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bài 1
a) \(\frac{-3}{14}-\frac{5}{-14}=\frac{-3}{14}-\frac{-5}{14}\)
\(=\frac{-3}{14}+\frac{5}{14}\)
\(=\frac{2}{14}=\frac{1}{7}\)
b)\(\frac{-5}{4}-\frac{3}{4}=\frac{-5}{4}+\frac{-3}{4}\)
\(=\frac{-8}{4}=\frac{-2}{1}=-2\)
c)\(\frac{15}{6}-\frac{-10}{20}=\frac{15}{6}-\frac{-1}{2}\)
\(=\frac{15}{6}-\frac{-3}{6}=\frac{15}{6}+\frac{3}{6}\)
\(=\frac{18}{6}=3\)
Bài 2
a)\(\frac{36}{-35}-\frac{6}{35}=\frac{-36}{35}+\frac{-6}{35}\)
\(=\frac{-30}{35}=\frac{-6}{5}\)
b)\(\frac{-1}{21}-\frac{-1}{28}=\frac{-28}{588}-\frac{-21}{588}\)
\(=\frac{-28}{588}+\frac{21}{588}\)
\(=\frac{-7}{588}=\frac{-1}{84}\)
c)\(\frac{-3}{29}-\frac{16}{59}=\frac{-177}{1711}-\frac{464}{1711}\)
\(=\frac{-177}{1711}+\frac{464}{1711}\)
\(=\frac{-641}{1711}\)
k cho mk nha
theo mik phần c bài 1 bạn có thể rút gọn
15/6 thành 5/2 nha
nhưng các cách làm của bạn đều đúng nha
=\(\left(4-2+3\right)\cdot\frac{-1}{2}\)
=\(5\cdot\left(\frac{-1}{2}\right)\)
=\(\frac{5\cdot\left(-1\right)}{2}\)
=\(\frac{-6}{2}\)
\(=\left(-3\right)\)
Có 2 trg hợp nhé: Nếu x là dấu nhân thì thực hiện theo phép nhân
Nếu x là ẩn số thì ko làm đc nhé vì ko có kết quả
Nên làm theo trường hợp 1
\(4.\frac{-1}{2}-2.\frac{-1}{2}+3.\frac{-1}{2}\)\(=\)\(\left(\frac{-1}{2}\right).\left(4-2+3\right)=\left(\frac{-1}{2}\right).5=\frac{-1.5}{2}=\frac{-5}{2}\)
\(B=81.\left(\frac{12-\frac{12}{7}-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{5+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}{6+\frac{6}{13}+\frac{6}{169}+\frac{6}{91}}\right).\frac{158158158}{711711711}\)
\(\Leftrightarrow B=81.\left(\frac{12\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}:\frac{5\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{6\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}\right).\frac{158\left(1001001\right)}{711\left(1001001\right)}\)
\(\Leftrightarrow B=81\left(\frac{12}{3}:\frac{5}{6}\right).\frac{158}{711}\)
\(\Leftrightarrow B=81\left(3.\frac{6}{5}\right).\frac{2}{9}\)
\(\Leftrightarrow B=81.\frac{18}{5}.\frac{2}{9}\)
\(\Leftrightarrow B=\frac{324}{5}\)
Hok tốt!!
\(7.\left[\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right):2\right]\)
\(7.\left[\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right):2\right]\)
\(7.\left(\frac{1}{3}-\frac{1}{13}\right):2\)
\(7.\frac{10}{39}:2=\frac{35}{39}\)
\(\frac{7}{15}+\frac{7}{35}+\frac{7}{63}+\frac{7}{99}+\frac{7}{143}\)
\(=\frac{7}{2}\cdot\left(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\right)\)
\(=\frac{7}{2}\cdot\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(=\frac{7}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{7}{2}\cdot\left(\frac{1}{3}-\frac{1}{13}\right)\)
\(=\frac{7}{2}\cdot\frac{10}{39}\)
\(=\frac{35}{39}\)
a, \(\frac{2}{5}.\frac{1}{3}-\frac{2}{15}:\frac{1}{5}+\frac{3}{5}.\frac{1}{3}\)
\(=\frac{1}{3}.\left(\frac{2}{5}+\frac{3}{5}\right)-\frac{2}{15}.5\)
\(=\frac{1}{3}.1-\frac{2}{3}\)
\(=\frac{1}{3}-\frac{2}{3}\)
\(=\frac{-1}{3}\)
b, \(\left(6-2\frac{4}{5}\right).3\frac{1}{8}+1\frac{3}{8}:\frac{1}{4}\)
\(=\left(6-\frac{14}{5}\right).\frac{25}{8}+\frac{11}{8}.4\)
\(=\frac{16}{5}.\frac{25}{8}+\frac{11}{2}\)
\(=10+\frac{11}{2}\)
\(=\frac{31}{2}\)
1/3×(3/5+2/5)-2/15×1/5
1/3×1-2/15×1/5
1/3-2/15×1/5
1/3-2/75
25/75-2/75
23/75
(6-14/5)×25/8-11/8:4/1
16/5×25/8-11/8:4/1
10/1-11/8:4/1
10/1-11/8×1/4
10/1-11/32
320/32-11/32
309/32
x/3=1/2
x.2=3.1
x.2=3
x=3:2
x=3/2
vậy x=3/2
x/3=9/2
x.2=3.9
x.2=27
x=27:2
x=27/2
vậy x=27/2
b)=(2/3 +2/7 - 2/28)/(-3/3 -3/7 + 3/28)
=[2(1/3+1/7-1/28)]/[(-3)(1/3+1/7-1/28)]
=2/-3
=-2/3
Đặt \(B=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2014^2}\)
Ta có : \(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
\(\frac{1}{5^2}< \frac{1}{4.5}\)
...
\(\frac{1}{2014^2}< \frac{1}{2013.2014}\)
\(\Rightarrow B< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2013.2014}\)
\(\Rightarrow B< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(\Rightarrow B< \frac{1}{2}-\frac{1}{2014}< \frac{1}{2}\)
\(\Rightarrow A< \frac{1}{2^2}+\frac{1}{2}=\frac{3}{4}\)
Vậy A<\(\frac{3}{4}\)
A<\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\)=\(\frac{2013}{2014}\)<\(\frac{3}{4}\)
Rút gọn phân số rồi tính như thường :)
\(-\frac{9991}{9992}+\frac{1}{99}-\frac{2}{19984}-\frac{5}{495}\)
\(=-\frac{9991}{9992}+\frac{1}{99}-\frac{1}{9992}-\frac{1}{99}\)
\(=\left(-\frac{9991}{9992}-\frac{1}{9992}\right)+\left(\frac{1}{99}-\frac{1}{99}\right)\)
\(=-1+0\)
\(=-1\)