Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A=3/2-5/6+/12-9/20+11/30-13/42+15/56-17/72+19/90
A=11/10
hok tốt nha
a: \(\dfrac{33}{x}=\dfrac{y}{8}=\dfrac{z}{160}=\dfrac{45}{120}\)
=>33/x=y/8=z/160=3/8
=>x=88; y=33; z=60
b: \(\dfrac{x}{3}=\dfrac{14}{y}=\dfrac{z}{60}=\dfrac{-8}{12}=-\dfrac{2}{3}\)
nên x=-2; y=-21; z=-40
a)121212/424242=2/7
1999999999/9999999995=1/5
Sorry bạn mik chỉ bt làm câu a thôi!
HT~
Câu b:
\(\frac{a}{b}:\frac{c}{d}=\frac{ad}{bc}=\frac{6}{5}\Leftrightarrow5ad=6bc\)
\(\frac{a}{b}-\frac{c}{d}=\frac{ad-bc}{bd}=\frac{1}{15}\Leftrightarrow5\left(ad-bc\right)=\frac{bd}{3}\)
\(\Rightarrow5ad-5bc=\frac{bd}{3}\)
Thay vào ta có:
\(\frac{a}{b}-\frac{c}{d}=\frac{a}{b}-\frac{1}{3}=\frac{1}{15}\Leftrightarrow\frac{a}{b}=-\frac{4}{15}\)
a) -12 => -11 => -10 => -9 => -8.
b) -1 => -11 => -5 => -3 => -1
..............................................................................................................................................................................................
Có \(P=\frac{1}{2}\times\frac{3}{4}\times\frac{5}{6}\times...\times\frac{399}{400}< \frac{2}{3}\times\frac{4}{5}\times...\times\frac{400}{401}\)
=> \(P^2< \frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times...\times\frac{400}{401}=\frac{1}{401}< \frac{1}{400}=\frac{1}{20}\)
=> \(P< \frac{1}{20}\)(đpcm).
a)Quy đồng: \(\frac{5}{8}=\frac{5.3}{8.3}=\frac{15}{24}\)
Vì \(\frac{5}{24}< \frac{10+5}{24}=\frac{15}{24}\)
\(\Rightarrow\frac{5}{24}< \frac{5+10}{24}=\frac{5}{8}\)
b) Quy đồng:
\(\frac{4}{9}=\frac{4.6}{9.6}=\frac{24}{9.6}\)
\(\frac{2}{3}=\frac{2.18}{3.18}=\frac{36}{9.6}\)
Vì \(\frac{36}{9.6}>\frac{24}{9.6}>\frac{6+9}{9.6}\)
\(\Rightarrow\frac{2}{3}>\frac{4}{9}>\frac{6+9}{6.9}\)
241/210
A = \(\dfrac{3}{2}-\dfrac{5}{6}+\dfrac{7}{12}-\dfrac{9}{20}+\dfrac{11}{30}-\dfrac{13}{42}+\dfrac{15}{56}-\dfrac{17}{72}+\dfrac{19}{90}\)
A \(=\dfrac{3}{1\cdot2}-\dfrac{5}{2\cdot3}+\dfrac{7}{3\cdot4}-\dfrac{9}{4\cdot5}+\dfrac{11}{5\cdot6}-\dfrac{13}{6\cdot7}+\dfrac{15}{7\cdot8}-\dfrac{17}{8\cdot9}+\dfrac{19}{9\cdot10}\)
A
\(=\left(1+\dfrac{1}{2}\right)+\left(\dfrac{1}{2}+\dfrac{2}{3}\right)+\left(\dfrac{1}{3}+\dfrac{3}{4}\right)+\left(\dfrac{1}{4}+\dfrac{1}{5}\right)+\left(\dfrac{1}{5}+\dfrac{1}{6}\right)+\left(\dfrac{1}{6}+\dfrac{1}{7}\right)+\left(\dfrac{1}{7}+\dfrac{7}{8}\right)+\left(\dfrac{1}{8}+\dfrac{1}{9}\right)+\left(\dfrac{1}{9}+\dfrac{1}{10}\right)\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
A = \(1-\dfrac{1}{10}=\dfrac{9}{10}\)