\(1:\left(\dfrac{x+2\sqrt{x}-2}{x\sqrt{x}+1}-\dfrac{\sqrt{x}-1}{x-\sqrt{x}+1}+\dfrac{1}{\sqrt{x}+1}\right)\)rút gọn bài này giùm mik vs mn ơi
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KV
3 tháng 6 2023
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 \(P=\dfrac{4\sqrt{x}-1}{\sqrt{x}}\)
2 tháng 6 2023
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
12 tháng 9 2021
\(a,M=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{2}{x}-\dfrac{2-x}{x\sqrt{x}+x}\right)\left(x>0;x\ne1\right)\\ M=\dfrac{x+\sqrt{x}+x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{2\sqrt{x}+2-2+x}{x\left(\sqrt{x}+1\right)}\\ M=\dfrac{2x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{x\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\\ M=\dfrac{2x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(b,M=-\dfrac{1}{2}\Leftrightarrow\dfrac{2x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=-\dfrac{1}{2}\\ \Leftrightarrow-4x=x+\sqrt{x}-2\\ \Leftrightarrow5x+\sqrt{x}-2=0\)
Đặt \(\sqrt{x}=t\)
\(\Leftrightarrow5t^2+t-2=0\\ \Delta=1^2-4\cdot5\left(-2\right)=41\\ \Leftrightarrow\left[{}\begin{matrix}t=\dfrac{-1-\sqrt{41}}{10}\\t=\dfrac{-1+\sqrt{41}}{10}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\left(1+\sqrt{41}\right)^2}{100}=\dfrac{-42-2\sqrt{41}}{100}\\x=\dfrac{\left(\sqrt{41}-1\right)^2}{100}=\dfrac{42-2\sqrt{41}}{100}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-21-\sqrt{41}}{50}\left(L\right)\\x=\dfrac{21-\sqrt{41}}{50}\left(N\right)\end{matrix}\right.\\ \Leftrightarrow x=\dfrac{21-\sqrt{41}}{50}\)
12 tháng 9 2021
a: Ta có: \(M=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{2}{x}+\dfrac{x-2}{x\sqrt{x}+x}\right)\)
\(=\dfrac{x+\sqrt{x}+x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{2\sqrt{x}+2+x-2}{x\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2x}{\sqrt{x}-1}\cdot\dfrac{x}{\sqrt{x}\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2x\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
2 tháng 8 2018
\(\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2=x-\sqrt{x}+1\)
11 tháng 8 2021
nãy đăng ảnh nhưng không hiện, lại phải mất công đánh lại :Đ
11 tháng 8 2021
a: Ta có: \(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)
\(=\dfrac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x-\sqrt{x}-\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\cdot\dfrac{2}{x+\sqrt{x}+1}\)
\(=\dfrac{2}{x+\sqrt{x}+1}\)
b: Ta có: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right)\cdot\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{2}{x-1}\right)\)
\(=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
10 tháng 11 2021
\(ĐK:x\ge0;x\ne1;x\ne4\\ P=\dfrac{\sqrt{x}+1-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\\ P=\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{x-1-x+4}\\ P=\dfrac{2\left(\sqrt{x}-2\right)}{3\left(\sqrt{x}+1\right)}=\dfrac{2\sqrt{x}-4}{3\sqrt{x}+3}\)
27 tháng 8 2021
a) \(P=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right).\left(\dfrac{1}{2\sqrt{x}}-\dfrac{\sqrt{x}}{2}\right)^2\left(đk:x>0\right)\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2-\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\left(\dfrac{1-x}{2\sqrt{x}}\right)^2=\dfrac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{x-1}.\dfrac{\left(x-1\right)^2}{4x}=\dfrac{-4\sqrt{x}\left(x-1\right)}{4x}=\dfrac{1-x}{\sqrt{x}}\)
b) \(P-\left(-2\sqrt{x}\right)=\dfrac{1-x}{\sqrt{x}}+2\sqrt{x}=\dfrac{1-x+2x}{\sqrt{x}}=\dfrac{1+x}{\sqrt{x}}>0\)
\(\Rightarrow P>-2\sqrt{x}\)
HP
27 tháng 8 2021
a, ĐK: \(x\ge0;x\ne1\)
\(P=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right)\left(\dfrac{1}{2\sqrt{x}}-\dfrac{\sqrt{x}}{2}\right)^2\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2-\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\dfrac{\left(2-2x\right)^2}{16x}\)
\(=\dfrac{-4\sqrt{x}}{x-1}.\dfrac{4\left(x-1\right)^2}{16x}\)
\(=-\dfrac{x-1}{\sqrt{x}}\)
\(1:\left(\dfrac{x+2\sqrt{x}-2}{x\sqrt{x}+1}-\dfrac{\sqrt{x}-1}{x-\sqrt{x}+1}+\dfrac{1}{\sqrt{x}+1}\right)=1:\dfrac{x+2\sqrt{x}-2-x+1+x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)