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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
\(B=\frac{-2a\sqrt{a}+2a^2}{\left(\sqrt{a}-\right)\left(a-1\right)}\)
\(C=-x\sqrt{x}+x+\sqrt{x}-1\)
\(D=x-\sqrt{x}+1\)
\(E=\left(\frac{\sqrt{\sqrt{x}-1}}{\sqrt{\sqrt{x}+1}}+\frac{\sqrt{\sqrt{x}+1}}{\sqrt{\sqrt{x}-1}}\right):\sqrt{\frac{1}{x-1}}\) \(ĐKXĐ:x>1\)
\(E=\left(\frac{\left(\sqrt{\sqrt{x}-1}\right)^2}{\left(\sqrt{\sqrt{x}+1}\right)\left(\sqrt{\sqrt{x}-1}\right)}+\frac{\left(\sqrt{\sqrt{x}+1}\right)^2}{\left(\sqrt{\sqrt{x}-1}\right)\left(\sqrt{\sqrt{x}+1}\right)}\right)\cdot\sqrt{\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{1}}\)
\(E=\left(\frac{\sqrt{x}-1}{\left(\sqrt{\sqrt{x}+1}\right)\left(\sqrt{\sqrt{x}-1}\right)}+\frac{\sqrt{x}+1}{\left(\sqrt{\sqrt{x}-1}\right)\left(\sqrt{\sqrt{x}+1}\right)}\right)\cdot\sqrt{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(E=\frac{\sqrt{x}-1+\sqrt{x}+1}{\left(\sqrt{\sqrt{x}+1}\right)\left(\sqrt{\sqrt{x}-1}\right)}\cdot\sqrt{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(E=\frac{2\sqrt{x}}{\sqrt{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}}\cdot\sqrt{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=2\sqrt{x}\)
Ta có:\(x=19-8\sqrt{3}=16-2.4\sqrt{3}+3=\left(4-\sqrt{3}\right)^2\)
\(\Rightarrow2\sqrt{x}=2.\sqrt{\left(4-\sqrt{3}\right)^2}=2.\left(4-\sqrt{3}\right)=8-2\sqrt{3}\)
