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Xét \(A^2=\left(\sqrt{x-1}+\sqrt{2x^2-5x+7}\right)^2\)
\(A^2=x-1+2x^2-5x+7+2\sqrt{\left(x-1\right)\left(2x^2-5x+7\right)}\)
\(A^2=2x^2-4x+6+2\sqrt{\left(x-1\right)\left(2x^2-5x+7\right)}\)
\(A^2=2\left(x-1\right)^2+4+2\sqrt{\left(x-1\right)\left(2x^2-5+7\right)}\)
\(A^2\ge4\Rightarrow A\ge2\)
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) \(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{4+4\sqrt{3}+3}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{25-10\sqrt{3}+3}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}\)
\(=\sqrt{5\sqrt{3}+25-5\sqrt{3}}\)
\(=\sqrt{25}=5\)
\(\dfrac{\sqrt{\dfrac{-\left(2\right)^5}{5^3.5^2}.\dfrac{-\left(5\right)^3}{2^9}.5^2}}{\sqrt[3]{\dfrac{-\left(3\right)^3}{2^6}.\dfrac{\left(5\right)^2}{3^2.2^5}.\dfrac{\left(5\right)^4}{3^4}}}=\dfrac{\sqrt{\dfrac{1}{2^4}}}{\sqrt[3]{\dfrac{-\left(5\right)^6}{2^{12}.3^3}}}=\dfrac{\dfrac{1}{4}}{\sqrt[3]{\left(\dfrac{-5^2}{2^4.3}\right)^3}}=\dfrac{\dfrac{1}{4}}{\dfrac{-25}{48}}=\dfrac{-12}{25}\)
;v Đề tuyển sinh là theo mỗi tỉnh ;v searrch gg tỉnh nào mà chẳng có =))
\(\sqrt{1-x-2x^2}=\sqrt{\left(1+x\right)\left(1-2x\right)}\le\dfrac{1+x-2x+1}{2}=\dfrac{-x+2}{2}\)
(AM-GM)
do đó \(A\le\dfrac{x}{2}+\dfrac{-x+2}{2}=1\)
Dấu = xảy ra khi 1+x=1-2x <=> x=0 (tmđk)
m=-5/4 đó bạn
bạn giải như thế nào vậy