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b: \(=\dfrac{\sqrt{20}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}-\dfrac{8}{\sqrt{5}-1}\)
\(=2\sqrt{5}-2-2\sqrt{5}\)
=-2
c: \(=\dfrac{\sqrt{4}\left(2\sqrt{2}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{3}-2\sqrt{2}\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\)
\(=\dfrac{-3}{\sqrt{6}}=-\dfrac{\sqrt{6}}{2}\)
Bài 1: Sửa \(a+b+c\le\dfrac{3}{2}\)
Áp dụng BĐT AM-GM ta có:
\(\dfrac{3}{2}\ge a+b+c\ge3\sqrt[3]{abc}\Rightarrow\sqrt[3]{abc}\le\dfrac{1}{2}\)
Áp dụng BĐT Holder ta có:
\(VT=\left(3+\dfrac{1}{a}+\dfrac{1}{b}\right)\left(3+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(3+\dfrac{1}{c}+\dfrac{1}{a}\right)\)
\(\ge\left(3+\dfrac{1}{\sqrt[3]{abc}}+\dfrac{1}{\sqrt[3]{abc}}\right)^3\ge\left(3+2+2\right)^3=343\)
Khi \(a=b=c=\dfrac{1}{2}\)
Bài 1: Một cách thuần túy Am-Gm
Biến đổi:
\(P=\frac{(3ab+a+b)(3bc+b+c)(3ac+a+c)}{(abc)^2}\)
Áp dụng BĐT AM-GM:
\(3ab+a+b=ab+ab+ab+\frac{a+b}{4}+\frac{a+b}{4}+\frac{a+b}{4}+\frac{a+b}{4}\geq 7\sqrt[7]{\frac{(ab)^3(a+b)^4}{4^4}}\)
Tương tự với các biểu thức còn lại và nhân theo vế:
\(P\geq \frac{343\sqrt[7]{\frac{(abc)^6(a+b)^4(b+c)^4(c+a)^4}{4^{12}}}}{(abc)^2}\)
Mặt khác: \((a+b)(b+c)(c+a)\geq 2\sqrt{ab}.2\sqrt{bc}.2\sqrt{ca}=8abc\) (AM-GM)
\(\Rightarrow P\geq \frac{343\sqrt[7]{\frac{(abc)^{10}}{4096}}}{(abc)^2}=343\sqrt[7]{\frac{1}{4096(abc)^4}}\)
AM-GM: \(\frac{3}{2}\geq a+b+c\geq 3\sqrt[3]{abc}\Rightarrow abc\leq \frac{1}{8}\)
Do đó mà \(P\geq 343\) khi \(a=b=c=\frac{1}{2}\)
\(a,Sửa:\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\\ =\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\dfrac{8\left(1+\sqrt{5}\right)}{-4}\\ =2\sqrt{5}-2-2\sqrt{5}=-2\\ b,=\dfrac{\sqrt{32}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\\ =\dfrac{\sqrt{2}\left(4-\sqrt{6}\right)}{\sqrt{3}\left(\sqrt{6}-4\right)}-\dfrac{1}{\sqrt{6}}=\dfrac{\sqrt{6}}{3}-\dfrac{\sqrt{6}}{6}=\dfrac{2\sqrt{6}-\sqrt{6}}{6}=\dfrac{\sqrt{6}}{6}\)
Đặt \(f\left(x\right)=10x\)
Khi đó ta có \(f\left(1\right)=10=P\left(1\right)\), \(f\left(2\right)=20=P\left(2\right)\), \(f\left(3\right)=30=P\left(3\right)\)
Do đó \(P\left(x\right)-f\left(x\right)=g\left(x\right).\left(x-1\right)\left(x-2\right)\left(x-3\right)\)
\(\Rightarrow P\left(x\right)=10+g\left(x\right).\left(x-1\right)\left(x-2\right)\left(x-3\right)\)
Vì \(P\left(x\right)\)là đa thức bậc 4 mà \(\left(x-1\right)\left(x-2\right)\left(x-3\right)\)là đa thức bậc 3 nên \(g\left(x\right)\)là đa thức bậc 1 hay \(g\left(x\right)=x+n\)
Vậy \(P\left(x\right)=\left(x+n\right)\left(x-1\right)\left(x-2\right)\left(x-3\right)+10\)
\(\Rightarrow P\left(12\right)=\left(12+n\right)\left(12-1\right)\left(12-2\right)\left(12-3\right)=\left(n+12\right).11.10.9=990\left(n+12\right)\)
\(=990n+11880\)
Và \(P\left(-8\right)=\left(-8+n\right)\left(-8-1\right)\left(-8-2\right)\left(-8-3\right)=\left(n-8\right)\left(-9\right)\left(-10\right)\left(-11\right)\)\(=-990\left(n-8\right)=-990n+7920\)
Vậy \(\frac{P\left(12\right)+P\left(-8\right)}{10}+25=\frac{990n+11880-990n+7920}{10}+25=\frac{19800}{10}+25=2005\)
\(a.\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}=\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}=2\sqrt{5}+\dfrac{8}{1-\sqrt{5}}=\dfrac{2\sqrt{5}-2}{1-\sqrt{5}}=\dfrac{-2\left(1-\sqrt{5}\right)}{1-\sqrt{5}}=-2\) \(b.\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}=\dfrac{4\sqrt{2}-2\sqrt{3}}{3\sqrt{2}-4\sqrt{3}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6.5}+\sqrt{27.6}}=\dfrac{\sqrt{2}\left(4-\sqrt{6}\right)}{\sqrt{3}\left(\sqrt{6}-4\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{17}\right)}=-\dfrac{\sqrt{2}}{\sqrt{3}}-\dfrac{1}{\sqrt{6}}=\dfrac{-2-1}{\sqrt{6}}=-\dfrac{\sqrt{3}}{\sqrt{2}}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2005}-\frac{1}{2006}\)
=> \(A=\frac{1}{1}-\frac{1}{2006}=\frac{2005}{2006}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2005.2006}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2005}-\frac{1}{2006}\)
\(A=1-\frac{1}{2006}\)
\(A=\frac{2005}{2006}\)
\(A=\sqrt{\dfrac{2\sqrt{10}+\sqrt{30}-2\sqrt{2}-\sqrt{6}}{2\sqrt{10}-2\sqrt{2}}}:\dfrac{2}{\sqrt{3}-1}\)
\(=\sqrt{\dfrac{\sqrt{10}\left(2+\sqrt{3}\right)-\sqrt{2}\left(2+\sqrt{3}\right)}{2\sqrt{2}\left(\sqrt{5}-1\right)}}.\dfrac{\sqrt{3}-1}{2}\)
\(=\sqrt{\dfrac{\left(2+\sqrt{3}\right)\sqrt{2}\left(\sqrt{5}-1\right)}{2\sqrt{2}\left(\sqrt{5}-1\right)}}.\dfrac{\sqrt{3}-1}{2}\)
\(=\dfrac{\left(2+\sqrt{3}\right)\left(\sqrt{3}-1\right)}{4}\)
\(=\dfrac{2\sqrt{3}-2+3-\sqrt{3}}{4}\)
\(=\dfrac{1+\sqrt{3}}{4}\)