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a) \(M=\frac{x+1+\sqrt{x}}{x+1}:\left(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{\sqrt{x}\left(x+1\right)-\left(x+1\right)}\right)\)
\(=\frac{x+\sqrt{x}+1}{x+1}:\left(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{x+\sqrt{x}+1}{x+1}:\frac{x+1-2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\)\(=\frac{x+\sqrt{x}+1}{x+1}.\frac{\left(x+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)^2}=\frac{x+\sqrt{x}+1}{\sqrt{x}-1}\)
b) \(M>3\Rightarrow\frac{x+\sqrt{x}+1}{\sqrt{x}-1}>3\Leftrightarrow\frac{x+\sqrt{x}+1}{\sqrt{x}-1}-3>0\)
\(\Leftrightarrow\frac{x+\sqrt{x}+1-3\left(\sqrt{x}-1\right)}{\sqrt{x}-1}>0\Leftrightarrow\frac{x+\sqrt{x}+1-3\sqrt{x}+3}{\sqrt{x}-1}>0\)\(\Leftrightarrow\frac{x-2\sqrt{x}+4}{\sqrt{x}-1}>0\)
Ta có: \(x-2\sqrt{x}+4=x-2\sqrt{x}+1+3=\left(\sqrt{x}-1\right)+3>0\)\(\Rightarrow\sqrt{x}-1>0\Leftrightarrow\sqrt{x}>1\Leftrightarrow x>1\)
Vậy x>1
c) \(M=7\Rightarrow\frac{x+\sqrt{x}+1}{\sqrt{x}-1}=7\Rightarrow x+\sqrt{x}+1=7\left(\sqrt{x}-1\right)\)
\(\Leftrightarrow x+\sqrt{x}+1=7\sqrt{x}-7\Leftrightarrow x-6\sqrt{x}+8=0\)\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-2=0\\\sqrt{x}-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=2\\\sqrt{x}=4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=4\\x=16\end{cases}\left(tm\right)}}\)
Vậy \(x\in\text{{}4;16\)
a. \(=\sqrt{2}.\left(\sqrt{7}+\sqrt{8}\right)\sqrt{5-\sqrt{3}\sqrt{7}}\)
\(=\left(\sqrt{7}+\sqrt{8}\right)\sqrt{3-2\sqrt{3}.\sqrt{7}+7}\)
\(=\left(\sqrt{7}+\sqrt{8}\right)\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)
\(=\left(\sqrt{7}+\sqrt{8}\right)\left(\sqrt{7}-\sqrt{3}\right)\)
Rồi nhân ra. bạn làm tiếp nhé. Tuy nhiên minh nghĩ bạn bị nhầm đề. là \(\sqrt{6}\) chứ không phải căn 16
b. \(=\frac{5\left(\sqrt{21}+1\right)}{21-16}+\frac{\sqrt{3}.\sqrt{7}\left(\sqrt{3}-\sqrt{7}\right)}{-\left(\sqrt{3}-\sqrt{7}\right)}\)
\(=\sqrt{21}+4-\sqrt{21}=4\)
\(\left(3+\frac{3-\sqrt{3}}{1-\sqrt{3}}\right)\left(\frac{\sqrt{21}+\sqrt{7}}{\sqrt{7}}+2\right)\)
\(=\left(3-\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right)\left(\frac{\sqrt{7}\left(\sqrt{3}+1\right)}{\sqrt{7}}+2\right)\)
\(=\left(3-\sqrt{3}\right)\left(\sqrt{3}+3\right)=9-3=6\)
\(3+\frac{3-\sqrt{3}}{1-\sqrt{3}}=3+\frac{1-\sqrt{3}+2}{1-\sqrt{3}}=3+1+\frac{2}{1-\sqrt{3}}=4+\frac{2}{1-\sqrt{3}}\)
\(=4+\frac{3-1}{1-\sqrt{3}}=4+\frac{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}{1-\sqrt{3}}=4-\sqrt{3}-1=-\sqrt{3}-3\)
\(\frac{\sqrt{21}+\sqrt{7}}{\sqrt{7}}+2=\frac{\sqrt{7}\left(\sqrt{3}+1\right)}{\sqrt{7}}+2=\left(\sqrt{3}+3\right)\)
Khi đó \(\left(3+\frac{3-\sqrt{3}}{1-\sqrt{3}}\right)\left(\frac{\sqrt{21}+\sqrt{7}}{\sqrt{7}}+2\right)=-\left(\sqrt{3}+3\right)^2=-12-6\sqrt{3}\)