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a) đk: \(x\ne0;4\); \(x>0\)
P = \(\left[\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{1}{\sqrt{x}-2}\right]\times\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)\)
= \(\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\times\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)\)
= \(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)}.\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
b) Để P < \(\dfrac{1}{2}\)
<=> \(\dfrac{\sqrt{x}-1}{\sqrt{x}}< \dfrac{1}{2}\)
<=> \(1-\dfrac{1}{\sqrt{x}}< \dfrac{1}{2}\)
<=> \(\dfrac{1}{\sqrt{x}}>\dfrac{1}{2}\)
<=> \(\sqrt{x}< 2\)
<=> x < 4
<=> 0 < x < 4
a: \(B=\dfrac{x+4\sqrt{x}+4-\left(x-4\sqrt{x}+4\right)+4x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}\left(4-x\right)}{\left(3-3\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)
\(=\dfrac{5x+4\sqrt{x}+4-x+4\sqrt{x}-4}{1}\cdot\dfrac{-\sqrt{x}}{\left(3-3\sqrt{x}\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{4x+8\sqrt{x}}{\sqrt{x}+2}\cdot\dfrac{\sqrt{x}}{3\left(\sqrt{x}-1\right)}\)
\(=\dfrac{4\sqrt{x}}{3\sqrt{x}-3}\)
b: Để B>0 thì \(3\sqrt{x}-3>0\)
hay x>1
c: Để B là số nguyên thì \(4\sqrt{x}⋮3\sqrt{x}-3\)
\(\Leftrightarrow12\sqrt{x}-12+12⋮3\sqrt{x}-3\)
\(\Leftrightarrow3\sqrt{x}-3\in\left\{1;-1;2;-2;3;-3;4;6;12\right\}\)
hay \(x\in\left\{\dfrac{16}{9};\dfrac{4}{9};\dfrac{25}{9};\dfrac{1}{9};\dfrac{49}{9};9;25\right\}\)
a, Với \(x>0;x\ne4;x\ne9\)
\(A=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{4-x}\right):\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)
\(=\left(\frac{4\sqrt{x}\left(2-\sqrt{x}\right)+8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(=\left(\frac{8\sqrt{x}-4x+8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{8\sqrt{x}+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}:\frac{-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{4\sqrt{x}\left(2+\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\frac{3-\sqrt{x}}{\sqrt{x}\left(2-\sqrt{x}\right)}=\frac{4\sqrt{x}}{2-\sqrt{x}}.\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{3-\sqrt{x}}=\frac{4x}{3-\sqrt{x}}\)
b, Ta có : A = -2 hay
\(\frac{4x}{3-\sqrt{x}}=-2\Rightarrow4x=-6+2\sqrt{x}\)
\(\Leftrightarrow4x+6-2\sqrt{x}=0\Leftrightarrow2\left(2x+3-\sqrt{x}\right)=0\)
\(\Leftrightarrow2x+3-\sqrt{x}=0\Leftrightarrow\sqrt{x}=2x+3\)
bình phương 2 vế ta có :
\(x=\left(2x+3\right)^2=4x^2+12x+9\)
\(\Leftrightarrow-4x^2-11x-9=0\)giải delta ta thu được : \(x=-\frac{11\pm\sqrt{23}i}{8}\)
\(a,A=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{4-x}\right):\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)
\(=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{2}{\sqrt{x}}\right)\)
\(=\frac{4\sqrt{x}.\left(2-\sqrt{x}\right)+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\frac{\sqrt{x}-1-2.\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{8\sqrt{x}-4x+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-1-2\sqrt{x}+4}\)
\(=\frac{\left(4x+8\sqrt{x}\right)\left(\sqrt{x}\right)\left(\sqrt{x}-2\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)\left(-\sqrt{x}+3\right)}\)
