Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(1-\left(\frac{12}{5}+y=\frac{8}{9}\right):\frac{16}{9}=0\)
\(1-\left(\frac{12}{5}+y-\frac{8}{9}\right)=0\times\frac{16}{9}\)
\(1-\left(\frac{12}{5}+y-\frac{8}{9}\right)=0\)
\(\frac{12}{5}+y-\frac{8}{9}=1-0\)
\(\frac{12}{5}-y+\frac{8}{9}=1\)
\(\frac{12}{5}-y=1-\frac{8}{9}\)
\(\frac{12}{5}-y=\frac{1}{9}\)
\(y=\frac{12}{5}-\frac{1}{9}\)
\(y=\frac{108}{45}-\frac{5}{45}\)
\(y=\frac{103}{45}\)
a: 4/9+3/7=28/63+27/63=55/63
3/4+7/24=18/24+7/24=25/24
1/3+2/9+4/27=9/27+6/27+4/27=19/27
b: 5/6-3/8=20/24-9/24=11/24
7/15-11/30=14/30-11/30=3/30=1/10
2/3+1/6-7/12
=8/12+2/12-7/12
=3/12=1/4
c: 18/25*15/6=15/25*18/6=3*3/5=9/5
30/49:6/7=30/49*7/6=210/294=5/7
1/2*3/4:6/5=3/8*5/6=15/48=5/16
d: 8*3/5:12/5=24/5*5/12=2
4:9/5:10/3=4*5/9*3/10=2/3
a) \(\dfrac{12}{20}\)
b)\(\dfrac{6}{21}\)
c)\(\dfrac{3}{4}\)
d) 6 ; \(\dfrac{3}{4}\)
a) \(\dfrac{3}{4}-\dfrac{1}{8}=\dfrac{6}{8}-\dfrac{1}{8}=\dfrac{6-1}{8}=\dfrac{5}{8}\)
b) \(\dfrac{2}{6}-\dfrac{5}{18}=\dfrac{6}{18}-\dfrac{5}{18}=\dfrac{6-5}{18}=\dfrac{1}{18}\)
c) \(\dfrac{2}{5}-\dfrac{3}{20}=\dfrac{8}{20}-\dfrac{3}{20}=\dfrac{8-3}{20}=\dfrac{5}{20}=\dfrac{1}{4}\)
a: Ta có:
\(\left(\dfrac{2}{5}+\dfrac{1}{5}\right)+\dfrac{1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
\(\dfrac{2}{5}+\left(\dfrac{1}{5}+\dfrac{1}{5}\right)=\dfrac{2}{5}+\dfrac{2}{5}=\dfrac{4}{5}\)
\(\dfrac{4}{5}=\dfrac{4}{5}\). Vậy \(\left(\dfrac{2}{5}+\dfrac{1}{5}\right)+\dfrac{1}{5}=\dfrac{2}{5}+\left(\dfrac{1}{5}+\dfrac{1}{5}\right)\)
Ta có:
\(\left(\dfrac{2}{9}+\dfrac{5}{9}\right)+\dfrac{1}{9}=\dfrac{7}{9}+\dfrac{1}{9}=\dfrac{8}{9}\)
\(\dfrac{2}{9}+\left(\dfrac{5}{9}+\dfrac{1}{9}\right)=\dfrac{2}{9}+\dfrac{6}{9}=\dfrac{8}{9}\)
\(\dfrac{8}{9}=\dfrac{8}{9}\). Vậy \(\left(\dfrac{2}{9}+\dfrac{5}{9}\right)+\dfrac{1}{9}=\dfrac{2}{9}+\left(\dfrac{5}{9}+\dfrac{1}{9}\right)\)
b: \(\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\dfrac{4}{3}=\dfrac{3}{3}+\dfrac{4}{3}=\dfrac{7}{3}\)
\(\dfrac{1}{3}+\left(\dfrac{2}{3}+\dfrac{4}{3}\right)=\dfrac{1}{3}+\dfrac{6}{3}=\dfrac{7}{3}\)
\(\dfrac{7}{3}=\dfrac{7}{3}\). Vậy \(\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\dfrac{4}{3}=\dfrac{1}{3}+\left(\dfrac{2}{3}+\dfrac{4}{3}\right)\)
Đề của anh bị sai mới đúng chứ ạ? Anh Đạt ghi là \(\left(\dfrac{2}{9}+\dfrac{5}{9}\right)+\dfrac{1}{9}\) chứ có phải \(\dfrac{2}{5}\) đâu ạ?
\(2:\dfrac{2}{7}+\dfrac{12}{5}=2.\dfrac{7}{2}+\dfrac{12}{5}=7+\dfrac{12}{5}=\dfrac{47}{5}\)
\(\dfrac{5}{16}:\left[\dfrac{5}{8}+\dfrac{3}{24}\right]=\dfrac{5}{16}:\dfrac{3}{4}=\dfrac{5}{16}.\dfrac{4}{3}=\dfrac{5}{12}\)
\(2:\dfrac{2}{7}+\dfrac{12}{5}=7\dfrac{12}{5}=\dfrac{47}{5}\)
\(\dfrac{5}{16}:\left(\dfrac{5}{8}+\dfrac{3}{24}\right)=\dfrac{5}{16}:\dfrac{3}{4}=\dfrac{5}{16}\cdot\dfrac{4}{3}=\dfrac{20}{48}=\dfrac{5}{12}\)
\(\dfrac{ }{ }\)21/16
-21/16