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a,\(P=\dfrac{\sqrt{x}-1-2}{\sqrt{x}-1}=1-\dfrac{2}{\sqrt{x-1}}\)
P<\(\dfrac{1}{2}\)\(\Leftrightarrow1-\dfrac{2}{\sqrt{x}-1}< \dfrac{1}{2}
\)
\(\Leftrightarrow\dfrac{1}{2}< \dfrac{2}{\sqrt{x}-1}\)\(\Leftrightarrow\dfrac{2}{4}< \dfrac{2}{\sqrt{x}-1}\)
\(\Rightarrow4>\sqrt{x}-1
\Leftrightarrow5>\sqrt{x}\)
\(\Leftrightarrow25>x\)
b, x=\(\sqrt{4+2.2.\sqrt{3}+3}+\sqrt{4-2.2.\sqrt{3}+3}\)
= \(\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
= \(|2+\sqrt{3}|+|2-\sqrt{3}|\)
= \(2+\sqrt{3}+2-\sqrt{3}=4\)
suy ra P=\(\dfrac{\sqrt{4}-3}{\sqrt{4}-1}=\dfrac{-1}{1}=-1\)
Bài 2: a) Ta có: Q=\(\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\) -\(\left(\dfrac{x+2}{\left(\sqrt{x}\right)^3-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\right)\) =\(\dfrac{1}{\sqrt{x}-1}\) -\(\left(\dfrac{x+2+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\) =\(\dfrac{1}{\sqrt{x}-1}-\left(\dfrac{x+2+x-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\) =\(\dfrac{1}{\sqrt{x}-1}-\dfrac{2x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\) =
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) Điều kiện cần có: \(\left\{{}\begin{matrix}x>0\\\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{1-\sqrt{x}}{x+\sqrt{x}}\ne0\end{matrix}\right.\)
Ta có:\(\left\{{}\begin{matrix}\sqrt{x}-\dfrac{1}{\sqrt{x}}=\dfrac{x-1}{\sqrt{x}}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\\\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{1-\sqrt{x}}{x+\sqrt{x}}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}
=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\end{matrix}\right.\)Từ đó thế vào bài cho ta:
\(P=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}:\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
b) Ta có: \(x=\dfrac{2}{2+\sqrt{3}}=\dfrac{2\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=4-2\sqrt{3}\)
\(\Leftrightarrow\sqrt{x}=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
\(\Leftrightarrow\dfrac{1}{\sqrt{x}}=\dfrac{1}{\sqrt{3}-1}=\dfrac{\sqrt{3}+1}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}=\dfrac{\sqrt{3}+1}{2}\)
Đưa vào bài, ta có: \(P=\left(\sqrt{3}-1+1\right)^2.\dfrac{\sqrt{3}+1}{2}=\dfrac{3\sqrt{3}+3}{2}\)
Bài 2:
a: \(A=\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{4}-\dfrac{3\sqrt{5}\left(3-\sqrt{5}\right)}{4}\)
\(=-5+3\sqrt{5}+\dfrac{5+\sqrt{5}-9\sqrt{5}+15}{4}\)
\(=-5+3\sqrt{5}+5-2\sqrt{5}=\sqrt{5}\)
b: \(B=\left(\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\right):\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+3\sqrt{x}+6-2\sqrt{x}-6}=1\)
điều kiện xác định : \(x\ge0;x\ne1\)
a) ta có : \(A=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
\(\Leftrightarrow A=\left(\dfrac{2\sqrt{x}+x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}-1}\right)\left(\dfrac{x+\sqrt{x}+1}{\sqrt{x}+2}\right)\) \(\Leftrightarrow A=\left(\dfrac{2\sqrt{x}+x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\left(\dfrac{x+\sqrt{x}+1}{\sqrt{x}+2}\right)\) \(\Leftrightarrow A=\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\left(\dfrac{x+\sqrt{x}+1}{\sqrt{x}+2}\right)=\dfrac{1}{\sqrt{x}+2}\)b) ta có : khi \(x=4+2\sqrt{3}\Rightarrow\sqrt{x}=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
\(\Rightarrow\sqrt{A}=\sqrt{\dfrac{1}{\sqrt{3}+1+2}}=\sqrt{\dfrac{1}{3+\sqrt{3}}}\)
a: \(A=\dfrac{3\sqrt{x}-3-\sqrt{x}-1-\sqrt{x}+3}{x-1}\)
\(=\dfrac{\sqrt{x}-1}{x-1}=\dfrac{1}{\sqrt{x}+1}\)
b: Thay \(x=3-2\sqrt{2}\) vào A,ta được:
\(A=\dfrac{1}{\sqrt{2}-1+1}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\)
a: \(A=\dfrac{\sqrt{x-1}+\sqrt{x}}{x-1-x}+\dfrac{\sqrt{x-1}-\sqrt{x}}{x-1-x}+\dfrac{x\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(=-\sqrt{x}-\sqrt{x-1}-\sqrt{x-1}+\sqrt{x}+x=x-2\sqrt{x-1}\)
b: Khi \(x=7-2\sqrt{3}\) thì \(A=7-2\sqrt{3}-2\sqrt{6-2\sqrt{3}}\)
\(\simeq0,351\)