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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
1, a, ĐKXĐ: x > 0
\(\Rightarrow P=\frac{\sqrt{x}\left(x\sqrt{x}+1\right)}{x-\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+1\)
\(\Rightarrow P=\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}-2\sqrt{x}-1+1\)
\(\Rightarrow P=\sqrt{x}\left(\sqrt{x}+1\right)-2\sqrt{x}\)
\(\Rightarrow P=x+\sqrt{x}-2\sqrt{x}\)
\(\Rightarrow P=x-\sqrt{x}\)
b, Thay x=100 vào biểu thức P, ta có:
P= 100 - \(\sqrt{100}\)
\(\Rightarrow P=100-10=90\)
Vậy với x=100 thì P=90
c, Ta có: P= \(x-\sqrt{x}=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)
Dấu "=" xảy ra khi...
2, a, ĐKXĐ: x \(\ge\) 0, x \(\ne\) 1
\(\Rightarrow A=\left(\frac{x+3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}+2}\right):\frac{1}{x-1}\)
\(\Rightarrow A=\left(\frac{x+3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right).\frac{x-1}{1}\)
\(\Rightarrow A=\left(\frac{x+3\sqrt{x}-1-\sqrt{x}-2-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right).\frac{x-1}{1}\)
\(\Rightarrow\)A= \(\frac{x+\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}.\frac{x-1}{1}\)= x-1
b, Để \(\frac{1}{A}\)là số tự nhiên (x \(\ge0\), \(x\ne1\))
\(\Rightarrow x-1=1\)
\(\Rightarrow x=2\)
Vậy x=2 thì \(\frac{1}{A}\) là số tự nhiên.
\(P=\frac{x+2}{\sqrt{x}^3-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(P=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)
\(P=\frac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
2,
\(A=\frac{5\left(\sqrt{7}-\sqrt{2}\right)}{\left(\sqrt{7}-\sqrt{2}\right)\left(\sqrt{7}+\sqrt{2}\right)}+\frac{\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}-\frac{7\sqrt{7}}{7}\)
\(A=\frac{5\left(\sqrt{7}-\sqrt{2}\right)}{7-2}+\frac{\left(\sqrt{2}+1\right)}{2-1}-\sqrt{7}\)
\(A=\sqrt{7}-\sqrt{2}+\sqrt{2}+1-\sqrt{7}=1\)
\(P=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)