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\(A=15+12+4\sqrt{45}+12\sqrt{5}=27+24\sqrt{5}\)
\(B=\left(2\sqrt{3}+6\sqrt{3}\right).\frac{\sqrt{3}}{2}-5\sqrt{6}=\frac{8\sqrt{3}.\sqrt{3}}{2}-5\sqrt{6}=12-5\sqrt{6}\)
\(C=4\sqrt{3}+\frac{4}{\sqrt{3}}+10\sqrt{5}-\frac{10}{\sqrt{5}}=\frac{16}{\sqrt{3}}+8\sqrt{5}\)
\(\left(\frac{2}{\sqrt{3}+1}+\frac{3}{\sqrt{3}-2}+\frac{15}{3-\sqrt{3}}\right).\frac{1}{\sqrt{3}+5}\)
= \(\left[\frac{2\left(\sqrt{3}+1\right)}{2}+\frac{2\left(\sqrt{3}+2\right)}{1}+\frac{15\left(3+\sqrt{3}\right)}{6}\right].\frac{1}{\sqrt{3}+5}\)
= \(\left[\frac{2\left(\sqrt{3}+1\right)-6\left(\sqrt{3}+2\right)+15\left(\sqrt{3}+3\right)}{2}\right].\frac{1}{\sqrt{3}+5}\)
= \(\left[\frac{2\sqrt{3}+2-6\sqrt{3}-12+5\sqrt{3}+15}{2}\right]\).\(\frac{1}{\sqrt{3}+5}\)
= \(\frac{\sqrt{3}+5}{2}.\frac{1}{\sqrt{3}+5}\)
= \(\frac{1}{2}\)
\(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\\ < =>\frac{1-\sqrt{2}}{1+\sqrt{2}\left(1-\sqrt{2}\right)}+\frac{\sqrt{2}-\sqrt{3}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}+...+\frac{\sqrt{99}-\sqrt{100}}{\left(\sqrt{99}+\sqrt{100}\right)\sqrt{99}-\sqrt{100}}\\ < =>\frac{1-\sqrt{2}}{1-2}+\frac{\sqrt{2}-\sqrt{3}}{2-3}+...+\frac{\sqrt{99}-\sqrt{100}}{99-100}\)
\(=\frac{1-\sqrt{2}}{-1}+\frac{\sqrt{2}-\sqrt{3}}{-1}+...+\frac{\sqrt{99}-\sqrt{100}}{-1}\\ =\frac{1-\sqrt{2}+\sqrt{2}-\sqrt{3}+\sqrt{3}-\sqrt{4}+...+\sqrt{99}-10}{-1}\\ =\frac{1-10}{-1}\\ =\frac{-9}{-1}\\ =9\)
P/s: Chuyền hết dấu tương đương ở trên thành bằng nhé, mình bị nhầm
1) ĐKXĐ: \(x>0;x\ne4;x\ne9\)
(*lười lắm, ko chép lại đề nha :V*)
\(P=\frac{\left(2+\sqrt{x}\right)^2+\sqrt{x}\left(2-\sqrt{x}\right)+4x+2\sqrt{x}-4}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\frac{2\sqrt{x}-\left(\sqrt{x}+3\right)}{\sqrt{x}\left(2-\sqrt{x}\right)}\\ =\frac{4+4\sqrt{x}+x+2\sqrt{x}-x+4x+2\sqrt{x}-4}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\cdot\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}\\ =\frac{4x+8\sqrt{x}}{2+\sqrt{x}}\cdot\frac{\sqrt{x}}{\sqrt{x}-3}\\ =\frac{4\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}+2}\cdot\frac{\sqrt{x}}{\sqrt{x}-3}=\frac{4x}{\sqrt{x}-3}\)
2) Để P>0 thì
\(\frac{4x}{\sqrt{x}-3}>0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}4x>0\\\sqrt{x}-3>0\end{matrix}\right.\\\left\{{}\begin{matrix}4x< 0\\\sqrt{x}-3< 0\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\\sqrt{x}>3\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\\sqrt{x}< 3\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x>9\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x< 9\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>9\\x< 0\left(ktm\right)\end{matrix}\right.\)
Vậy với \(x>9\) thì \(P>0\).
Chúc bạn học tốt nha
.
Bạn giải thêm cho mk câu này đi
c) tìm giá trị của x để P = -1
\(=\frac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}+\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+1}\left(\frac{\sqrt{3}\left(2+\sqrt{6}\right)+\sqrt{3}\left(2-\sqrt{6}\right)}{\left(2-\sqrt{6}\right)\left(2+\sqrt{6}\right)}\right)-\frac{1}{\sqrt{2}}\)
\(=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{2}\left(\sqrt{2}+\sqrt{3}\right)}-\frac{1}{2+\sqrt{6}}+\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+1}\left(-2\sqrt{3}\right)-\frac{1}{\sqrt{2}}\)
\(=\frac{1}{\sqrt{2}}-\frac{2-\sqrt{6}}{\left(2-\sqrt{6}\right)\left(2+\sqrt{6}\right)}+\frac{\left(\sqrt{2}-1\right)\left(-2\sqrt{6}+6\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}-\frac{1}{\sqrt{2}}\)
\(=\frac{2-\sqrt{6}}{2}-4\sqrt{3}+6\sqrt{2}+2\sqrt{6}-6\)
\(=6\sqrt{2}-4\sqrt{3}+\frac{3\sqrt{6}}{2}-5\)
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