\(=\frac{-4\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}\right)\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)\left(-\sqrt{x}+3\right)}\)
\(=\frac{4x}{\sqrt{x}-3}\)
a) \(P=\left(3-\dfrac{3}{\sqrt{x}-1}\right):\left(\dfrac{x+2}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\)
\(=\left(\dfrac{3\left(\sqrt{x}-1\right)-3}{\sqrt{x}-1}\right):\left[\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x+2}\right)}-\dfrac{\sqrt{x}}{\sqrt{x}+2}\right]\)
\(=\dfrac{3\sqrt{x}-3-3}{\sqrt{x}-1}:\dfrac{x+2-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3\sqrt{x}-6}{\sqrt{x}-1}:\dfrac{x+2-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3\sqrt{x}-6}{\sqrt{x}-1}:\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3\sqrt{x}-6}{\sqrt{x}-1}:\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{3\sqrt{x}-6}{\sqrt{x}-1}.\left(\sqrt{x}-1\right)\)
\(=3\sqrt{x}-6\)
b) \(P=\dfrac{4\sqrt{x}-1}{\sqrt{x}}\)
\(\Leftrightarrow3\sqrt{x}-6=\dfrac{4\sqrt{x}-1}{\sqrt{x}}\) (1)
ĐKXĐ: \(x>0\)
\(\left(1\right)\Leftrightarrow3x-6\sqrt{x}=4\sqrt{x}-1\)
\(\Leftrightarrow3x-6\sqrt{x}-4\sqrt{x}+1=0\)
\(\Leftrightarrow3x-10\sqrt{x}+1=0\) (2)
Đặt \(t=\sqrt{x}\ge0\)
\(\left(2\right)\Leftrightarrow3t^2-10t+1=0\)
\(\Delta'=25-4=22\)
Phương trình có hai nghiệm phân biệt:
\(t_1=\dfrac{5+\sqrt{22}}{3}\) (nhận)
\(t_2=\dfrac{5-\sqrt{22}}{3}\) (nhận)
Với \(t=\dfrac{5+\sqrt{22}}{3}\) \(\Leftrightarrow\sqrt{x}=\dfrac{5+\sqrt{22}}{3}\Leftrightarrow x=\dfrac{47+10\sqrt{22}}{9}\) (nhận)
Với \(t=\dfrac{5-\sqrt{22}}{3}\Leftrightarrow\sqrt{x}=\dfrac{5-\sqrt{22}}{3}\Leftrightarrow x=\dfrac{47-10\sqrt{22}}{9}\) (nhận)
Vậy \(x=\dfrac{47+10\sqrt{22}}{9};x=\dfrac{47-10\sqrt{22}}{9}\) thì \(P=\dfrac{4\sqrt{x}-1}{\sqrt{x}}\)
a: \(P=\dfrac{3\sqrt{x}-3-3}{\sqrt{x}-1}:\dfrac{x+2-x+\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{3\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}=3\sqrt{x}-6\)
b: P=(4căn x-1)/căn x
=>3x-6căn x-4căn x+1=0
=>3x-10căn x+1=0
=>x=(47+10căn 22)/9 hoặc x=(47-10căn 22)/9
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: \(B=\left(\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\right):\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+3\sqrt{x}+6-2\sqrt{x}-6}=1\)
a) \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right).\dfrac{x-4}{\sqrt{4x}}=\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{x-4}.\dfrac{x-4}{2\sqrt{x}}=\dfrac{2x}{2\sqrt{x}}=\sqrt{x}\)
b) \(P=\sqrt{x}>3\Leftrightarrow x>9\)
a: Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\cdot\dfrac{x-4}{2\sqrt{x}}\)
\(=\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{2\sqrt{x}}\)
\(=\sqrt{x}\)
b: Để P>3 thì x>9
\(B=\left(\dfrac{x}{x-4}+\dfrac{2}{2-\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\right)\\ ĐKXĐ:x\ge0;x\ne4\\ =\left(\dfrac{x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right):\left(\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}+\dfrac{10-x}{\sqrt{x}+2}\right)\\ =\dfrac{x-2\sqrt{x}-4+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{x-4+10-x}{\sqrt{x}+2}\\ =\dfrac{x-\sqrt{x}-6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{6}{\sqrt{x}+2}\\ =\dfrac{x-3\sqrt{x}+2\sqrt{x}-6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}+2}{6}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)+2\left(\sqrt{x}-3\right)}{6\left(\sqrt{x}-2\right)}\\ =\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}{6\left(\sqrt{x}-2\right)}\)
