Câu 1: Tính giá trị biểu thức: \(A=\frac{\sqrt{x}+1}{\sqrt{x}-1}\) khi x = 9
Câu 2: Cho biểu thức P = \(\left(\frac{x-2}{x+2\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right).\frac{\sqrt{x}+1}{\sqrt{x}-1}\) với \(x>0,x\ne1\)
a) Chứng minh \(P=\frac{\sqrt{x}+1}{\sqrt{x}}\)
b) Tìm giá trị của x để \(2P=2\sqrt{x}+5\)
1/ ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
Thay \(x=9\) vào biểu thức A ta có :
\(A=\frac{\sqrt{9}+1}{\sqrt{9}-1}=\frac{3+1}{3-1}=2\)
Vậy...
2/ ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
Ta có :
\(P=\left(\frac{x-2}{x+2\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right).\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\left(\frac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\right).\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\frac{x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}}\)
Vậy....
b/ Ta có :
\(2P=2\sqrt{x}+5\)
\(\Leftrightarrow\frac{2\left(\sqrt{x}+1\right)}{\sqrt{x}}=2\sqrt{x}+5\)
\(\Leftrightarrow2\sqrt{x}+2=2x+5\sqrt{x}\)
\(\Leftrightarrow2x+3\sqrt{x}-2=0\)
\(\Leftrightarrow\left(2\sqrt{x}-1\right)\left(\sqrt{x}+2\right)=0\)
\(\Leftrightarrow2\sqrt{x}-1=0\)
\(\Leftrightarrow x=\frac{1}{4}\)
Vậy..
Câu 1 :
\(A=\frac{\sqrt{x}+1}{\sqrt{x-1}}\) khi x = 9
tại x = 9 thay vào A ta được : \(\frac{\sqrt{9}+1}{\sqrt{9}-1}\) = \(\frac{3+1}{3-1}=\frac{4}{2}=2\)
Câu 2 :
a, Ta có : P = \(\left(\frac{x-2}{x+2\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right).\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
= \(\left(\frac{x-2}{\sqrt{x}.\sqrt{x}+2\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right).\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\left(\frac{x-2}{\sqrt{x}\left(\sqrt{x+2}\right)}+\frac{1}{\sqrt{x}+2}\right)\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
= \(\left(\frac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\right).\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
= \(\left(\frac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\right).\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
=\(\frac{x-2+2\sqrt{x}-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}}\) => đpcm