Chương III : Thống kê

\(\left\{{}\begin{matrix}\dfrac{2a-9}{a+3}=\dfrac{2a+6-15}{a+3}=\dfrac{2a+6}{a+3}-\dfrac{15}{a+3}=2-\dfrac{15}{a+3}\\\dfrac{5a+17}{a+3}=\dfrac{5a+15+2}{a+3}=\dfrac{5a+15}{a+3}+\dfrac{2}{a+3}=5+\dfrac{2}{a+3}\\\dfrac{3a}{a+3}=\dfrac{3a+9-9}{a+3}=\dfrac{3a+9}{a+3}-\dfrac{9}{a+3}=3-\dfrac{9}{a+3}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}15⋮a+3\\2⋮a+3\\9⋮a+3\end{matrix}\right.\) \(\Rightarrow a+3\inƯC\left(15;2;9\right)=\pm1\)

Vậy thỏa mãn khi \(a=\pm1\)

1)\(A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}\)

\(A< \dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{99.100}\)

\(A< \dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(A< \dfrac{1}{4}-\dfrac{1}{100}\)

\(A< \dfrac{1}{4}\)(1)

\(A>\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{100.101}\)

\(A>\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{100}-\dfrac{1}{101}\)

\(A>\dfrac{1}{5}-\dfrac{1}{101}\)

\(A>\dfrac{96}{505}>\dfrac{1}{6}\)

\(A>\dfrac{1}{6}\)(2)

Từ (1) và (2)

\(\dfrac{1}{6}< A< \dfrac{1}{4}\)

2)

\(A=\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}+\dfrac{1}{95.98}\)

\(A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}+\dfrac{1}{95}-\dfrac{1}{98}\right)\)\(A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{98}\right)\)

\(A=\dfrac{1}{3}.\dfrac{24}{49}=\dfrac{8}{49}\)

b: \(\left|x-\dfrac{3}{5}\right|< \dfrac{1}{3}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{3}{5}>-\dfrac{1}{3}\\x-\dfrac{3}{5}< \dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\dfrac{4}{15}< x< \dfrac{14}{15}\)

c: \(\left|x+\dfrac{11}{2}\right|>-5.5\)

mà \(\left|x+\dfrac{11}{2}\right|\ge0\forall x\)

nên \(x\in R\)

Áp dụng bđt (1) /a-b/=/b-a/

(2) /a/+/b/ >=/a+b/ Dấu "=" xảy ra khi và chỉ khi a.b>=0

Ta có A= /x-2017/ + /x-1/= /2017-x/ + /x-1/>= /2017-x+x-1/=/2016/=2016

Dấu "=" xảy ra khi và chỉ khi (2017-x)(x-1)>=0

<=> 2017-x>=0 vs x-1>=0 hoặc 2017-x<=0 vs x-1<=0

<=> 1<= x <=2017 (t/hợp còn lại loại)

sai đề r tui làm ko đc :)

Ta có :

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)

Áp dụng t/c dãy tỉ số bawg nhau ta có :

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)

\(\Leftrightarrow\left(\dfrac{a}{b}\right)^3=\left(\dfrac{b}{c}\right)^3=\left(\dfrac{c}{d}\right)^3=\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{b}.\dfrac{a}{b}.\dfrac{a}{b}=\dfrac{b}{c}.\dfrac{b}{c}.\dfrac{b}{c}=\dfrac{c}{d}.\dfrac{c}{d}.\dfrac{c}{d}=\dfrac{a}{d}\)

\(\Leftrightarrow\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\left(đpcm\right)\)

giúp j