Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có :
\(B=\left(\frac{1}{x-4}-\frac{1}{x+4\sqrt{x}+4}\right).\frac{x+2\sqrt{x}}{\sqrt{x}}\)
\(=\left(\frac{1}{\left(\sqrt{x}+2\right)\left(\sqrt{x-2}\right)}-\frac{1}{\left(\sqrt{x}+2\right)^2}\right).\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}}\)
\(=\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}-\frac{\sqrt{x}-2}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}\right).\left(\sqrt{x}+2\right)\)
\(=\frac{\sqrt{x}+2-\sqrt{x}+2}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}.\left(\sqrt{x}+2\right)\)
\(=\frac{4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(\left(\frac{2}{\sqrt{3}+1}+\frac{3}{\sqrt{3}-2}+\frac{15}{3-\sqrt{3}}\right).\frac{1}{\sqrt{3}+5}\)
= \(\left[\frac{2\left(\sqrt{3}+1\right)}{2}+\frac{2\left(\sqrt{3}+2\right)}{1}+\frac{15\left(3+\sqrt{3}\right)}{6}\right].\frac{1}{\sqrt{3}+5}\)
= \(\left[\frac{2\left(\sqrt{3}+1\right)-6\left(\sqrt{3}+2\right)+15\left(\sqrt{3}+3\right)}{2}\right].\frac{1}{\sqrt{3}+5}\)
= \(\left[\frac{2\sqrt{3}+2-6\sqrt{3}-12+5\sqrt{3}+15}{2}\right]\).\(\frac{1}{\sqrt{3}+5}\)
= \(\frac{\sqrt{3}+5}{2}.\frac{1}{\sqrt{3}+5}\)
= \(\frac{1}{2}\)
C =\(\left(1+\frac{x+\sqrt{x}}{\sqrt{x}+1}\right).\left(1-\frac{x-\sqrt{x}}{\sqrt{x}-1}\right)\)
=\(\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)\)
=1-x
C=\(\left(1-\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\).\(\left(1-\frac{x-\sqrt{x}}{\sqrt{x}-1}\right)\)
=\(\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)\)
=\(1-x\)
\(A=\left(\frac{1+\sqrt{3}}{\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}-\frac{1-\sqrt{3}}{\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}\right).\sqrt{3}\)
\(=\left(\frac{1+\sqrt{3}-1+\sqrt{3}}{-2}\right).\sqrt{3}=-3\)
\(B=\frac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}}\)
Để \(A=\frac{B}{6}\Leftrightarrow B=6A\Rightarrow\frac{\sqrt{x}-1}{\sqrt{x}}=-18\)
\(\Rightarrow\sqrt{x}-1=-18\sqrt{x}\Rightarrow\sqrt{x}=\frac{1}{19}\Rightarrow x=\frac{1}{361}\)
Ta có:
\(P=\sqrt{\frac{15}{2}}\cdot\sqrt{\frac{10\left(a-1\right)^2}{3}}\\ =\sqrt{\frac{15}{2}\cdot\frac{10\left(a-1\right)^2}{3}}\\ =\sqrt{25\left(a-1\right)^2}\\ =5\left|a-1\right|\\ =\left[{}\begin{matrix}5\left(a-1\right)\left(a=1\right)\\5\left(1-a\right)\left(a< 1\right)\end{matrix}\right.\\ =\left[{}\begin{matrix}5a-5\\5-5a\end{matrix}\right.\)
P.s: Ko chắc lắm nha :v