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A=\(\sqrt{\left(\sqrt{7}-2\right)^2}\)+\(\frac{25\sqrt{7}-63}{3\sqrt{7}-7}\)=\(\frac{12\sqrt{7}-28}{3\sqrt{7}-7}\)=4
Lời giải:
\(A=\frac{3}{\sqrt{7}+2}+\frac{4}{3-\sqrt{7}}-\frac{21}{\sqrt{7}}=\frac{3(\sqrt{7}-2)}{(\sqrt{7}+2)(\sqrt{7}-2)}+\frac{4(3+\sqrt{7})}{(3-\sqrt{7})(3+\sqrt{7})}-\frac{21\sqrt{7}}{7}\)
\(=\frac{3(\sqrt{7}-2)}{7-2^2}+\frac{4(3+\sqrt{7})}{3^2-7}-3\sqrt{7}\)
\(=\sqrt{7}-2+2(3+\sqrt{7})-3\sqrt{7}=4\)
\(P=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\left(\frac{2x-6\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{x+3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\frac{3x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+9\right)}\right).\frac{\sqrt{x}+3}{2\left(\sqrt{x}-1\right)}\)
\(=\frac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}+3}{2\sqrt{x}-2}=\frac{-3\sqrt{x}-3}{2x-8\sqrt{x}+6}\)
Nếu đề ko sai thì đấy là kết quả
\(\frac{\left(\sqrt{x}-3\right)^2+12\sqrt{x}}{3+\sqrt{x}}=\) \(\frac{x-6\sqrt{x}+9+12\sqrt{x}}{3+\sqrt{x}}\)
\(=\frac{x+6\sqrt{x}+9}{3+\sqrt{x}}\)
\(=\frac{\left(3+\sqrt{x}\right)^2}{3+\sqrt{x}}\)
\(=3+\sqrt{x}\)
\(\frac{\left(\sqrt{x}-3\right)^2+12\sqrt{x}}{3+\sqrt{x}}\left(x\ge0\right)=\frac{x-6\sqrt{x}+9+12\sqrt{x}}{3+\sqrt{x}}\)
\(=\frac{x+\sqrt{6}+9}{3+\sqrt{x}}=\frac{\left(\sqrt{x}+3\right)^2}{3+\sqrt{x}}=3+\sqrt{x}\left(x\ge0\right)\)