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)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: \(\dfrac{33}{x}=\dfrac{y}{8}=\dfrac{z}{160}=\dfrac{45}{120}\)
=>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
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).
Câu 1:
Ta có: \(A=\frac{12n+1}{2n+3}=\frac{\left(12n+18\right)-17}{2n+3}=6-\frac{17}{2n+3}\)
Để A là một số nguyên thì \(2n+3\inƯ\left(17\right)=\left\{\pm1;\pm17\right\}\)
\(\Leftrightarrow2n\in\left\{-20;-4;-2;14\right\}\Rightarrow n\in\left\{-10;-2;-1;7\right\}\)
Để A là một phân số thì \(n\notin\left\{-10;-2;-1;7;-\frac{3}{2}\right\}\)
Vậy ...
=\(\dfrac{-10}{36}+\dfrac{20}{36}-\dfrac{11}{36}=\dfrac{-10+20-11}{36}=\dfrac{-1}{36}\)
Bài 1:
a) \(\dfrac{-5}{18}+\dfrac{5}{9}-\dfrac{11}{36}=\dfrac{-10}{36}+\dfrac{20}{36}-\dfrac{11}{36}\)\(=\dfrac{-1}{36}\)
b) \(\dfrac{-39}{44}:1\dfrac{2}{11}=\dfrac{-39}{44}:\dfrac{13}{11}=\dfrac{-39}{44}.\dfrac{11}{13}=\dfrac{-3}{4}\)
c) \(\dfrac{-7}{11}.\dfrac{11}{19}+\dfrac{-7}{11}.\dfrac{8}{19}+\dfrac{-4}{11}=\dfrac{-7}{11}.\left(\dfrac{11}{19}+\dfrac{8}{19}\right)+\dfrac{-4}{11}=\dfrac{-7}{11}.1+\dfrac{-4}{11}=-1\)
Bài 2:
a) \(x+\dfrac{2}{5}=-\dfrac{11}{15}\)
\(\rightarrow x=-\dfrac{11}{15}-\dfrac{2}{5}=-\dfrac{11}{15}-\dfrac{6}{15}=\dfrac{-17}{15}\)
b) \(\left(x-\dfrac{7}{18}\right).\dfrac{18}{29}=-\dfrac{12}{29}\)
\(x-\dfrac{7}{18}=-\dfrac{12}{29}:\dfrac{18}{29}\)
\(x-\dfrac{7}{18}=-\dfrac{12}{29}.\dfrac{29}{18}=-\dfrac{12}{18}\)
\(x=\dfrac{-12}{18}+\dfrac{7}{18}=\dfrac{-5}{18}\)