A=\(\frac{\sqrt{x}+1}{\sqrt{x}-1}\)Tính A khi X=\(4+2\sqrt{3}\)
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CH
Cô Hoàng Huyền
Admin
VIP
21 tháng 6 2017
a) ĐK: x > 1
\(P=\left(\frac{\sqrt{x-1}}{3+\sqrt{x-1}}+\frac{x+8}{9-\left(x-1\right)}\right):\left(\frac{3\sqrt{x-1}+1}{\left(x-1\right)-3\sqrt{x-1}}-\frac{1}{\sqrt{x-1}}\right)\)
\(P=\frac{\sqrt{x-1}\left(3-\sqrt{x-1}\right)+x+8}{9-\left(x-1\right)}:\frac{3\sqrt{x-1}+1-\left(\sqrt{x-1}-3\right)}{\sqrt{x-1}\left(\sqrt{x-1}-3\right)}\)
\(P=\frac{3\sqrt{x-1}-x+1+x+8}{10-x}:\frac{2\sqrt{x-1}+4}{\sqrt{x-1}\left(\sqrt{x-1}-3\right)}\)
\(P=\frac{3\left(\sqrt{x-1}+3\right)}{10-x}.\frac{\sqrt{x-1}\left(\sqrt{x-1}-3\right)}{2\sqrt{x-1}+4}\)
\(P=\frac{-3\sqrt{x-1}}{2\sqrt{x-1}+4}\)
b) \(x=\sqrt[4]{\frac{17+12\sqrt{2}}{1}}-\sqrt[4]{\frac{17-12\sqrt{2}}{1}}=1+\sqrt{2}-\left(\sqrt{2}-1\right)=2\)
Vậy \(P=\frac{-3\sqrt{2-1}}{2\sqrt{2-1}+4}=-\frac{1}{2}\)
21 tháng 6 2017
cô Hoàng Thị Thu Huyền làm rõ cho em ý b đc ko ạ chỗ biến đổi x
14 tháng 5 2021
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
NM
Nguyễn Minh Quang
Giáo viên
19 tháng 7 2021
a. ĐKXĐ:
\(\hept{\begin{cases}\sqrt{x}-1\ne0\\x-\sqrt{x}\ne0\\x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x>0\\x\ne1\end{cases}}}\)
b. ta có \(A=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{x-1}{\sqrt{x}.\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}}\)
c. khi \(x=\frac{1}{4}\Rightarrow\sqrt{x}=\frac{1}{2}\Rightarrow A=\frac{\frac{1}{2}+1}{\frac{1}{2}}=3\)
khi \(x=3+2\sqrt{2}\Rightarrow\sqrt{x}=\sqrt{2}+1\Rightarrow A=\frac{\sqrt{2}+1+1}{\sqrt{2}+1}=\sqrt{2}\)
19 tháng 7 2021
\(a,ĐKXĐ:A=x\ge0;x\ne1\)
\(b,A=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\)
\(A=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(A=\frac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(A=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(A=\frac{\sqrt{x}+1}{\sqrt{x}}< =>ĐPCM\)
c,thay \(x=\frac{1}{4}\)vào A
\(c,A=\frac{\sqrt{\frac{1}{4}}+1}{\sqrt{\frac{1}{4}}}\)
\(A=\frac{\frac{1}{2}+1}{\frac{1}{2}}\)
\(A=3\)
\(x=3+2\sqrt{2}\)
\(x=\sqrt{2}^2+2\sqrt{2}+1\)
\(x=\left(\sqrt{2}+1\right)^2\)thay x vào A
\(A=\frac{\sqrt{\left(\sqrt{2}+1\right)^2}+1}{\sqrt{\left(\sqrt{2}+1\right)^2}}\)
\(A=\frac{\sqrt{2}+1+1}{\sqrt{2}+1}\)
\(A=\frac{2+\sqrt{2}}{\sqrt{2}+1}\)
\(A=\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}=\sqrt{2}\)
NV
14 tháng 7 2016
a/ \(A=\left(\frac{2\sqrt{x}+x}{\sqrt{x}^3-1}-\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+2}{x+\sqrt{x}+1}\)
\(=\left[\frac{2\sqrt{x}+x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}-1}\right]:\frac{\sqrt{x}+2}{x+\sqrt{x}+1}\)
\(=\frac{2\sqrt{x}+x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{x+\sqrt{x}+1}{\sqrt{x}+2}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}-1}.\frac{1}{\sqrt{x}+2}=\frac{1}{\sqrt{x}+2}\)
b/ Thay \(x=4+2\sqrt{3}\) vào A ta được:
\(A=\frac{1}{\sqrt{4+2\sqrt{3}}+2}=\frac{1}{\sqrt{\left(\sqrt{3}+1\right)^2}+2}=\frac{1}{\sqrt{3}+3}\)
\(\Rightarrow\sqrt{A}=\frac{1}{\sqrt{\sqrt{3}+3}}\)
\(x=4+2\sqrt{3}\)
=>\(\sqrt{x}=\sqrt{4+2\sqrt{3}}=\sqrt{3+2.\sqrt{3}.1+1}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
\(A=\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{\sqrt{3}+1+1}{\sqrt{3}+1-1}=\frac{\sqrt{3}+2}{\sqrt{3}}=\frac{3+2\sqrt{3}}{3}\)