Tính 3/4−[(−5/3)−(1/12+2/9)]
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a] 4/12 ; 5/12 ; 11/2 ; 1/4
b] 1 ; 9/12; 12/5 ; 11/3
c] 8/21; 6/11;8/7
d] 2/3 ;2/7 15/2
a]1/3 5/12 11/2 1/4
b]1 3/4 12/5 33/9
c]8/21 6/11 8/7
d]2/3 2/7 15/2
1 / 9 + 2 / 9 + 4 / 9 = 7 / 9
5 / 5 + 6 / 5 + 1 = 15 / 5 = 3
12 / 2 + 4 / 2 + 1 + 3 / 2 = 21 / 2
7 / 3 + 6 / 3 + 1 + 1 = 19 / 3
1.a,=(54+45+1).113
=100.113
=11300
b,=(3/7+8/14)+(4/9+10/18)
=1+1
=2
2.a,=13/10+1/3
=49/30
b,=12/9.(1/12+1/6)
=12/9.1/4
=1/3
c,=3/4.3/2
=9/8
d,=3/2-1/3
=7/6
1:tính bằng cách thuận tiện nhất:
a)54 x 113 + 45 x 113 + 113
= 54 x 113 + 45 x 113 + 113x1
=113 x(54+45+1)
= 113x100
=1300
b)3/7 + 4/9 + 8/14 + 10/18
=(3/7+8/14)+(4/9+10/18)
= 1 + 1
=2
Ta có: \(\dfrac{3}{4}-\left[\left(-\dfrac{3}{5}\right)-\left(\dfrac{1}{12}+\dfrac{2}{9}\right)-\left(-\dfrac{2}{9}-\dfrac{5}{3}\right)\right]\)
\(=\dfrac{3}{4}-\left(-\dfrac{3}{5}-\dfrac{1}{12}-\dfrac{2}{9}+\dfrac{2}{9}+\dfrac{5}{3}\right)\)
\(=\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{1}{12}-\dfrac{5}{3}\)
\(=\dfrac{45}{60}+\dfrac{36}{60}+\dfrac{5}{60}-\dfrac{100}{60}\)
\(=\dfrac{-14}{60}=\dfrac{-7}{30}\)
\(\dfrac{1}{2}\times\dfrac{3}{4}:\dfrac{6}{5}=\dfrac{3}{8}:\dfrac{6}{5}=\dfrac{5}{16}\)
\(\dfrac{2}{3}+\dfrac{1}{6}:\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{2}{7}=\dfrac{14}{21}+\dfrac{6}{21}=\dfrac{20}{21}\)
\(8\times\dfrac{3}{5}:\dfrac{12}{5}=\dfrac{24}{5}:\dfrac{12}{5}=2\)
\(\dfrac{4}{9}+\dfrac{3}{4}-\dfrac{1}{3}=\dfrac{16}{36}+\dfrac{27}{36}-\dfrac{12}{36}=\dfrac{31}{36}\)
Bài 1:
\(A=\dfrac{-1}{3}+1+\dfrac{1}{3}=1\)
\(B=\dfrac{2}{15}+\dfrac{5}{9}-\dfrac{6}{9}=\dfrac{2}{15}-\dfrac{1}{9}=\dfrac{18-15}{135}=\dfrac{3}{135}=\dfrac{1}{45}\)
\(C=\dfrac{-1}{5}+\dfrac{1}{4}-\dfrac{3}{4}=\dfrac{-1}{5}-\dfrac{1}{2}=\dfrac{-7}{10}\)
Bài 2:
a: \(=\dfrac{1}{5}+\dfrac{1}{2}+\dfrac{2}{5}-\dfrac{3}{5}+\dfrac{2}{21}-\dfrac{10}{21}+\dfrac{3}{20}\)
\(=\left(\dfrac{1}{5}+\dfrac{2}{5}-\dfrac{3}{5}\right)+\left(\dfrac{2}{21}-\dfrac{10}{21}\right)+\left(\dfrac{1}{2}+\dfrac{3}{20}\right)\)
\(=\dfrac{-8}{21}+\dfrac{13}{20}=\dfrac{113}{420}\)
b: \(B=\dfrac{21}{23}-\dfrac{21}{23}+\dfrac{125}{93}-\dfrac{125}{143}=\dfrac{6250}{13299}\)
