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1) Khi x = 49 thì:
\(A=\frac{4\sqrt{49}}{\sqrt{49}-1}=\frac{4\cdot7}{7-1}=\frac{28}{6}=\frac{14}{3}\)
2) Ta có:
\(B=\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{x-1}\)
\(B=\frac{\sqrt{x}-1+x+\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
c) \(P=A\div B=\frac{4\sqrt{x}}{\sqrt{x}-1}\div\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{4\sqrt{x}}{\sqrt{x}+1}\)
Ta có: \(P\left(\sqrt{x}+1\right)=x+4+\sqrt{x-4}\)
\(\Leftrightarrow\frac{4\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}=x+4+\sqrt{x-4}\)
\(\Leftrightarrow4\sqrt{x}=x+4+\sqrt{x-4}\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+\sqrt{x-4}=0\)
Mà \(VT\ge0\left(\forall x\ge0,x\ne1\right)\)
\(\Rightarrow\hept{\begin{cases}\left(\sqrt{x}-2\right)^2=0\\\sqrt{x-4}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}=2\\x-4=0\end{cases}}\Rightarrow x=4\)
Vậy x = 4
\(A=\left(\dfrac{1}{\sqrt{x-1}}+\dfrac{1}{\sqrt{x-1}}\right)^2\cdot\dfrac{x^2-1}{2}-\sqrt{x^2-1}\) (ĐK: \(x>1\))
\(A=\left(\dfrac{2}{\sqrt{x-1}}\right)^2\cdot\dfrac{x^2-1}{2}-\sqrt{x^2-1}\)
\(A=\dfrac{4}{x-1}\cdot\dfrac{\left(x+1\right)\left(x-1\right)}{2}-\sqrt{x^2-1}\)
\(A=2\left(x+1\right)-\sqrt{\left(x+1\right)\left(x-1\right)}\)
\(A=\sqrt{x+1}\left(2\sqrt{x+1}-\sqrt{x-1}\right)\)
a: \(P=\frac{x\cdot\sqrt{x}+1}{x-\sqrt{x}+1}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}=\sqrt{x}+1\)
b: Khi x=100 thì \(P=\sqrt{100}+1=10+1=11\)
a)
\(P=\dfrac{x\sqrt{x+1}}{x-\sqrt{x+1}}-1\)
\(P=\dfrac{x\sqrt{x+1}-\left(x-\sqrt{x+1}\right)}{x-\sqrt{x+1}}\)
\(P=\dfrac{x\sqrt{x+1}-x+\sqrt{x+1}}{x-\sqrt{x+1}}\)
\(P=\dfrac{x(\sqrt{x+1}-1)+\sqrt{x+1}}{x-\sqrt{x+1}}\)
Vậy \(P=\dfrac{x(\sqrt{x+1}-1)+\sqrt{x+1}}{x-\sqrt{x+1}}\)
b)
Thay \(x=100\) vào biểu thức \(P=\dfrac{x\left(\sqrt{x+1}-1\right)+\sqrt{x+1}}{x-\sqrt{x+1}}\), ta được:
\(P=\dfrac{100(\sqrt{100+1}-1)+\sqrt{100+1}}{100-\sqrt{100+1}}\)
\(P=\dfrac{100\sqrt{101}-100+\sqrt{101}}{100-\sqrt{101}}\)
\(P=\dfrac{(100+1)\sqrt{101}-100}{100-\sqrt{101}}\)
\(P=\dfrac{101\sqrt{101}-100}{100-\sqrt{101}}\)
\(P=\dfrac{(101\sqrt{101}-100)(100+\sqrt{101})}{(100-\sqrt{101})(100+\sqrt{101})}\)
Vậy \(P=\dfrac{10000\sqrt{101}+201}{9899}\approx\dfrac{100700}{9899}\approx10,18\)
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