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Ta có:
\(S=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}\)
\(\Rightarrow S=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)
Nhận xét:
\(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}>\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}=\frac{1}{4}\)
\(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}>\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}=\frac{1}{5}\)
\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}=\frac{1}{6}\)
\(\Rightarrow S>\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{37}{60}>\frac{3}{5}\)
\(\Rightarrow S>\frac{3}{5}\left(1\right)\)
Lại có:
\(S=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)
Nhận xét:
\(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}< \frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{1}{3}\)
\(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}< \frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}=\frac{1}{4}\)
\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}< \frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}=\frac{1}{5}\)
\(\Rightarrow S< \frac{1}{3}+\frac{1}{4}+\frac{1}{5}=\frac{47}{60}< \frac{4}{5}\)
\(\Rightarrow S< \frac{4}{5}\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\)
\(\Rightarrow\frac{3}{5}< S< \frac{4}{5}\) (Đpcm)
dễ quá
ha ha ha
c) \(\dfrac{x+1}{35}+\dfrac{x+2}{34}+\dfrac{x+3}{33}=\dfrac{x+4}{32}+\dfrac{x+5}{31}+\dfrac{x+6}{30}\)
\(\Rightarrow\dfrac{x+1}{35}+1+\dfrac{x+2}{34}+1+\dfrac{x+3}{33}+1=\dfrac{x+4}{32}+1+\dfrac{x+5}{31}+1+\dfrac{x+6}{30}+1\)
\(\Rightarrow\dfrac{x+1+35}{35}+\dfrac{x+2+34}{34}+\dfrac{x+3+33}{33}=\dfrac{x+4+32}{32}+\dfrac{x+5+31}{31}+\dfrac{x+6+30}{30}\)
\(\Rightarrow\dfrac{x+36}{35}+\dfrac{x+36}{34}+\dfrac{x+36}{33}=\dfrac{x+36}{32}+\dfrac{x+36}{31}+\dfrac{x+36}{30}\)
\(\Rightarrow\dfrac{x+36}{35}+\dfrac{x+36}{34}+\dfrac{x+36}{33}-\dfrac{x+36}{32}-\dfrac{x+36}{31}-\dfrac{x+36}{30}=0\)
\(\Rightarrow\left(x+36\right)\left(\dfrac{1}{35}+\dfrac{1}{34}+\dfrac{1}{33}+\dfrac{1}{32}+\dfrac{1}{31}+\dfrac{1}{30}\right)=0\)
\(\Rightarrow x+36=0\left(\text{vì }\dfrac{1}{35}+\dfrac{1}{34}+\dfrac{1}{33}+\dfrac{1}{32}+\dfrac{1}{31}+\dfrac{1}{30}\ne0\right)\)
\(\Rightarrow x=-36\)
Vậy ...
a/ Ta có: \(-4\dfrac{3}{5}.2\dfrac{4}{3}\le x\le-2\dfrac{3}{5}:1\dfrac{6}{15}\)
\(\Rightarrow\dfrac{-23}{5}.\dfrac{10}{3}\le x\le\dfrac{-13}{5}:\dfrac{21}{15}\)
\(\Rightarrow\dfrac{-46}{3}\le x\le\dfrac{-13}{5}.\dfrac{15}{21}\)
\(\Rightarrow\dfrac{-46}{3}\le x\le\dfrac{-13}{7}\)
\(\Rightarrow-15,\left(3\right)\le x\le-1,\left(857142\right)\)
Vì x \(\in\) Z nên x \(\in\left\{-1;-2;-3;...;-15\right\}\)
Chúc bạn học tốt!!!
cau 1
de a dat gia tri lon nhat suy ra5a-17/4a-23 lon nhat
suy ra 4a-23 phai nho nhat khac 0 va la so nguyen duong
suy ra 4a-23=1
suy ra 4a=1+23=24
suy ra a=24 chia 4=6
vay de a nho nhat thi a=6
a: \(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}< x< \dfrac{1}{48}-\dfrac{1}{16}+\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{6}{12}-\dfrac{4}{12}-\dfrac{3}{12}< x< \dfrac{1}{48}-\dfrac{3}{48}+\dfrac{8}{48}\)
\(\Leftrightarrow\dfrac{-1}{12}< x< \dfrac{1}{8}\)
\(\Leftrightarrow-2< 24x< 3\)
=>x=0
b: \(\Leftrightarrow\dfrac{9-10}{12}< \dfrac{x}{12}< 1-\dfrac{8-3}{12}=\dfrac{7}{12}\)
=>-1<x<7
hay \(x\in\left\{0;1;2;3;4;5;6\right\}\)
9) \(\dfrac{x}{4}=\dfrac{9}{x}\)
Theo định nghĩa về hai phân số bằng nhau, ta có:
\(4\cdot9=x^2\\ 36=x^2\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
8)
\(x:\dfrac{5}{3}+\dfrac{1}{3}=-\dfrac{2}{5}\\ x:\dfrac{5}{3}=-\dfrac{2}{5}+\dfrac{1}{3}\\ x:\dfrac{5}{3}=-\dfrac{1}{15}\\ x=\dfrac{1}{15}\cdot\dfrac{5}{3}\\ x=\dfrac{1}{9}\)
7)
\(2x-16=40+x\\ 2x-x=40+16\\ x\left(2-1\right)=56\\ x=56\)
6)
\(1\dfrac{1}{2}+x=\dfrac{3}{2}-7\\ \dfrac{3}{2}+x=\dfrac{3}{2}-7\\ \dfrac{3}{2}-\dfrac{3}{2}=-7-x\\ -7-x=0\\ x=-7-0\\ x=-7\)
5)
\(3\dfrac{1}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\\ \dfrac{7}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\\ \dfrac{1}{2}x=\dfrac{7}{2}-\dfrac{2}{3}\\ \dfrac{1}{2}x=\dfrac{17}{6}\\ x=\dfrac{17}{6}:\dfrac{1}{2}\\ x=\dfrac{17}{3}\)
4)
\(x\cdot\left(x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
3)
\(\left(\dfrac{2x}{5}+2\right):\left(-4\right)=-1\dfrac{1}{2}\\ \left(\dfrac{2x}{5}+2\right):\left(-4\right)=-\dfrac{3}{2}\\ \dfrac{2x}{5}+2=-\dfrac{3}{2}\cdot\left(-4\right)\\ \dfrac{2x}{5}+2=6\\ \dfrac{2x}{5}=6-2\\ \dfrac{2x}{5}=4\\ 2x=4\cdot5\\ 2x=20\\ x=20:2\\ x=10\)
2)
\(\dfrac{1}{3}+\dfrac{1}{2}:x=-0,25\\ \dfrac{1}{3}+\dfrac{1}{2}:x=-\dfrac{1}{4}\\ \dfrac{1}{2}:x=-\dfrac{1}{4}-\dfrac{1}{3}\\ \dfrac{1}{2}:x=-\dfrac{7}{12}\\ x=\dfrac{1}{2}:-\dfrac{7}{12}\\ x=-\dfrac{6}{7}\)
1)
\(\dfrac{4}{3}+x=\dfrac{2}{15}\\ x=\dfrac{2}{15}-\dfrac{4}{3}x=-\dfrac{6}{5}\)