Bài 3:
\(\dfrac{7}{3}-\dfrac{1}{2}-\left(-\dfrac{3}{70}\right)=\dfrac{7}{3}-\dfrac{1}{2}+\dfrac{3}{70}=\dfrac{490}{210}-\dfrac{105}{210}+\dfrac{9}{210}=\dfrac{394}{210}=\dfrac{197}{105}\)
\(\dfrac{5}{12}-\dfrac{3}{-16}+\dfrac{3}{4}=\dfrac{5}{12}+\dfrac{3}{16}+\dfrac{3}{4}=\dfrac{20}{48}+\dfrac{9}{48}+\dfrac{36}{48}=\dfrac{65}{48}\)
Bài 4:
\(\dfrac{3}{4}-x=1\)
\(\Rightarrow-x=1-\dfrac{3}{4}\)
\(\Rightarrow x=-\dfrac{1}{4}\)
Vậy: \(x=-\dfrac{1}{4}\)
\(x+4=\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{1}{5}-4\)
\(\Rightarrow x=-\dfrac{19}{5}\)
Vậy: \(x=-\dfrac{19}{5}\)
\(x-\dfrac{1}{5}=2\)
\(\Rightarrow x=2+\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{11}{5}\)
Vậy: \(x=\dfrac{11}{5}\)
\(x+\dfrac{5}{3}=\dfrac{1}{81}\)
\(\Rightarrow x=\dfrac{1}{81}-\dfrac{5}{3}\)
\(\Rightarrow x=-\dfrac{134}{81}\)
Vậy: \(x=-\dfrac{134}{81}\)
a: \(=\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}=\dfrac{15}{48}=\dfrac{5}{16}\)
b: \(=\dfrac{2}{3}+\dfrac{12}{42}=\dfrac{2}{3}+\dfrac{2}{7}=\dfrac{20}{21}\)
c: \(=\dfrac{24}{5}\cdot\dfrac{5}{12}=2\)
d: \(=\dfrac{1}{9}+\dfrac{3}{4}=\dfrac{4+27}{36}=\dfrac{31}{36}\)
\(a,\dfrac{4}{5}+\dfrac{2}{3}+\dfrac{1}{9}=\dfrac{12}{15}+\dfrac{10}{15}+\dfrac{1}{9}=\dfrac{22}{15}+\dfrac{1}{9}=\dfrac{66}{45}+\dfrac{5}{45}=\dfrac{71}{45}\)
\(b,\dfrac{3}{7}+\dfrac{11}{14}+\dfrac{19}{28}=\dfrac{12}{28}+\dfrac{22}{28}+\dfrac{19}{28}=\dfrac{53}{28}\)
\(c,\dfrac{1}{2}+\dfrac{1}{7}+\dfrac{-1}{5}=\dfrac{7}{14}+\dfrac{2}{14}+\dfrac{-1}{5}=\dfrac{9}{14}+\dfrac{-1}{5}=\dfrac{45}{70}+\dfrac{-14}{70}=\dfrac{31}{70}\)
\(d,\dfrac{7}{8}+\dfrac{5}{16}+\dfrac{-3}{4}=\dfrac{14}{16}+\dfrac{5}{16}+\dfrac{-12}{16}=\dfrac{7}{16}\)
\(e,\dfrac{1}{4}+\dfrac{5}{12}+\dfrac{-1}{13}=\dfrac{3}{12}+\dfrac{5}{12}+\dfrac{-1}{13}=\dfrac{8}{12}+\dfrac{-1}{13}=\dfrac{2}{3}+\dfrac{-1}{13}=\dfrac{26}{39}+\dfrac{-3}{39}=\dfrac{23}{39}\)
\(g,\dfrac{2}{3}+\dfrac{3}{8}+\dfrac{-5}{12}=\dfrac{16}{24}+\dfrac{9}{24}+\dfrac{-5}{12}=\dfrac{25}{24}+\dfrac{-5}{12}=\dfrac{25}{24}+\dfrac{-10}{24}=\dfrac{15}{24}\)
\(=\dfrac{3}{4}-\left(-\dfrac{5}{3}-\dfrac{11}{36}\right)=\dfrac{3}{4}+\dfrac{71}{36}=\dfrac{49}{18}\)