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\(\dfrac{\sqrt{\dfrac{-\left(2\right)^5}{5^3.5^2}.\dfrac{-\left(5\right)^3}{2^9}.5^2}}{\sqrt[3]{\dfrac{-\left(3\right)^3}{2^6}.\dfrac{\left(5\right)^2}{3^2.2^5}.\dfrac{\left(5\right)^4}{3^4}}}=\dfrac{\sqrt{\dfrac{1}{2^4}}}{\sqrt[3]{\dfrac{-\left(5\right)^6}{2^{12}.3^3}}}=\dfrac{\dfrac{1}{4}}{\sqrt[3]{\left(\dfrac{-5^2}{2^4.3}\right)^3}}=\dfrac{\dfrac{1}{4}}{\dfrac{-25}{48}}=\dfrac{-12}{25}\)
;v Đề tuyển sinh là theo mỗi tỉnh ;v searrch gg tỉnh nào mà chẳng có =))
1a) \(\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)
\(=\sqrt{4+\sqrt{8}}.\sqrt{\left(2+\sqrt{2+\sqrt{2}}\right)\left(\sqrt{2-\sqrt{2+\sqrt{2}}}\right)}\)
\(=\sqrt{4+\sqrt{8}}.\sqrt{4-2-\sqrt{2}}\)
\(=\sqrt{4+\sqrt{8}}.\sqrt{2-\sqrt{2}}=\sqrt{\left(4+\sqrt{8}\right)\left(2-\sqrt{2}\right)}\)
\(=\sqrt{8-4\sqrt{2}-\sqrt{16}+2\sqrt{8}}\)
\(=\sqrt{8-4\sqrt{2}-4+4\sqrt{2}}\)
\(=\sqrt{4}=2\)
1b) \(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{4+4\sqrt{3}+3}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{25-10\sqrt{3}+3}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}\)
\(=\sqrt{5\sqrt{3}+25-5\sqrt{3}}\)
\(=\sqrt{25}=5\)
ĐẶT x = \(\sqrt{3}\)
\(\dfrac{\sqrt{3}}{1-\sqrt{\sqrt{3}+1}}+\dfrac{\sqrt{3}}{1+\sqrt{\sqrt{3}+1}}\)
\(\Leftrightarrow\dfrac{x}{1-\sqrt{x+1}}+\dfrac{x}{1+\sqrt{x+1}}\)
\(\Leftrightarrow\dfrac{x+x\sqrt{x+1}+x-x\sqrt{x+1}}{\left(1-\sqrt{x+1}\right).\left(1+\sqrt{x+1}\right)}\)
\(\Leftrightarrow\dfrac{2x}{1-x-1}\)
\(\Leftrightarrow\dfrac{2x}{-x}\) = -2
Mình mới làm quen toán 9, có gì sai sót mong bạn thông cảm. Chúc bạn học tốt :))
\(\dfrac{2+\sqrt{2}}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}\)
\(=\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}+\dfrac{\sqrt{3}-\sqrt{2}}{3-2}\)
\(=\sqrt{2}+\sqrt{3}-\sqrt{2}=\sqrt{3}\)
c.ơn bn nhiều lắm