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Bấm mt là ra mà bạn :v
Ta có: \(\left\{{}\begin{matrix}7A=\dfrac{7^{2013}+7}{7^{2013}+1}=1+\dfrac{6}{7^{2013}+1}\\7B=\dfrac{7^{2014}+7}{7^{2014}+1}=1+\dfrac{6}{7^{2014}+1}\end{matrix}\right.\Leftrightarrow7A>7B\Leftrightarrow A>B\)
\(^{P\left(x\right)=x^{2018}-100x^{2017}+100x^{2016}-...+100x+2016}\) \(^{P\left(99\right)=x^{2018}-\left(99+1\right)x^{2017}+\left(99+1\right)x^{2016}-...+\left(99+1\right)x+2016}\) \(^{P\left(99\right)=x^{2018}-x^{2018}-x^{2017}+x^{2017}+x^{2016}-...+x^2+x+2016}\) \(^{P\left(99\right)=x+2016=99+2016=2115}\)
\(A\left(x\right)=2x^2-11x+5=2\left(x^2-\dfrac{11}{2}x+\dfrac{5}{2}\right)=2\left(x^2-\dfrac{11}{2}x+\dfrac{121}{8}-\dfrac{101}{8}\right)=2\left(x-\dfrac{11}{4}\right)^2-\dfrac{101}{4}\ge-\dfrac{101}{4}\)
1 cách giải khác:v \(x^2-1=2y^2\Leftrightarrow x^2=2y^2+1\) Suy ra \(x^2\) là số lẻ \(\Leftrightarrow x\) lẻ.Đặt: \(x=2l+1\) ta được: \(\left(2l+1\right)^2=2y^2+1\Leftrightarrow4l^2+4l+1=2y^2+1\) \(\Rightarrow4l^2+4l=2y^2\Leftrightarrow2\left(2l^2+2l\right)=2y^2\Leftrightarrow2\left(l^2+l\right)=y^2\).Suy ra \(y^2\) chẵn suy ra y là số nguyên tố chẵn. Hay y=2( snt chẵn duy nhất là 2). Thay vào tìm được x=3
\(a^3b^3+2b^3c^3+3a^3c^3\) \(=a^3b^3+2b^3c^3+2a^3c^3+a^3c^3\) \(=a^3\left(b^3+c^3\right)+2c^3\left(a^3+b^3\right)\) \(=-a^6-2c^6\le0\) (đúng) .Dấu "=" khi: \(a=b=c=0\)
Áp dụng bất đẳng thức Cauchy-Schwarz: \(S=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{4}{c}+\dfrac{16}{d}\ge\dfrac{\left(1+1+2+4\right)^2}{a+b+c+d}=\dfrac{64}{8}=8\)
\(NL=x^2-x+1=x^2-x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)Dấu "=" xảy ra khi: \(x=\dfrac{1}{2}\)
Áp dụng bất đẳng thức Cauchy-Schwarz ta có: \(NL=\dfrac{x}{y+1}+\dfrac{y}{x+1}=\dfrac{x^2}{xy+x}+\dfrac{y^2}{xy+y}\ge\dfrac{\left(x+y\right)^2}{2xy+x+y}=\dfrac{1}{2xy+1}\) Mặt khác theo AM-GM: \(xy\le\dfrac{\left(x+y\right)^2}{4}\Leftrightarrow2xy\le\dfrac{\left(x+y\right)^2}{2}\Leftrightarrow2xy\le\dfrac{1}{2}\) hay \(NL\ge\dfrac{1}{2xy+1}\ge\dfrac{1}{\dfrac{1}{2}+1}=\dfrac{1}{\dfrac{3}{2}}=\dfrac{2}{3}\).Dấu "=" xảy ra khi: \(x=y=\dfrac{1}{2}\)
Áp dụng liên tiếp Bunyakovsky và Cauchy-Schwarz ta được: \(NL^2=\left(\sqrt{\dfrac{a}{2a+b+c}}+\sqrt{\dfrac{b}{2b+a+c}}+\sqrt{\dfrac{c}{2c+a+b}}\right)^2\) \(\le\left(1^2+1^2+1^2\right)\left(\dfrac{a}{2a+b+c}+\dfrac{b}{2b+a+c}+\dfrac{c}{2c+a+b}\right)=3\left(\dfrac{a}{2a+b+c}+\dfrac{b}{2b+a+c}+\dfrac{c}{2c+a+b}\right)\) \(=3\left(\dfrac{a}{\left(a+b\right)+\left(a+c\right)}+\dfrac{b}{\left(a+b\right)+\left(b+c\right)}+\dfrac{c}{\left(a+c\right)+\left(b+c\right)}\right)\le\dfrac{3}{4}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}+\dfrac{b}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{a+c}+\dfrac{c}{b+c}\right)=\dfrac{3}{4}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{a+c}{a+c}\right)=\dfrac{3}{4}.3=\dfrac{9}{4}\)\(NL^2\le\dfrac{9}{4}\Leftrightarrow NL\le\dfrac{3}{2}\).Dấu "=" khi \(a=b=c\)