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\(A=\frac{\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(A=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{4}{x-1}\)
b) \(\frac{4}{x-1}=7\)
\(\Leftrightarrow4=7.\left(x-1\right)\)
\(\Leftrightarrow\frac{4}{7}=x-1\)
\(\Leftrightarrow\frac{4}{7}+1=x\)
\(\Leftrightarrow\frac{11}{7}=x\)
\(\Rightarrow x=\frac{11}{7}\)
a. 2\(\sqrt{3.16}\)+\(\sqrt{3.9}\)+\(\sqrt{3}\)
=2.4.\(\sqrt{3}\)+3\(\sqrt{3}\)+\(\sqrt{3}\)
12\(\sqrt{3}\)
\(C^3=2-\sqrt{5}+2+\sqrt{5}+3\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\left(\sqrt[3]{2-\sqrt{5}}+\sqrt[3]{2+\sqrt{5}}\right)\)
\(=4+3\sqrt[3]{4-5}.C=4-3C\Leftrightarrow C^3+3C-4=0\Leftrightarrow\left(C-1\right)\left(C^2+C+4\right)=0\)
\(\Leftrightarrow C-1=0\Leftrightarrow C=1\)
Ta có: \(A=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)
\(\Rightarrow A^3=\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)^3\)
\(=9+4\sqrt{5}+9-4\sqrt{5}+3\sqrt[3]{\left(9+4\sqrt{5}\right)\left(9-4\sqrt{5}\right)}\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)\)
\(=18+3\sqrt[3]{81-80}\cdot A\)
\(=18+3A\)
\(\Rightarrow A^3-3A-18=0\)
\(\Leftrightarrow\left(A^3-3A^2\right)+\left(3A^2-9A\right)+\left(6A-18\right)=0\)
\(\Leftrightarrow\left(A-3\right)\left(A^2+3A+6\right)=0\)
Mà \(A^2+3A+6=\left(A+\frac{3}{2}\right)^2+\frac{15}{4}\left(\forall A\right)\)
\(\Rightarrow A=3\)
Vậy A = 3