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a) Tại x=16 thì A = \(\dfrac{\sqrt{16}-1}{\sqrt{16}+2}=\dfrac{4-1}{4+2}=\dfrac{1}{2}\)
b) B = \(\dfrac{\sqrt{x}+1+\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\div\dfrac{\sqrt{x}}{x+\sqrt{x}}\)
= \(\dfrac{\sqrt{x}+1+x-\sqrt{x}}{x+\sqrt{x}}\times\dfrac{x+\sqrt{x}}{\sqrt{x}}\)
= \(\dfrac{x+1}{\sqrt{x}}\)
B = \(\dfrac{x+1}{\sqrt{x}}\)= 2
⇒ x + 1 = 2\(\sqrt{x}\)
⇒ x - \(2\sqrt{x}\) +1 = 0
⇒ \(\left(\sqrt{x}-1\right)^2\) = 0
⇒ \(\sqrt{x}-1=0\)
⇒ x = 1
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
a) ĐKXĐ: \(x>0,x\ne1\)
\(M=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right).\dfrac{1}{\sqrt{x}}=\dfrac{\left(\sqrt{x}-1\right)^2-\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{1}{\sqrt{x}}\)
\(=\dfrac{-4\sqrt{x}}{x-1}.\dfrac{1}{\sqrt{x}}=-\dfrac{4}{x-1}\)
b) \(M=\dfrac{x}{-3}\Rightarrow\dfrac{-4}{x-1}=\dfrac{x}{-3}\Rightarrow x^2-x=12\Rightarrow x^2-x-12=0\)
\(\Rightarrow\left(x-4\right)\left(x+3\right)=0\) mà \(x>0\Rightarrow x=4\) (thỏa)
a) \(M=\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{6\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\left(x\ge0,x\ne1\right)\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)-6\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{x-4\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\)
b) \(M=\dfrac{\sqrt{x}-3}{\sqrt{x}+2}=1-\dfrac{5}{\sqrt{x}+2}\in Z\)
\(\Rightarrow\sqrt{x}+2\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)
Do \(\sqrt{x}\ge0\forall x\)
\(\Rightarrow\sqrt{x}\in\left\{3\right\}\Rightarrow x=9\left(tm\right)\)
a: \(A=\dfrac{-\left(\sqrt{x}-2\right)}{\sqrt{x}-1}:\dfrac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-\left(\sqrt{x}-2\right)^2}{3}\)
Đề bạn gõ sai, mình có sửa lại r nha
\(a,A=\dfrac{1-\sqrt{x}+1}{\sqrt{x}-1}:\dfrac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{3}\\ x=5\Leftrightarrow A=\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{3}=\dfrac{5-2\sqrt{5}}{3}\\ c,A=-\dfrac{1}{3}\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)=-1\Leftrightarrow x-2\sqrt{x}+1=0\\ \Leftrightarrow\left(\sqrt{x}-1\right)^2=0\Leftrightarrow x=1\left(ktm\right)\Leftrightarrow x\in\varnothing\)
ĐKXĐ: \(x\ge0;x\ne1\)
\(M=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{\left(x-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\left(\dfrac{x-1-2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(x-1\right)}\right):\left(\dfrac{\sqrt{x}+1-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)^2}.\left(\sqrt{x}+1\right)=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
b.
