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C = \(\left(1-\frac{x-3\sqrt{x}}{x-9}\right):\)\(\left(\frac{-\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}-2}{\sqrt{x}+3}-\frac{9-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right)\)( \(x\ge0\) , \(x\ne9;4\))
= \(\frac{x-9-x+3\sqrt{x}}{x-9}\): \(\frac{9-x+\left(\sqrt{x}-2\right)^2-9+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
= \(\frac{3\sqrt{x}-9}{x-9}\): \(\frac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
= \(\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)\(:\frac{\sqrt{x}-2}{\sqrt{x}+3}\)
= \(\frac{3}{\sqrt{x}+3}.\frac{\sqrt{x}+3}{\sqrt{x}-2}\)
= \(\frac{3}{\sqrt{x}-2}\)
#mã mã#
Tik
\(A=\frac{1}{\sqrt{x-1}-\sqrt{x}}+\frac{1}{\sqrt{x-1}+\sqrt{x}}+\frac{\sqrt{x^3}-x}{\sqrt{x}-1}\) \(ĐKXĐ:x\ne\pm1\)
\(=\frac{\sqrt{x-1}+\sqrt{x}+\sqrt{x-1}-\sqrt{x}}{\left(\sqrt{x-1}-\sqrt{x}\right)\left(\sqrt{x-1}+\sqrt{x}\right)}+\frac{x\sqrt{x}-x}{\sqrt{x}-1}\)
\(=\frac{2\sqrt{x-1}}{x-1-x}+\frac{x\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(=x-2\sqrt{x-1}\)
Câu c mình ko làm được
a) ĐKXĐ: \(x\ge0;x\ne9\)
\(P=\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{2}{\sqrt{x}+3}+\frac{5\sqrt{x}+3}{x-9}\)
\(=\frac{\sqrt{x}.\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{5\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{2\sqrt{x}-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{5\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{\left(\sqrt{x}+3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+3}{\sqrt{x}-3}\)
a) ĐKXĐ: \(x\ge0,x\ne9,x\ne4\)
b) C= \(\left(1-\frac{x-3\sqrt{x}}{x-9}\right):\left(\frac{\sqrt{x}-3}{2-\sqrt{x}}+\frac{\sqrt{x}-2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{x}-6}\right)\)
=\(\frac{x-9-x+3\sqrt{x}}{x-9}:\left(\frac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)+\left(\sqrt{x}-2\right)^2-9+x}{\left(\sqrt{x}-2\right)\left(3+\sqrt{x}\right)}\right)\)
= \(\frac{3\sqrt{x}-9}{x-9}:\frac{9-x-x+4\sqrt{x}-4-9+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
= \(\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{-\left(\sqrt{x}-2\right)^2}\)
= \(\frac{-3}{\sqrt{x}-2}\)
Vậy C= \(\frac{-3}{\sqrt{x}-2}\)
c) Ta có C=4 =>\(\frac{-3}{\sqrt{x}-2}=4\)
\(\Leftrightarrow-3=4\sqrt{x}-8\)
\(\Leftrightarrow4\sqrt{x}=5\)
\(\Leftrightarrow\sqrt{x}=\frac{5}{4}\)
\(\Leftrightarrow x=\frac{25}{16}\left(tmđk\right)\)
Vậy với x= \(\frac{25}{16}\) thì C=4