Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\frac{\sqrt{4mn^2}}{\sqrt{20m}}=\sqrt{\frac{4mn^2}{20m}}=\sqrt{\frac{n^2}{5}}=\frac{n}{\sqrt{5}}\)
b) \(\frac{\sqrt{16a^4b^6}}{\sqrt{12a^6b^6}}=\sqrt{\frac{16a^4b^6}{12a^6b^6}}=\sqrt{\frac{4}{3a^2}}=\frac{2}{\sqrt{3}.\left|a\right|}=-\frac{2}{a\sqrt{3}}\)
d) \(\frac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}=x+\sqrt{xy}+y\)
e) \(\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\frac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)
a) P=\((\frac{x-1}{\sqrt{x}})\):\([\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{\sqrt{x}(\sqrt{x}+1)}]\) (x>0;x≠1)
P=\(\frac{x-1}{\sqrt{x}}\):\([\)\(\frac{(\sqrt{x}-1).(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)}\)\(]\)
P=\(\frac{x-1}{\sqrt{x}}\):\(\frac{x-1}{\sqrt{x}(\sqrt{x}+1)}\)
P=\(\frac{x-1}{\sqrt{x}}\cdot\frac{\sqrt{x}(\sqrt{x}+1)}{x-1}\)
P=\(\sqrt{x}+1\)
b,Có x=\(\frac{2}{2+\sqrt{3}}=\frac{2\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=4-2\sqrt{3}=\left(1-\sqrt{3}\right)^2\)
=>\(\sqrt{x}=\sqrt{\left(1-\sqrt{3}\right)^2}=\left|1-\sqrt{3}\right|=\sqrt{3}-1\)
Có P=\(\sqrt{x}+1=\sqrt{3}-1+1=\sqrt{3}\)
c, Có P\(\sqrt{x}=6\sqrt{x}-3\)
<=>\(\sqrt{x}\left(P-6\right)+3=0\) <=> \(\sqrt{x}\left(\sqrt{x}+1-6\right)+3=0\) <=> \(\sqrt{x}\left(\sqrt{x}-5\right)+3=0\)
<=> \(x-5\sqrt{x}+3=0\) <=> \(\left(x-\frac{5+\sqrt{13}}{2}\right)\left(x-\frac{5-\sqrt{13}}{2}\right)=0\) => \(\left[{}\begin{matrix}x=\frac{5+\sqrt{13}}{2}\left(tm\right)\\x=\frac{5-\sqrt{13}}{2}\left(tm\right)\end{matrix}\right.\)
\(M=\frac{\sqrt{x}}{\sqrt{x}+1}\left(x\ge0\right)\)
Khi \(M=\sqrt{x}-2\)
\(\Rightarrow\frac{\sqrt{x}}{\sqrt{x}+1}=\sqrt{x}-2\)
\(\Leftrightarrow\sqrt{x}=\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)\)
\(\Leftrightarrow\sqrt{x}=x+\sqrt{x}-2\sqrt{x}-2\)
\(\Leftrightarrow\sqrt{x}=x-\sqrt{x}-2\)
\(\Leftrightarrow x-\sqrt{x}-\sqrt{x}-2=0\)
\(\Leftrightarrow x-2\sqrt{x}+1-3=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2=3\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2=\left(\pm\sqrt{3}\right)^2\)
\(\Leftrightarrow\sqrt{x}-1=\pm\sqrt{3}\)
\(\Leftrightarrow\sqrt{x}=\pm\sqrt{3}+1\)
\(\Leftrightarrow\orbr{\begin{cases}x=\left(\sqrt{3}+1\right)^2\\x=\left(-\sqrt{3}+1\right)^2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3+2\sqrt{3}+1\\1-2\sqrt{3}+3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4+2\sqrt{3}\\x=4-2\sqrt{3}\end{cases}}\)
Vậy \(x\in\left\{4\pm2\sqrt{3}\right\}\)khi \(M=\sqrt{x}-2\)