a, \(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2015.2017}\)

\(\Rightarrow\) \(2S=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2015.2017}\)

\(\Rightarrow\) \(2S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\)

\(\Rightarrow\) \(2S=1-\frac{1}{2017}\)

\(\Rightarrow\) \(2S=\frac{2016}{2017}\)

\(\Rightarrow\) \(S=\frac{1008}{2017}\)

\(P=\left(b+c+d\right)\left(\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)=1+\frac{b}{c}+\frac{b}{d}+\frac{c}{b}+1+\frac{c}{d}+\frac{d}{b}+\frac{d}{c}+1\)

\(=3+\frac{b}{c}+\frac{b}{d}+\frac{c}{d}+\frac{c}{b}+\frac{d}{b}+\frac{d}{c}\)

Mặt khác do \(b\le c\le d\Rightarrow\left(d-c\right)\left(c-b\right)\ge0\)

\(\Leftrightarrow cd-bd-c^2+bc\ge0\Leftrightarrow bc+cd\ge c^2+bd\)

\(\Leftrightarrow\frac{bc+cd}{cd}\ge\frac{c^2+bd}{cd}\Leftrightarrow\frac{b}{d}+1\ge\frac{c}{d}+\frac{b}{c}\)

\(\frac{bc+cd}{bc}\ge\frac{c^2+bd}{bc}\Leftrightarrow\frac{d}{b}+1\ge\frac{c}{b}+\frac{d}{c}\)

\(\Leftrightarrow\frac{b}{d}+\frac{d}{b}+2\ge\frac{b}{c}+\frac{c}{d}+\frac{c}{b}+\frac{d}{c}\)

\(\Leftrightarrow2\left(\frac{b}{d}+\frac{d}{b}\right)+2\ge\frac{b}{c}+\frac{b}{d}+\frac{c}{d}+\frac{c}{b}+\frac{d}{b}+\frac{d}{c}=P\)

Mà \(a\le b\le d\le2a\Rightarrow\left\{{}\begin{matrix}\frac{1}{2}\le\frac{b}{d}\le1\\1\le\frac{d}{b}\le2\end{matrix}\right.\)

\(\Rightarrow\left(\frac{b}{d}-1\right)\left(\frac{d}{b}-2\right)\ge0\Leftrightarrow1-2\frac{b}{d}-\frac{d}{b}+2\ge0\)

\(\Leftrightarrow\frac{b}{d}+\frac{d}{b}\le3-\frac{b}{d}\le3-\frac{1}{2}=\frac{5}{2}\)

\(\Rightarrow P\le2.\frac{5}{2}+2=7\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}b=c=a\\d=2a\end{matrix}\right.\)

S = a+b+c + (1/a + 1/b + 1/c)

   >= (a+b+c) + 9/a+b+c

    = [ (a+b+c) + 9/4.(a+b+c) ] + 27/4.(a+b+c)

   >= \(2\sqrt{\left(a+b+c\right).\frac{9}{4.\left(a+b+c\right)}}\)   +    27/(4.3/2)

     = 3 + 9/2

     = 15/2

Dấu "=" xảy ra <=> a=b=c=1/2

Vậy ......

Tk mk nha

bài này còn có thể theo phương pháp chọn điểm rơi trong bài toán cực trị, bạn thử tìm hiểu nhé!!!!

câu 1:

\(CTHH:Na_2SO_3\)

câu 2:

\(CTHH:FeSO_4\)

a) ta có:

\(\frac{-1}{2}-1\le x\le\frac{1}{2}.3\)

hay \(-1,5\le x\le1,5\)

vì x\(\in Z\) nên ta chọn x=-1,0,1

ta có:

3S=\(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\)

3S-S=\(\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^9}\right)\)

2S=1-\(\frac{1}{3^9}\)

s=\(\left(1-\frac{1}{3^9}\right):2\)

dau= xay ra khi (ab=2acbd+bd^2=bd*(2ac+bd))

sao ab phải = 2abd+(bd)^2

chứ hông phải \(2abd+\left(bd\right)^2=0\)hả

why???