Question

Bài 1: Tìm x biết: \(\left(-2\right)\cdot\left(x+1\right)-3\cdot\left(1-x\right)=4\)

Bài 2: Chứng minh rằng: \(\frac{3}{5}< \frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}< \frac{4}{5}\)

Answer 1

Bài 1 :

\(\left(-2\right)\left(x+1\right)-3\left(1-x\right)=4\)

\(\Leftrightarrow-2x-2-3+3x=4\)

\(\Leftrightarrow x=4+2+3=9\)

Bài 2 :

Cho \(S=\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}\)

\(\Leftrightarrow S=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)\)

\(+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)

\(\Rightarrow S< \left(\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\right)+\left(\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\right)\)

\(+\left(\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\right)\)

\(\Leftrightarrow S< \frac{10}{30}+\frac{10}{40}+\frac{10}{50}=\frac{47}{60}< \frac{48}{60}=\frac{4}{5}\)⁽¹⁾

Lại có :

\(S=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)\)

\(+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)

\(\Leftrightarrow S>\left(\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\right)+\left(\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\right)\)

\(+\left(\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\right)\)

\(\Leftrightarrow S>\frac{10}{40}+\frac{10}{50}+\frac{10}{60}=\frac{37}{60}>\frac{36}{60}=\frac{3}{5}\)⁽²⁾

Từ ⁽¹⁾ và ⁽²⁾ , ta có :

\(\frac{3}{5}< S< \frac{4}{5}hay\frac{3}{5}< \frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}< \frac{4}{5}\)

Answer 2

Nguyen Ribi Nkok Ngok Khôi Bùi nguyễn ngọc dinh Phùng Tuệ Minh Akai Haruma buithianhtho ?Amanda? Nguyễn Thành Trương Nguyễn Ngô Minh Trí

Answer 3

Bài 1 : x=9