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Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé
1.\(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}=\frac{\left(5+\sqrt{5}\right)\left(5+\sqrt{5}\right)}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}+\frac{\left(5-\sqrt{5}\right)\left(5-\sqrt{5}\right)}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\)
\(=\frac{25+10\sqrt{5}+5}{25-5}+\frac{25-10\sqrt{5}+5}{25-5}\)
\(=\frac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{20}\)
\(=\frac{60}{20}=3\)
2.
a) \(\sqrt{45x}-2\sqrt{20x}+2\sqrt{80x}=21\)
ĐK : x ≥ 0
<=> \(\sqrt{5x\cdot9}-2\sqrt{5x\cdot4}+2\sqrt{5x\cdot16}=21\)
<=> \(\sqrt{5x\cdot3^2}-2\sqrt{2^2\cdot5x}+2\sqrt{5x\cdot4^2}=21\)
<=> \(\left|3\right|\sqrt{5x}-2\cdot\left|2\right|\sqrt{5x}+2\cdot\left|4\right|\sqrt{5x}=21\)
<=> \(\sqrt{5x}\cdot\left(3-4+8\right)=21\)
<=> \(\sqrt{5x}\cdot7=21\)
<=> \(\sqrt{5x}=3\)
<=> \(5x=9\)
<=> \(x=\frac{9}{5}\left(tm\right)\)
ơ đang làm lại bấm " Gửi trả lời " ._.
2b) \(\sqrt{x^2-10x+25}=4\)
<=> \(\sqrt{\left(x-5\right)^2}=4\)
<=> \(\left|x-5\right|=4\)
<=> \(\orbr{\begin{cases}x-5=4\\x-5=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=9\\x=1\end{cases}}\)
3. \(A=\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right)\div\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
ĐK : \(\hept{\begin{cases}x>0\\x\ne1\\x\ne4\end{cases}}\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x-1}\right)}\right)\div\left(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\left(\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\div\left(\frac{x-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\frac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\left(\frac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\frac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\times\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}\)
\(=\frac{\sqrt{x}-2}{3\sqrt{x}}\)
Bài 1:
a) \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{2}-\frac{2\left(\sqrt{3}-1\right)}{2}\)
\(=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)
b) \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{2}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\)
\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)
c) ĐK: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)
\(=1-a+a=1\)
\(P=\dfrac{5\sqrt{x}}{x-1}+\dfrac{3}{2\sqrt{x}+2}-\dfrac{5}{2\sqrt{x}-2}\)
\(P=\dfrac{5\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{3}{2\left(\sqrt{x}+1\right)}-\dfrac{5}{2\left(\sqrt{x}-1\right)}\)
\(P=\dfrac{2.5\sqrt{x}+3\left(\sqrt{x}-1\right)-5\left(\sqrt{x}+1\right)}{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(P=\dfrac{10\sqrt{x}+3\sqrt{x}-3-5\sqrt{x}-5}{2\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}\)
\(P=\dfrac{8\sqrt{x}-8}{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(P=\dfrac{8\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(P=\dfrac{4}{\sqrt{x}+1}\)