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\(S=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1+1-\frac{1}{50}\)
\(=2-\frac{1}{50}< 2\)
\(S=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
\(=1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
\(\Rightarrow S< 1+\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...\frac{1}{49\cdot50}\right)\)
\(S< 1+\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(S< 1+\left(1-\frac{1}{50}\right)\)
Mà \(1-\frac{1}{50}< 1\Rightarrow1+\left(1-\frac{1}{50}\right)< 2\)( ĐPCM )
Bài 1 :
Ta có : \(S=\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\)
\(=\frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b}\)
\(=\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\)
Ta chứng minh BĐT \(\frac{x}{y}+\frac{y}{x}\ge2,\forall x,y>0\)
Thật vậy : BĐT \(\Leftrightarrow\frac{x}{y}+\frac{y}{x}-2=\frac{\left(x-y\right)^2}{xy}\ge0\) ( đúng )
Vậy \(\frac{x}{y}+\frac{y}{x}\ge2,\forall x,y>0\)
Áp dụng vào bài toán ta có : \(S\ge2+2+2=6\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)
Vậy min \(S=6\) tại \(a=b=c\)
\(\frac{1}{4^2}>0;\frac{1}{5^2}>0;...;\frac{1}{50^2}>0\Rightarrow S>0\)
\(\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{50^2}< \frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{49\cdot50}\)
\(\Leftrightarrow\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{50^2}< \frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{49}-\frac{1}{50}\)
\(\Leftrightarrow\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{50^2}< \frac{1}{3}-\frac{1}{50}\)
\(\Leftrightarrow\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{50^2}< \frac{47}{150}< 1\)
=> 0 < S < 1 => S không phải số nguyên
TA CÓ Vế trái <\(\frac{1}{1}+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}\)\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{49}-\frac{1}{50}\)
\(=2-\frac{1}{50}< 2\)
do đó VT <2(dpcm)
Đề: Chứng minh S < 2
\(S=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{50^2}\)
\(S=\frac{1}{1\times1}+\frac{1}{2\times2}+\frac{1}{3\times3}+\frac{1}{4\times4}+.....+\frac{1}{50\times50}\)
\(S< \frac{1}{1\times1}+\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+.....+\frac{1}{49\times50}\)
\(S< 1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{49}-\frac{1}{50}\)
\(S< 2-\frac{1}{50}< 2\)
Chúc bạn học tốt
p=\(\frac{1}{49}+\frac{2}{48}+\frac{3}{47}+...+\frac{48}{2}+49\)
=\(\left(\frac{1}{49}+1\right)+\left(\frac{2}{48}+1\right)+\left(1+\frac{3}{47}\right)+...+\left(1+\frac{48}{2}\right)+\frac{50}{50}\)
=\(\frac{50}{50}+\frac{50}{49}+\frac{50}{48}+...+\frac{50}{2}\)
=\(50\left(\frac{1}{50}+\frac{1}{49}+\frac{1}{48}+...+\frac{1}{2}\right)\)
p=50*S
\(\frac{S}{\text{p}}=\frac{1}{50}\)
Giải:
\(S=\dfrac{1}{50}+\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{98}+\dfrac{1}{99}\)
\(S=\left(\dfrac{1}{50}+\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{74}\right)+\left(\dfrac{1}{75}+...+\dfrac{1}{98}+\dfrac{1}{99}\right)\)
\(\Rightarrow S>\left(\dfrac{1}{50}+\dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}\right)+\left(\dfrac{1}{75}+...+\dfrac{1}{75}+\dfrac{1}{75}\right)\)
\(\Rightarrow S>\dfrac{1}{2}+\dfrac{1}{3}>\dfrac{1}{2}\)
\(\Rightarrow S>\dfrac{1}{2}\left(đpcm\right)\)
\(S=\frac{1}{4}+\left(\frac{1}{3^2}+\frac{1}{4^2}+..+\frac{1}{50^2}\right)\)
Xét \(A=\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\)
\(A< \frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A< \frac{1}{2}-\frac{1}{50}< \frac{1}{2}\)
\(=>A< \frac{1}{2}\)
=>\(S=\frac{1}{4}+A< \frac{1}{4}+\frac{1}{2}=\frac{3}{4}\)
vậy S<3/4