sqrt{2}\text{ hoặc }x^2=8-4\sqrt{2}\)
\(a=\sqrt{2+\sqrt{\frac{4+2\sqrt{3}}{2}}}-\sqrt{6-3\sqrt{\frac{4+2\sqrt{3}}{2}}}\)\(=\sqrt{2+\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}}}-\sqrt{6-3\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}}}\)
\(=\sqrt{2+\frac{\sqrt{3}+1}{\sqrt{2}}}-\sqrt{6-3\frac{\sqrt{3}+1}{\sqrt{2}}}=\sqrt{\frac{4+\sqrt{6}+\sqrt{2}}{2}}-\sqrt{3}\sqrt{\frac{4-\sqrt{6}-\sqrt{2}}{2}}\)
\(a^2=\frac{4+\sqrt{6}+\sqrt{2}}{2}+3.\frac{4-\sqrt{6}-\sqrt{2}}{2}-2\sqrt{3}\sqrt{\frac{\left(4+\sqrt{6}+\sqrt{2}\right)\left(4-\sqrt{6}-\sqrt{2}\right)}{2.2}}\)
\(=8-\left(\sqrt{6}+\sqrt{2}\right)-2\sqrt{3}.\frac{1}{2}.\sqrt{4^2-\left(\sqrt{6}+\sqrt{2}\right)^2}\)
\(=8-\sqrt{6}-\sqrt{2}-\sqrt{3}\sqrt{8-4\sqrt{3}}\)
\(=8-\sqrt{2}-\sqrt{6}-\sqrt{\left(3\sqrt{2}-\sqrt{6}\right)^2}\)
\(=8-\sqrt{2}-\sqrt{6}-\left(3\sqrt{2}-\sqrt{6}\right)\)
\(=8-4\sqrt{2}\)
\(\Rightarrow a\text{ là nghiệm phương trình }x^4-16x^2+32=0\)
\(x^2=2+\sqrt{2+\sqrt{3}}+6-3\sqrt{2+\sqrt{3}}-2.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{6-3\sqrt{2+\sqrt{3}}}\)
\(x^2=8-2\sqrt{2+\sqrt{3}}-2.\sqrt{3.\left(2+\sqrt{2+\sqrt{3}}\right).\left(2-\sqrt{2+\sqrt{3}}\right)}\)
\(x^2=8-2\sqrt{2+\sqrt{3}}-2.\sqrt{3.\left(4-\left(2+\sqrt{3}\right)\right)}=8-2\sqrt{2+\sqrt{3}}-2.\sqrt{3.\left(2-\sqrt{3}\right)}\)
\(x^2=8-\sqrt{2}\sqrt{4+2.\sqrt{3}}-\sqrt{6}.\sqrt{4-2.\sqrt{3}}=8-\sqrt{2}.\sqrt{\left(1+\sqrt{3}\right)^2}-\sqrt{6}.\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(x^2=8-\sqrt{2}.\left(1+\sqrt{3}\right)-\sqrt{6}.\left(\sqrt{3}-1\right)=8-\sqrt{2}-\sqrt{6}-3\sqrt{2}+\sqrt{6}=8-4\sqrt{2}\)
=> \(x^4=\left(x^2\right)^2=\left(8-4\sqrt{2}\right)^2=\left(4\sqrt{2}\right)^2.\left(\sqrt{2}-1\right)^2=32.\left(2-2\sqrt{2}+1\right)=96-64\sqrt{2}\)
=> \(x^4-16x^2+32=96-64\sqrt{2}-16.\left(8-4\sqrt{2}\right)+32=\left(96-96\right)-64\sqrt{2}+64\sqrt{2}=0\)
=> đpcm