\(\dfrac{3\sqrt{x}+1}{x+2\sqrt{x}-3}-\dfrac{2}{\sqrt{x}+3}\)
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Những câu hỏi liên quan
7 tháng 2 2021
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
Ta có: \(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}-\left(2x-2\sqrt{x}+3\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-3x+6\sqrt{x}-11-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+5\sqrt{x}-8}{\left(\sqrt{x}+3\right)\cdot\left(\sqrt{x}-1\right)}\)
NT
Nguyễn thành Đạt
CTVHS
15 tháng 8 2023
1) ĐKXĐ của phân thức là : \(\left\{{}\begin{matrix}\sqrt{x}\ge0\\\sqrt{x}-3\ne0\\x-9\ne0\\\sqrt{x}+3\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\ne3\\\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\ne0\\\sqrt{x}\ne-3\left(LĐ\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)
Ta có : \(P=\dfrac{\sqrt{x}}{\sqrt{x}-3}:\left(\dfrac{x+3}{x-9}+\dfrac{1}{\sqrt{x}+3}\right)\)
\(P=\dfrac{\sqrt{x}}{\sqrt{x}-3}:\left(\dfrac{x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right)\)
\(P=\dfrac{\sqrt{x}}{\sqrt{x}-3}:\dfrac{x+3+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(P=\dfrac{\sqrt{x}}{\sqrt{x}-3}:\dfrac{x+\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(P=\dfrac{\sqrt{x}}{\sqrt{x}-3}.\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}.\left(\sqrt{x}+1\right)}\)
\(P=\dfrac{\sqrt{x}+3}{\sqrt{x}+1}\)
2) Với \(x=4-2\sqrt{3}=3-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\)
\(\Rightarrow\sqrt{x}=\sqrt{3}-1\)
Do đó : \(P=\dfrac{\sqrt{3}-1+3}{\sqrt{3}-1+1}\)
\(P=\dfrac{\sqrt{3}+2}{\sqrt{3}}=\dfrac{3+2\sqrt{3}}{3}\)
3) Xét hiệu của : P với 3
\(\dfrac{\sqrt{x}+3}{\sqrt{x}+1}-3\)
\(=\dfrac{-2\sqrt{x}}{\sqrt{x}+1}\)
Ta thấy : \(\sqrt{x}+1\ge1;-2\sqrt{x}\le0\)
\(\Rightarrow\dfrac{-2\sqrt{x}}{\sqrt{x}+1}\le0\)
\(\Rightarrow P\le3\)
Dấu bằng xảy ra : \(\Leftrightarrow x=0\). Thế lại ta thấy ktm nên P<3
23 tháng 9 2021
a) \(ĐK:x\ge0,x\ne1\)
\(=\dfrac{3x+3\sqrt{x}-3-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3x+3\sqrt{x}-3-x+4+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{2x+4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{2\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{2\sqrt{x}}{\sqrt{x}-1}\)
b) \(P=\dfrac{2\sqrt{x}}{\sqrt{x}-1}< 0\Leftrightarrow\sqrt{x}-1< 0\Leftrightarrow\sqrt{x}< 1\)
Kết hợp với đk:
\(\Rightarrow0\le x< 1\)
CT
18 tháng 8 2021
a. \(N=\left(\dfrac{x+2}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right).\dfrac{4\sqrt{x}}{3}\) \(\left(ĐKXĐ:x\ge0\right)\)
\(N=\left(\dfrac{x+2}{x\sqrt{x}+1}-\dfrac{x-\sqrt{x}+1}{x\sqrt{x}+1}\right).\dfrac{4\sqrt{x}}{3}\)
\(\text{}\text{}N=\dfrac{\sqrt{x}+1}{x\sqrt{x}+1}.\dfrac{4\sqrt{x}}{3}\)
\(N=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
b.\(N=\dfrac{8}{9}\Leftrightarrow\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\dfrac{8}{9}\)
\(\Leftrightarrow3\sqrt{x}=2x-2\sqrt{x}+2\)
\(\Leftrightarrow\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=4\end{matrix}\right.\)
c.\(\dfrac{1}{N}>\dfrac{3\sqrt{x}}{4}\Leftrightarrow\dfrac{3\left(x-\sqrt{x}+1\right)}{4\sqrt{x}}>\dfrac{3\sqrt{x}}{4}\)
\(\Leftrightarrow x-\sqrt{x}+1>x\)
\(\Leftrightarrow x< 1\)
18 tháng 8 2021
a: ĐKXĐ: \(x\ge0\)
Ta có: \(N=\left(\dfrac{x+2}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right)\cdot\dfrac{4\sqrt{x}}{3}\)
\(=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\)
\(=\dfrac{4\sqrt{x}}{3x-3\sqrt{x}+3}\)
TH
2
TG
12 tháng 7 2021
a) a ≠ 1; a ≥ 0
\(\dfrac{a-5\sqrt{a}+4}{a-1}=\dfrac{a-\sqrt{a}-4\sqrt{a}+4}{a-1}=\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)-4\left(\sqrt{a}-1\right)}{a-1}=\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-4\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}=\dfrac{\sqrt{a}-4}{\sqrt{a}+1}\)
b) a ≥ 0; \(x\ne\pm\sqrt{3}\)
\(\dfrac{\sqrt{x^2+2\sqrt{3x}+3}}{x^2-3}=\dfrac{x+\sqrt{3}}{\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)}=\dfrac{1}{x-\sqrt{3}}\)
12 tháng 7 2021
1) ĐKXĐ: \(\left\{{}\begin{matrix}a\ge0\\a\ne1\end{matrix}\right.\)
Ta có: \(\dfrac{a-5\sqrt{a}+4}{a-1}\)
\(=\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-4\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(=\dfrac{\sqrt{a}-4}{\sqrt{a}+1}\)
2) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne\sqrt{3}\end{matrix}\right.\)
Ta có: \(\dfrac{\sqrt{x^2+2\sqrt{3x}+3}}{x^2-3}\)
\(=\dfrac{x+\sqrt{3}}{\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)}\)
\(=\dfrac{1}{x-\sqrt{3}}\)
26 tháng 7 2023
a: ĐKXĐ: x>0; x<>4
\(P=\left(2-\sqrt{x}+2\right)\cdot\dfrac{1}{\sqrt{x}-2}=\dfrac{4-\sqrt{x}}{\sqrt{x}-2}\)
b: P=2/3
=>(4-căn x)/(căn x-2)=2/3
=>2căn x-4=12-3căn x
=>5căn x=16
=>x=256/25
c: Khi x=8-2căn 7 thì \(P=\dfrac{4-\sqrt{7}+1}{\sqrt{7}-1-2}=\dfrac{5-\sqrt{7}}{\sqrt{7}-3}=-4-\sqrt{7}\)
AT
14 tháng 7 2021
\(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{x-\sqrt{x}+6}{x+\sqrt{x}-2}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{x-\sqrt{x}-2}{x+\sqrt{x}-2}\right)\left(x\ge0,x\ne1\right)\)
\(=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{x-\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{\sqrt{x}+2+x-\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}:\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+x-\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}:\dfrac{2x-\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{2x-\sqrt{x}-3}=\dfrac{x+8}{2x-\sqrt{x}-3}\)
17 tháng 10 2023
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
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
Ta có: \(\dfrac{3\sqrt{x}+1}{x+2\sqrt{x}-3}-\dfrac{2}{\sqrt{x}+3}\)
\(=\dfrac{3\sqrt{x}+1-2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{3\sqrt{x}+1-2\sqrt{x}+2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{1}{\sqrt{x}-1}\)