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Lời giải:
Bình phương biểu thức $a$ ta có:
\(a^2=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\sqrt{4^2-(10+2\sqrt{5})}\)
\(=8+2\sqrt{6-2\sqrt{5}}=8+2\sqrt{5+1-2\sqrt{5}}\)
\(=8+2\sqrt{(\sqrt{5}-1)^2}=8+2(\sqrt{5}-1)=6+2\sqrt{5}\)
\(=[\pm (\sqrt{5}+1)]^2\)
Mà $a>0$ nên \(a=\sqrt{5}+1\)
Xét thêm 1 số \(1-\sqrt{5}\)
Ta thấy \(\left\{\begin{matrix} \sqrt{5}+1+1-\sqrt{5}=2\\ (\sqrt{5}+1)(1-\sqrt{5})=-4\end{matrix}\right.\) Do đó, theo định lý Viete đảo thì $a$ là nghiệm của pt \(x^2-2x-4=0\), tức là $a^2-2a-4=0$
Do đó:
\(T=\frac{a^2(a^2-2a-4)-2a(a^2-2a-4)+a^2-2a-4+8}{a^2-2a-4-10a+16}\)
\(=\frac{8}{-10a+16}=\frac{8}{-10(\sqrt{5}+1)+16}=\frac{8}{6-10\sqrt{5}}=\frac{4}{3-5\sqrt{5}}\)
Bài 3:
a: \(A=\dfrac{x+5\sqrt{x}-10\sqrt{x}-5\sqrt{x}+25}{x-25}\)
\(=\dfrac{x-10\sqrt{x}+25}{x-25}=\dfrac{\sqrt{x}-5}{\sqrt{x}+5}\)
b: \(B=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{x-9}\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}=\dfrac{3}{\sqrt{x}+3}\)
Đặt \(D=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(\Leftrightarrow D^2=8+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}\)
\(\Leftrightarrow D^2=8+2\sqrt{16-10-2\sqrt{5}}\)
\(\Leftrightarrow D^2=8+2\sqrt{6-2\sqrt{5}}\)
\(\Leftrightarrow D^2=8+2\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(\Leftrightarrow D^2=8+2\left(\sqrt{5}-1\right)\)
\(\Leftrightarrow D^2=6+2\sqrt{5}\)
\(\Leftrightarrow D^2=\left(\sqrt{5}+1\right)^2\)
\(\Rightarrow D=\sqrt{5}+1\)
Thay vào ta tính được: \(A=\sqrt{5}+1-\sqrt{5}=1\)
Vậy A = 1
1/ Rút gọn: \(a)3\sqrt{2a}-\sqrt{18a^3}+4\sqrt{\dfrac{a}{2}}-\dfrac{1}{4}\sqrt{128a}\left(a\ge0\right)=3\sqrt{2a}-3a\sqrt{2a}+2\sqrt{2a}-2\sqrt{2a}=3\sqrt{2a}\left(1-a\right)\)b)\(\dfrac{\sqrt{2}-1}{\sqrt{2}+2}-\dfrac{2}{2+\sqrt{2}}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-1-2}{\sqrt{2}+2}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-3}{\sqrt{2}+2}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-3+2+1+2\sqrt{2}}{\sqrt{2}\left(1+\sqrt{2}\right)}=\dfrac{3\sqrt{2}}{\sqrt{2}\left(1+\sqrt{2}\right)}=\dfrac{3}{1+\sqrt{2}}\)c)\(\dfrac{2+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{2-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{\left(\sqrt{2}+\sqrt{3+\sqrt{5}}\right)\sqrt{2}}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{\sqrt{2}\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right)}=\dfrac{2\sqrt{2}+\sqrt{10}}{2+\sqrt{6+2\sqrt{5}}}+\dfrac{2\sqrt{2}-\sqrt{10}}{2-\sqrt{6-2\sqrt{5}}}=\dfrac{2\sqrt{2}+\sqrt{10}}{2+\sqrt{\left(\sqrt{5}+1\right)^2}}+\dfrac{2\sqrt{2}-\sqrt{10}}{2-\sqrt{\left(\sqrt{5}-1\right)^2}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{2+\sqrt{5}+1}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{2-\sqrt{5}+1}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{3+\sqrt{5}}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{3-\sqrt{5}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)\left(3-\sqrt{5}\right)+\sqrt{2}\left(2-\sqrt{5}\right)\left(3+\sqrt{5}\right)}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\dfrac{\sqrt{2}\left(6-2\sqrt{5}+3\sqrt{5}-5+6+2\sqrt{5}-3\sqrt{5}-5\right)}{9-5}=\dfrac{2\sqrt{2}}{4}=\dfrac{1}{\sqrt{2}}\)
Làm nốt nè :3
\(2.a.P=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}}{x-2\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x}=\dfrac{x-1}{x}\left(x>0;x\ne1\right)\)\(b.P>\dfrac{1}{2}\Leftrightarrow\dfrac{x-1}{x}-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{x-2}{2x}>0\)
\(\Leftrightarrow x-2>0\left(do:x>0\right)\)
\(\Leftrightarrow x>2\)
\(3.a.A=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{a}}\right):\dfrac{\sqrt{a}+1}{a-1}=\dfrac{\sqrt{a}-1}{\sqrt{a}-1}.\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}+1}=\sqrt{a}-1\left(a>0;a\ne1\right)\)
\(b.Để:A< 0\Leftrightarrow\sqrt{a}-1< 0\Leftrightarrow a< 1\)
Kết hợp với DKXĐ : \(0< a< 1\)
\(a^2=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\cdot\sqrt{16-10-2\sqrt{5}}\)
\(=8+2\left(\sqrt{5}-1\right)=6+2\sqrt{5}\)
hay \(a=\sqrt{5}+1\)
\(T=\dfrac{\left(6+2\sqrt{5}\right)^2-4\cdot\left(16+8\sqrt{5}\right)+6+2\sqrt{5}+6\sqrt{5}+6+4}{6+2\sqrt{5}-2\sqrt{5}-2+12}\)
\(=\dfrac{56+24\sqrt{5}-50-24\sqrt{5}}{16}=\dfrac{6}{16}=\dfrac{3}{8}\)