b) Với \(x\ge0;x\ne4\)
\(\Rightarrow\) Để B>0
thì \(\Rightarrow\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}{6\left(\sqrt{x}-2\right)}>0\)
\(\Rightarrow\dfrac{\sqrt{x}-3}{\sqrt{x}-2}>0\left(Vì\text{ }\sqrt{x}+2;6>0\right)\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}-3>0\\\sqrt{x}-2>0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}-3< 0\\\sqrt{x}-2< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-3>0\\\sqrt{x}-2< 0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}>3\\\sqrt{x}< 2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>9\\x< 4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ge9\\0< x< 4\end{matrix}\right.\)
b) \(B>0\Leftrightarrow\dfrac{-1}{\sqrt{x}-2}>0,do-1< 0\Rightarrow B>0\Leftrightarrow\sqrt{x}-2< 0\Leftrightarrow\sqrt{x}< 2\Leftrightarrow0\le x< 4\)
\(a)C=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\dfrac{x-4}{\sqrt{4x}}\\ =\left(\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{x-4}+\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{x-4}\right)\dfrac{x-4}{2\sqrt{x}}\\ =\left(\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{x-4}\right)\dfrac{x-4}{2\sqrt{x}}\\ =\dfrac{2x}{x-4}\cdot\dfrac{x-4}{2\sqrt{x}}\\ =\dfrac{2x\left(x-4\right)}{2\sqrt{x}\left(x-4\right)}\\ =\sqrt{x}\)
b) C>3
\(\Rightarrow\sqrt{x}>3\\ \Leftrightarrow x>9\)
a: Sửa đề: \(P=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right):\dfrac{2}{x^2-2x+1}\)
\(=\left(\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right)\cdot\dfrac{\left(x-1\right)^2}{2}\)
\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)^2}\cdot\dfrac{\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}+1\right)^2}{2}\)
\(=\dfrac{x-\sqrt{x}-2-\left(x+\sqrt{x}-2\right)}{\sqrt{x}-1}\cdot\dfrac{1}{2}\)
\(=\dfrac{-\sqrt{x}}{\sqrt{x}-1}\)
b: Để P>0 thì \(-\dfrac{\sqrt{x}}{\sqrt{x}-1}>0\)
=>\(\dfrac{\sqrt{x}}{\sqrt{x}-1}< 0\)
=>\(\sqrt{x}< 1\)
=>\(0< =x< 1\)
c: Thay \(x=7-4\sqrt{3}=\left(2-\sqrt{3}\right)^2\) vào P, ta được:
\(P=\dfrac{-\sqrt{\left(2-\sqrt{3}\right)^2}}{\sqrt{\left(2-\sqrt{3}\right)^2}-1}\)
\(=\dfrac{-\left(2-\sqrt{3}\right)}{2-\sqrt{3}-1}=\dfrac{-2+\sqrt{3}}{1-\sqrt{3}}=\dfrac{2-\sqrt{3}}{\sqrt{3}-1}\)
\(=\dfrac{\sqrt{3}-1}{2}\)
a) P=\(\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right).\dfrac{x-4}{\sqrt{4x}}\) với x > 0 và x≠4
=\(\left(\dfrac{\sqrt{x}.\left(\sqrt{x}+2\right)}{x-4}+\dfrac{\sqrt{x}.\left(\sqrt{x}-2\right)}{x-4}\right).\dfrac{x-4}{2\sqrt{x}}\)
=\(\left(\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{x-4}\right)\).\(\dfrac{x-4}{2\sqrt{x}}\)
=\(\dfrac{2x}{x-4}.\dfrac{x-4}{2\sqrt{x}}\)
=\(\dfrac{x}{\sqrt{x}}\)
b) \(\dfrac{x}{\sqrt{x}}\) >3
<=> x> \(3\sqrt{x}\)
<=> x>9
a: \(=\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{x-4}\cdot\dfrac{x-4}{2\sqrt{x}}\)
\(=\dfrac{2x}{2\sqrt{x}}=\sqrt{x}\)
b: Để P>3 thì \(\sqrt{x}>3\)
hay x>9