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K
Khách
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KL
2
NN
16 tháng 10 2020
\(ĐKXĐ:x>3\)
\(\frac{\sqrt{x-3}}{\sqrt{\sqrt{x}+\sqrt{3}}}.\frac{\sqrt{3}}{\sqrt{\sqrt{x}-\sqrt{3}}}=\frac{\sqrt{x-3}.\sqrt{3}}{\sqrt{\sqrt{x}+\sqrt{3}}.\sqrt{\sqrt{x}-\sqrt{3}}}\)
\(=\frac{\sqrt{x-3}.\sqrt{3}}{\sqrt{\left(\sqrt{x}+\sqrt{3}\right)\left(\sqrt{x}-\sqrt{3}\right)}}=\frac{\sqrt{x-3}.\sqrt{3}}{\sqrt{x-3}}=\sqrt{3}\)
15 tháng 9 2020
\(A=\frac{x\sqrt{x}-3}{x-2\sqrt{x}-3}-\frac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}+\frac{\sqrt{x}+3}{3-\sqrt{x}}\)
\(=\frac{x\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}-\frac{2\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x\sqrt{x}-3-2x+12\sqrt{x}-18-x+4\sqrt{x}+3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x\sqrt{x}-3x+8\sqrt{x}-24}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\frac{\left(\sqrt{x}-3\right)\left(x+8\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}=\frac{x+8}{\sqrt{x}+1}\)
HK
2
14 tháng 7 2019
\(\frac{x-3\sqrt{x}}{\sqrt{x}-3}-\frac{x-4\sqrt{x}+3}{\sqrt{x}+3}=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\sqrt{x}-3}-\frac{x-3\sqrt{x}-\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\sqrt{x}-\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+3}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+3\right)-\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+3}\)
\(=\frac{x+3\sqrt{x}-x+4\sqrt{x}-3}{\sqrt{x}+3}\)
\(=\frac{7\sqrt{x}-3}{\sqrt{x}+3}\)
B
14 tháng 7 2019
\(\frac{x-3\sqrt{x}}{\sqrt{x}-3}-\frac{x+4\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\sqrt{x}-3}-\frac{x+\sqrt{x}+3\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\sqrt{x}-3}-\frac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}+1\right)}{\sqrt{x}+3}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\sqrt{x}-3}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}\)
\(=\sqrt{x}-\sqrt{x}-1=-1\)
21 tháng 5 2016
\(A=\frac{x\sqrt{x}-3}{x-2\sqrt{x}-3}-\frac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}+\frac{\sqrt{x}+3}{3-\sqrt{x}}\left(DK:9\ne x\ge0\right)\)
\(=\frac{x\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}-\frac{2\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x\sqrt{x}-3-2\left(x-6\sqrt{x}+9\right)-\left(x+4\sqrt{x}+3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x\sqrt{x}-3x+8\sqrt{x}-24}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\frac{x\left(\sqrt{x}-3\right)+8\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\frac{\left(x+8\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\frac{x+8}{\left(\sqrt{x}+1\right)}\)
29 tháng 1 2022
\(P=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{-\left(\sqrt{x}-3\right)}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3}{\sqrt{x}+3}:\dfrac{-\left(x-9\right)+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{-x+9+2x-4\sqrt{x}-5}\)
\(=\dfrac{3\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}=\dfrac{3}{\sqrt{x}-2}\)
VL
1
29 tháng 10 2016
\(ĐKXĐ:x\ge0;x\ne9\)
rút gọn : \(M=\left(\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{3}{\sqrt{x}-3}\right).\frac{\sqrt{x}+3}{\sqrt{x}+9}\)
\(M=\left(\frac{\sqrt{x}\left(\sqrt{x}-3\right)+3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right).\frac{\sqrt{x}+3}{\sqrt{x}+9}\)
\(M=\left(\frac{x-3\sqrt{x}+3\sqrt{x}+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right).\frac{\sqrt{x}+3}{\sqrt{x}+9}\)
\(M=\frac{x+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}+3}{\sqrt{x}+9}\)
\(M=\frac{x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+9\right)}\)
\(M=\frac{x+9}{x+6\sqrt{x}-27}\)
4 tháng 8 2017
ĐK \(x\ne\left\{2;3\right\}\)
Ta có \(A=\frac{\sqrt{x}+2}{\sqrt{x}-3}-\frac{\sqrt{x}+1}{\sqrt{x}-2}-3.\frac{\sqrt{x}-1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-4-\left(x-2\sqrt{x}-3\right)-3\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-4-x+2\sqrt{x}+3-3\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\frac{-\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=-\frac{1}{\sqrt{x}-3}\)
16 tháng 9 2018
ĐKXĐ: \(x\ne3;x\ne-3\)
Biểu thức = \(\frac{2\sqrt{x}.\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)-3+11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
=\(\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\) = \(\frac{3x+9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\) =\(\frac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
=\(\frac{3\sqrt{x}}{\sqrt{x}-3}\)
ĐK \(x\ge0\)
Ta có \(\frac{x-\sqrt{3}x+3}{x\sqrt{x}+3\sqrt{3}}=\frac{x-\sqrt{3}x+3}{\left(\sqrt{x}+\sqrt{3}\right)\left(x-\sqrt{3}x+3\right)}=\frac{1}{\sqrt{x}+\sqrt{3}}\)