Bạn cần bài nào thì nên ghi chú bài đó ra nhé. Nếu cần nhiều bài thì nên tách lẻ mỗi bài mỗi post để được hỗ trợ tốt hơn nhé.
3:
1: \(M=\sqrt{8-2\sqrt{15}}+\dfrac{\sqrt{5}}{\sqrt{5}-\sqrt{3}}-\dfrac{2}{\sqrt{4-2\sqrt{3}}}\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}+\dfrac{\sqrt{5}\left(\sqrt{5}+\sqrt{3}\right)}{2}-\dfrac{2}{\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\sqrt{5}-\sqrt{3}+\dfrac{5+\sqrt{15}}{2}-\dfrac{2}{\sqrt{3}-1}\)
\(=\sqrt{5}-\sqrt{3}+2,5+\dfrac{1}{2}\sqrt{15}-\dfrac{2\left(\sqrt{3}+1\right)}{2}\)
\(=\sqrt{5}-\sqrt{3}+2,5+\dfrac{1}{2}\sqrt{15}-\sqrt{3}-1\)
\(=\dfrac{1}{2}\sqrt{15}+\sqrt{5}+1,5\)
2:
ĐKXĐ: \(\left\{{}\begin{matrix}x^2-6x+9>0\\16-x^2>=0\\2x-1>=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(x-3\right)^2>0\\x^2< =16\\x>=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-3< >0\\-4< =x< =4\\x>=\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< >3\\\dfrac{1}{2}< =x< =4\end{matrix}\right.\)
1:
1: \(B=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{5}{\sqrt{x}+1}-\dfrac{8\sqrt{x}-6}{x-1}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+5\left(\sqrt{x}-1\right)-8\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+5\sqrt{x}-5-8\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
2:
ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >1\end{matrix}\right.\)
C>7
=>C-7>0
=>\(\dfrac{2\sqrt{x}+3}{\sqrt{x}-1}-7>=0\)
=>\(\dfrac{2\sqrt{x}+3-7\sqrt{x}+7}{\sqrt{x}-1}>=0\)
=>\(\dfrac{-5\sqrt{x}+10}{\sqrt{x}-1}>=0\)
=>\(\dfrac{\sqrt{x}-2}{\sqrt{x}-1}< =0\)
TH1: \(\left\{{}\begin{matrix}\sqrt{x}-2>=0\\\sqrt{x}-1< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\sqrt{x}>=2\\\sqrt{x}< 1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=4\\0< =x< 1\end{matrix}\right.\)
=>Loại
Th2: \(\left\{{}\begin{matrix}\sqrt{x}-2< =0\\\sqrt{x}-1>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\sqrt{x}< =2\\\sqrt{x}>1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =4\\x>1\end{matrix}\right.\)
=>\(1< x< =4\)