\(M=\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\ge1-\dfrac{2}{0+1}=-1\)
\(M_{min}=-1\) khi \(x=0\)
3:
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >9\end{matrix}\right.\)
\(M=\left(\dfrac{1}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x}+3}\right):\dfrac{3}{\sqrt{x}-3}\)
\(=\dfrac{\sqrt{x}+3-\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{3}\)
\(=\dfrac{6}{3\left(\sqrt{x}+3\right)}=\dfrac{2}{\sqrt{x}+3}\)
b: M>1/3
=>M-1/3>0
=>\(\dfrac{2}{\sqrt{x}+3}-\dfrac{1}{3}>0\)
=>\(\dfrac{6-\sqrt{x}-3}{3\left(\sqrt{x}+3\right)}>0\)
=>\(3-\sqrt{x}>0\)
=>\(\sqrt{x}< 3\)
=>0<=x<9
c: \(\sqrt{x}+3>=3\) với mọi x thỏa mãn ĐKXĐ
=>\(M=\dfrac{2}{\sqrt{x}+3}< =\dfrac{2}{3}\) với mọi x thỏa mãn ĐKXĐ
Dấu = xảy ra khi x=0
`P=(sqrtx/(sqrtx-1)+sqrtx/(x-1)):(2/x-(2-x)/(xsqrtx+x))`
`đk:x>0,x ne 1`
`P=((x+sqrtx+sqrtx)/(x-1)):(2/x+(x-2)/(x(sqrtx+1)))`
`=(x+2sqrtx)/(x-1):((2sqrtx+2+x-2)/(x(sqrtx+1)))`
`=(x+2sqrtx)/(x-1):(x+2sqrtx)/(x(sqrtx+1))`
`=(x+2sqrtx)/(x-1)*(x(sqrtx+1))/(x+2sqrtx)`
`=(x(sqrtx+1))/((sqrtx-1)(sqrtx+1))`
`=x/(sqrtx-1)`
`b)P>2`
`<=>x/(sqrtx-1)-2>0`
`<=>(x-2sqrtx+2)/(sqrtx-1)>0`
`<=>((sqrtx-1)^2+1)/(sqrtx-1)>0`
`<=>sqrtx-1>0`
`<=>x>1`
a) đk: x>0;x khác 1;0
P = \(\left[\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]:\left[\dfrac{2}{x}-\dfrac{2-x}{x\left(\sqrt{x}+1\right)}\right]\)
= \(\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\dfrac{2\left(\sqrt{x}+1\right)-2+x}{x\left(\sqrt{x}+1\right)}\)
= \(\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\dfrac{x+2\sqrt{x}}{x\left(\sqrt{x}+1\right)}\)
= \(\dfrac{x}{\sqrt{x}-1}\)
b) Để P > 2
<=> \(\dfrac{x}{\sqrt{x}-1}-2>0\)
<=> \(\dfrac{x-2\sqrt{x}+2}{\sqrt{x}-1}>0\)
<=> \(\dfrac{\left(\sqrt{x}-1\right)^2+1}{\sqrt{x}-1}>0\)
<=> \(\sqrt{x}-1>0\)
<=> x > 1
\(a,M=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{2}{x}-\dfrac{2-x}{x\sqrt{x}+x}\right)\left(x>0;x\ne1\right)\\ M=\dfrac{x+\sqrt{x}+x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{2\sqrt{x}+2-2+x}{x\left(\sqrt{x}+1\right)}\\ M=\dfrac{2x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{x\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\\ M=\dfrac{2x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(b,M=-\dfrac{1}{2}\Leftrightarrow\dfrac{2x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=-\dfrac{1}{2}\\ \Leftrightarrow-4x=x+\sqrt{x}-2\\ \Leftrightarrow5x+\sqrt{x}-2=0\)
Đặt \(\sqrt{x}=t\)
\(\Leftrightarrow5t^2+t-2=0\\ \Delta=1^2-4\cdot5\left(-2\right)=41\\ \Leftrightarrow\left[{}\begin{matrix}t=\dfrac{-1-\sqrt{41}}{10}\\t=\dfrac{-1+\sqrt{41}}{10}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\left(1+\sqrt{41}\right)^2}{100}=\dfrac{-42-2\sqrt{41}}{100}\\x=\dfrac{\left(\sqrt{41}-1\right)^2}{100}=\dfrac{42-2\sqrt{41}}{100}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-21-\sqrt{41}}{50}\left(L\right)\\x=\dfrac{21-\sqrt{41}}{50}\left(N\right)\end{matrix}\right.\\ \Leftrightarrow x=\dfrac{21-\sqrt{41}}{50}\)
a: Ta có: \(M=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{2}{x}+\dfrac{x-2}{x\sqrt{x}+x}\right)\)
\(=\dfrac{x+\sqrt{x}+x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{2\sqrt{x}+2+x-2}{x\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2x}{\sqrt{x}-1}\cdot\dfrac{x}{\sqrt{x}\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2x\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)