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câu 2
\(...=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{\left(2+\sqrt{5}\right)^2}=\left|2-\sqrt{5}\right|-\left|2+\sqrt{5}\right|=-4\)
câu 1
\(P=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{1}{\sqrt{x}}\right)\)
\(=\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{3}{\left(3-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\)
\(P< -1\Leftrightarrow\frac{-3\sqrt{x}}{2\sqrt{x}+4}+1< 0\Leftrightarrow-\sqrt{x}+4< 0\Leftrightarrow\sqrt{x}>4\Leftrightarrow x>16\)
c/ \(C'=\frac{1}{\frac{1}{3-2\sqrt{x}}}.\frac{1}{\frac{1}{\sqrt{3-2\sqrt{x}}}+1}=\frac{\sqrt{\left(3-2\sqrt{x}\right)^3}}{1+\sqrt{\left(3-2\sqrt{x}\right)}}\)
Đặt \(\sqrt{\left(3-2\sqrt{x}\right)}=a\)
\(\Rightarrow C'=\frac{a^3}{a+1}=a^2-a+1-\frac{1}{a+1}\)
Đế C' nguyên thì a + 1 là ước của 1
\(\Rightarrow a=0\)
\(\Rightarrow\sqrt{\left(3-2\sqrt{x}\right)}=0\)
\(\Rightarrow x=\frac{9}{4}\left(l\right)\)
Vậy không có x.
Không biết có nhầm chỗ nào không nữa. Lam biếng kiểm tra lại quá. You kiểm tra lại hộ nhé. Thanks
a/ \(C=\left(\frac{2\sqrt{x}}{2x-5\sqrt{x}+3}-\frac{5}{2\sqrt{x}-3}\right):\left(3+\frac{2}{1-\sqrt{x}}\right)\)
\(=\left(\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-3\right)}-\frac{5}{2\sqrt{x}-3}\right):\left(\frac{3\sqrt{x}-5}{\sqrt{x}-1}\right)\)
\(=\left(\frac{2\sqrt{x}-5\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-3\right)}\right):\left(\frac{3\sqrt{x}-5}{\sqrt{x}-1}\right)\)
\(=\frac{5-3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-3\right)}.\frac{\sqrt{x}-1}{3\sqrt{x}-5}\)
\(=\frac{1}{3-2\sqrt{x}}\)
Câu b, c tự làm nhé
ĐKXĐ : \(x\ge0\)
\(A=\frac{2}{3}.\frac{2+\left(\frac{2\sqrt{x}-1}{\sqrt{3}}\right)^2+\left(\frac{2\sqrt{x}+1}{\sqrt{3}}\right)^2}{\left[1+\left(\frac{2\sqrt{x}+1}{\sqrt{3}}\right)^2\right]\left[1+\left(\frac{2\sqrt{x}-1}{\sqrt{3}}\right)^2\right]}.\frac{2010}{x+1}\)
\(A=\frac{2}{3}.\frac{2+\left(\frac{2\sqrt{x}-1}{\sqrt{3}}+\frac{2\sqrt{x}+1}{\sqrt{3}}\right)^2-2\left(\frac{2\sqrt{x}-1}{\sqrt{3}}\right)\left(\frac{2\sqrt{x}+1}{\sqrt{3}}\right)}{\left[1+\frac{\left(2\sqrt{x}+1\right)^2}{3}\right]\left[1+\frac{\left(2\sqrt{x}-1\right)^2}{3}\right]}.\frac{2010}{x+1}\)
\(A=\frac{2}{3}.\frac{2+\left(\frac{4\sqrt{x}}{\sqrt{3}}\right)^2-\frac{2\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}{3}}{\left(\frac{4x+4\sqrt{x}+4}{3}\right)\left(\frac{4x-4\sqrt{x}+4}{3}\right)}.\frac{2010}{x+1}\)
\(A=\frac{2}{3}.\frac{2+\frac{16x}{3}-\frac{2\left(4x-1\right)}{3}}{\frac{16\left(x+1+\sqrt{x}\right)\left(x+1-\sqrt{x}\right)}{9}}.\frac{2010}{x+1}\)
\(A=\frac{2}{3}.\frac{\frac{6+16x-8x+2}{3}}{\frac{16\left(x+1\right)^2-16x}{9}}.\frac{2010}{x+1}\)
\(A=\frac{x+1}{x^2+x+1}.\frac{2010}{x+1}=\frac{2010}{x^2+x+1}\le2010\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=0\)
...
Ta có :
\(B=\left(\frac{1}{x-4}-\frac{1}{x+4\sqrt{x}+4}\right).\frac{x+2\sqrt{x}}{\sqrt{x}}\)
\(=\left(\frac{1}{\left(\sqrt{x}+2\right)\left(\sqrt{x-2}\right)}-\frac{1}{\left(\sqrt{x}+2\right)^2}\right).\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}}\)
\(=\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}-\frac{\sqrt{x}-2}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}\right).\left(\sqrt{x}+2\right)\)
\(=\frac{\sqrt{x}+2-\sqrt{x}+2}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}.\left(\sqrt{x}+2\right)\)
\(=\frac{4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
C =\(\left(1+\frac{x+\sqrt{x}}{\sqrt{x}+1}\right).\left(1-\frac{x-\sqrt{x}}{\sqrt{x}-1}\right)\)
=\(\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)\)
=1-x
C=\(\left(1-\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\).\(\left(1-\frac{x-\sqrt{x}}{\sqrt{x}-1}\right)\)
=\(\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)\)
=\(1-x\)
\(ĐK:\left\{{}\begin{matrix}x\ge0\\x\ne9\\x\ne25\end{matrix}\right.\)
\(\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{2x}{9-x}-1\right):\left(\frac{\sqrt{x}-1}{x-3\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)-2x-\left(x-9\right)}{x-9}:\frac{\sqrt{x}-1-2\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{-2x-3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{-\sqrt{x}+5}\)
\(=\frac{-\left(2\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}\cdot\frac{\sqrt{x}}{5-\sqrt{x}}\)
\(=\frac{-\left(2\sqrt{x}-3\right)\cdot\sqrt{x}}{5-\sqrt{x}}=\frac{-2x+3\sqrt{x}}{5-\sqrt{x}}\)
ĐKXĐ :\(x\) > 0 , x\(\ne9\)
\(\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{2x}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}-1\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{2}{\sqrt{x}}\right)=\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)+2x-\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1-2\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)=\)\(\frac{3\sqrt{x}-x+2x-9+x}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x-3}\right)}{\sqrt{x}-1-2\sqrt{x}+6}=\frac{2x-3\sqrt{x}-9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{5-\sqrt{x}}=\frac{\left(\sqrt{x}+3\right)\left(2\sqrt{x}-3\right)}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{5-\sqrt{x}}=\frac{\sqrt{x}\left(2\sqrt{x}-3\right)}{\sqrt{x}-5}\)