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\(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+\sqrt{48}}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2-\sqrt{3}\right)^2}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-20+10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}\)
\(=\sqrt{5\sqrt{3}+25-5\sqrt{3}}\)
= 5
\(\dfrac{\sqrt{3}-\sqrt{5+\sqrt{24}}+\sqrt{\sqrt{72}+11}}{\sqrt{6+\sqrt{20}}+\sqrt{2}-\sqrt{7+\sqrt{40}}}\)
\(=\dfrac{\sqrt{3}-\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(3+\sqrt{2}\right)^2}}{\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{2}-\sqrt{\left(\sqrt{2}+\sqrt{5}\right)^2}}\)
\(=\dfrac{\sqrt{3}-\sqrt{2}-\sqrt{3}+3+\sqrt{2}}{\sqrt{5}+1+\sqrt{2}-\sqrt{2}-\sqrt{5}}\)
\(=3\)
\(A=\sqrt{\sqrt{7}-\sqrt{3}}-\sqrt{\sqrt{7}+\sqrt{3}}\)
=>\(A^2=\sqrt{7}-\sqrt{3}+\sqrt{7}+\sqrt{3}-2\sqrt{4}\)
=>A^2=2căn 7-4
=>A=2căn 7-4
=>\(M=\dfrac{2\left(\sqrt{7}-2\right)}{\sqrt{7}-2}=2\)
\(A=\sqrt[3]{\left(\frac{1}{2}+\frac{1}{2}\sqrt{13}\right)^3}+\sqrt[3]{\left(\frac{1}{2}-\frac{1}{2}\sqrt{13}\right)^3}\)
\(=\frac{1}{2}+\frac{\sqrt{13}}{2}+\frac{1}{2}-\frac{\sqrt{13}}{2}=1\)
\(B=\sqrt[3]{\left(2+\sqrt{2}\right)^3}+\sqrt[3]{\left(2-\sqrt{2}\right)^3}=2+\sqrt{2}+2-\sqrt{2}=4\)
Mk sửa lại đề nha
\(A=\left(\frac{x-5\sqrt{x}}{x-25}-1\right):\left(\frac{25-x}{x+2\sqrt{x}-15}-\frac{\sqrt{x}+3}{\sqrt{x}+5}+\frac{\sqrt{x}-5}{\sqrt{x}-3}\right)\left(ĐKXĐ:x\ne25\right)\)
\(A=\left(\frac{x-5\sqrt{x}-x+25}{x-25}\right):\left(\frac{25-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}-\frac{\sqrt{x}+3}{\sqrt{x}+5}+\frac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(A=\left(\frac{25-5\sqrt{x}}{x-25}\right):\left(\frac{25-x-x+9+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(A=\left(\frac{5.\left(5-\sqrt{x}\right)}{x-25}\right):\left(\frac{9-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
Rút gọn:
\(\dfrac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)
= \(\dfrac{2\sqrt{3+\sqrt{5-\sqrt{13+4\sqrt{3}}}}}{\sqrt{6}+\sqrt{2}}\)
= \(\dfrac{2\sqrt{3+\sqrt{5-\sqrt{\left(1+2\sqrt{3}\right)^2}}}}{\sqrt{6}+\sqrt{2}}\)
= \(\dfrac{2\sqrt{3+\sqrt{5-\left(1+2\sqrt{3}\right)}}}{\sqrt{6}+\sqrt{2}}\)
= \(\dfrac{\sqrt{3+\sqrt{5-\left(1+2\sqrt{3}\right)}}.\left(\sqrt{6}-\sqrt{2}\right)}{2}\)
= \(\dfrac{\sqrt{\left[3+\sqrt{5-\left(1+2\sqrt{3}\right)}\right].6}-\sqrt{\left[3+\sqrt{5-\left(1+2\sqrt{3}\right)}\right].2}}{2}\)
= \(\dfrac{\sqrt{\left(3+\sqrt{5-1-2\sqrt{3}}\right).6}-\sqrt{\left(3+\sqrt{5-1-2\sqrt{3}}\right).2}}{2}\)
= \(\dfrac{\sqrt{\left(3+\sqrt{4-2\sqrt{3}}\right).6}-\sqrt{\left(3+\sqrt{4-2\sqrt{3}}\right).2}}{2}\)
= \(\dfrac{\sqrt{\left[3+\sqrt{\left(1-\sqrt{3}\right)^2}\right].6}-\sqrt{\left[3+\sqrt{\left(1-\sqrt{3}\right)^2}\right].2}}{2}\)
= \(\dfrac{\sqrt{\left(3+\sqrt{3}-1\right).6}-\sqrt{\left(3+\sqrt{3}-1\right).2}}{2}\)
= \(\dfrac{\sqrt{\left(2+\sqrt{3}\right).6}-\sqrt{\left(2+\sqrt{3}\right).2}}{2}\)
= \(\dfrac{\sqrt{12+6\sqrt{3}}-\sqrt{4+2\sqrt{3}}}{2}\)
= \(\dfrac{\sqrt{\left(3+\sqrt{3}\right)^2}-\sqrt{\left(1+\sqrt{3}\right)^2}}{2}\)
= \(\dfrac{3+\sqrt{3}-\left(1+\sqrt{3}\right)}{2}\)
= \(\dfrac{3+\sqrt{3}-1-\sqrt{3}}{2}\)
= \(\dfrac{2}{2}\)
= \(1\)