R/gọn và tìm điều kiện xác định:
\(M=\left(\dfrac{2}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x+1}}\right).\dfrac{\sqrt{x}}{x+\sqrt{x}+2}\)
K
Khách
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Những câu hỏi liên quan
10 tháng 12 2020
ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\notin\left\{1;0\right\}\end{matrix}\right.\)
Sửa đề: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{1+\sqrt{x}}+\dfrac{2}{x-1}\right)\)
Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{1+\sqrt{x}}+\dfrac{2}{x-1}\right)\)
\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}-1}{1}\)
\(=\dfrac{x-1}{\sqrt{x}}\)
12 tháng 7 2021
a)ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
Ta có: \(D=\dfrac{\sqrt{x}-2}{\sqrt{x}+3}-\dfrac{5}{x+\sqrt{x}-6}+\dfrac{1}{2-\sqrt{x}}\)
\(=\dfrac{x-4\sqrt{x}+4-5-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-5\sqrt{x}-4}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
12 tháng 8 2021
a) ĐKXĐ: \(x\ge0;x\ne1\)
b) \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\dfrac{2}{\sqrt{x}+1}\left(x\ge0;x\ne1\right)\\ P=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{2}\\ P=\dfrac{x-\sqrt{x}-x-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{2}\\ P=\dfrac{-2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{2}\\ P=\dfrac{-\sqrt{x}}{\sqrt{x}-1}\)
2 tháng 9 2021
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-1}{\sqrt{x}}\)
b: Thay \(x=3+2\sqrt{2}\) vào P, ta được:
\(P=\dfrac{2\sqrt{2}+2}{\sqrt{2}+1}=2\)
DQ
31 tháng 5 2017
ĐKXĐ: \(x\ge0\)
\(K=\left(1+\dfrac{\sqrt{x}}{x+1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right)-1\)
\(K=\left[\dfrac{\left(\sqrt{x}-1\right)\left(x+1\right)}{\left(\sqrt{x}-1\right)\left(x+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+1\right)}\right]:\left[\dfrac{x+1}{\left(\sqrt{x}-1\right)\left(x+1\right)}-\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1\right)}\right]-1\)
\(K=\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+1\right)}:\dfrac{x+1-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1\right)}-1\)
\(K=\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x-2\sqrt{x}+1}-1\\K=\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}-1\\ K=\dfrac{x+\sqrt{x}+1}{\sqrt{x}-1}-1 \)
\(K=\dfrac{x+\sqrt{x}+1-\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{x+2}{\sqrt{x}-1}\)
6 tháng 2 2022
ĐKXĐ: \(x\ge0;x\ne9\)
\(B=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{3}{x-9}\right):\dfrac{1}{\sqrt{x}-3}\)
\(=\left[\dfrac{\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]:\dfrac{1}{\sqrt{x}-3}\)
\(=\dfrac{\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}-3}{1}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+3}\)
6 tháng 2 2022
ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >9\end{matrix}\right.\)
\(B=\dfrac{\sqrt{x}-3+3}{x-9}\cdot\left(\sqrt{x}-3\right)=\dfrac{\sqrt{x}}{\sqrt{x}+3}\)
10 tháng 12 2023
ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >1\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
26 tháng 9 2023
Lời giải:
a. ĐKXĐ: $x>0; x\neq 1$
b. \(P=\left[\frac{x}{\sqrt{x}(\sqrt{x}-1)}-\frac{1}{\sqrt{x}(\sqrt{x}-1)}\right]: \left[\frac{\sqrt{x}-1}{(\sqrt{x}-1)(\sqrt{x}+1)}+\frac{2}{(\sqrt{x}-1)(\sqrt{x}+1)}\right]\)
\(=\frac{x-1}{\sqrt{x}(\sqrt{x}-1)}:\frac{\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+1)}=\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)}:\frac{\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+1)} =\frac{\sqrt{x}+1}{\sqrt{x}}:\frac{1}{\sqrt{x}-1}=\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}}=\frac{x-1}{\sqrt{x}}\)
c.
$P<0\Leftrightarrow \frac{x-1}{\sqrt{x}}<0$
$\Leftrightarrow x-1<0$
$\Leftrightarrow x<1$. Kết hợp đkxđ suy ra $0< x<1 $
sửa đề:
\(M=\left(\dfrac{2}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right).\dfrac{\sqrt{x}}{x+\sqrt{x}+2}\)
ĐKXĐ: \(x\ge0\);\(x\ne\pm1\)
\(M=\left(\dfrac{2}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right).\dfrac{\sqrt{x}}{x+\sqrt{x}+2}\\M=\left[\dfrac{2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right].\dfrac{\sqrt{x}}{x+\sqrt{x}+2} \) \(M=\dfrac{x+\sqrt{x}+2}{x-1}.\dfrac{\sqrt{x}}{x+\sqrt{x}+2}=\dfrac{\sqrt{x}}{x-1}\)