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giúp với giúp với giúp e với ạ. e cảm ơn
TD
6 tháng 2 2022
a)\(-1,6:\left(1+\dfrac{2}{3}\right)=-1,6:\dfrac{5}{3}=-\dfrac{8}{5}.\dfrac{3}{5}=\dfrac{-24}{25}\)
b)\(\left(\dfrac{-2}{3}\right)+\dfrac{3}{4}-\left(-\dfrac{1}{6}\right)+\left(\dfrac{-2}{5}\right)=-\dfrac{2}{3}+\dfrac{3}{4}+\dfrac{1}{6}-\dfrac{2}{5}=\dfrac{-40+45+10-24}{60}=\dfrac{-9}{60}=\dfrac{-3}{20}\)
c)\(\left(\dfrac{-3}{7}:\dfrac{2}{11}+\dfrac{-4}{7}:\dfrac{2}{11}\right).\dfrac{7}{33}=\left(\dfrac{-3}{7}.\dfrac{11}{2}+\dfrac{-4}{7}.\dfrac{11}{2}\right).\dfrac{7}{33}=\left[\dfrac{11}{2}\left(\dfrac{-3}{7}+\dfrac{-4}{7}\right)\right].\dfrac{7}{33}=\dfrac{-11}{2}.\dfrac{7}{33}=\dfrac{-7}{6}\)
d)\(\dfrac{-5}{8}+\dfrac{4}{9}:\left(\dfrac{-2}{3}\right)-\dfrac{7}{20}.\left(\dfrac{-5}{14}\right)=\dfrac{-5}{8}-\dfrac{4}{9}.\dfrac{3}{2}+\dfrac{1}{8}=\dfrac{-5}{8}+\dfrac{1}{8}-\dfrac{2}{3}=-\dfrac{7}{6}\)
HT
2
15 tháng 12 2021
a. Chu vi là \(\left(12+5\right).2=34\left(m\right)\)
Diện tích là \(12.5=60\left(m^2\right)=600000\left(cm^2\right)\)
b. Cần lát \(600000:\left(40.40\right)=375\) viên gạch
CD
1
Y
31 tháng 7 2023
\(a,-\left(m+n-k\right)+\left(m-k\right)-\left(-m+n\right)\\ =-m-n+k+m-k+m-n\\ =\left(-m+m+m\right)+\left(-n-n\right)+\left(k-k\right)\\ =m-2n\)
\(b,\left(x-y\right)-\left(x+y\right)-\left(2x-3y\right)\\ =x-y-x-y-2x+3y\\ =\left(x-x-2x\right)+\left(-y-y+3y\right)\\ =-2x+y\)
HT
12
MT
3
12 tháng 8 2016
\(\left|x+\frac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)
=> \(\left|x+\frac{4}{15}\right|-3,75=-2,15\)
=> \(\left|x+\frac{4}{15}\right|=-2,15+3,75\)
=> \(\left|x+\frac{4}{15}\right|=1,6=\frac{8}{5}\)
=> \(\orbr{\begin{cases}x+\frac{4}{15}=\frac{8}{5}\\x+\frac{4}{15}=\frac{-8}{5}\end{cases}}\)=> \(\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{-28}{15}\end{cases}}\)
PD
14
7 tháng 9 2019
Bài làm
12 . 59 + 59 . 137 - 59 . 49
= 59 . ( 12 + 137 - 49 )
= 59 . 100
= 5900
~ Dấu " . " là dấu nhân nha. ~
# Học tốt #
HV
3
10 tháng 9 2023
3d:
20<x<45
x chia 4 dư 1 nên x-1 thuộc B(4)
=>\(x-1\in\left\{0;4;...;44;48\right\}\)
=>\(x\in\left\{1;5;...;45;49\right\}\)
mà 20<x<45
nên x thuộc {21;26;31;35;41}
4:
a: A={x∈N|51<=x<=127}
b: B={x∈N|100<=x<=999}
c: C={x∈N|x=7k+5; 0<=k<=8}
\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{\left(n-1\right)n}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\\ =1-\dfrac{1}{n}\\ =\dfrac{n-1}{n}\)
\(B=\dfrac{49}{2\cdot9}+\dfrac{49}{9\cdot16}+\dfrac{49}{16\cdot23}+...+\dfrac{49}{65\cdot72}\\ =7\cdot\left(\dfrac{7}{2\cdot9}+\dfrac{7}{9\cdot16}+\dfrac{7}{16\cdot23}+...+\dfrac{7}{65\cdot72}\right)\\ =7\cdot\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{23}+...+\dfrac{1}{65}-\dfrac{1}{72}\right)\\ =7\cdot\left(\dfrac{1}{2}-\dfrac{1}{72}\right)\\ =7\cdot\dfrac{35}{72}\\ =\dfrac{245}{72}\)
\(E=\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+...+\dfrac{4}{95\cdot99}\)
\(=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{95}-\dfrac{1}{99}\)
\(=\dfrac{1}{3}-\dfrac{1}{99}=\dfrac{33}{99}-\dfrac{1}{99}=\dfrac{32}{99}\)
\(D=\dfrac{1}{7}+\dfrac{1}{91}+\dfrac{1}{247}+\dfrac{1}{475}+\dfrac{1}{775}+\dfrac{1}{1147}\)
\(=\dfrac{1}{1\cdot7}+\dfrac{1}{7\cdot13}+\dfrac{1}{13\cdot19}+\dfrac{1}{19\cdot25}+\dfrac{1}{25\cdot31}+\dfrac{1}{31\cdot37}\)
\(=\dfrac{1}{6}\left(\dfrac{6}{1\cdot7}+\dfrac{6}{7\cdot13}+...+\dfrac{6}{31\cdot37}\right)\)
\(=\dfrac{1}{6}\left(1-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+...+\dfrac{1}{31}-\dfrac{1}{37}\right)\)
\(=\dfrac{1}{6}\left(1-\dfrac{1}{37}\right)=\dfrac{1}{6}\cdot\dfrac{36}{37}=\dfrac{6}{37}\)
\(C=\dfrac{3}{1\cdot3}+\dfrac{3}{3\cdot5}+...+\dfrac{3}{49\cdot51}\)
\(=\dfrac{3}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{49\cdot51}\right)\)
\(=\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
\(=\dfrac{3}{2}\left(1-\dfrac{1}{51}\right)=\dfrac{3}{2}\cdot\dfrac{50}{51}=\dfrac{1}{17}\cdot25=\dfrac{25}{17}\)