Tính
\(-1,221.\dfrac{2}{-5}\)
K
Khách
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Những câu hỏi liên quan
23 tháng 2 2022
\(=\left(\dfrac{2}{5}+\dfrac{-2}{5}\right)+\left(\dfrac{5}{4}+\dfrac{5}{4}-\dfrac{5}{2}\right)\\ =0+0=0\)
23 tháng 2 2022
\(\dfrac{2}{5}+\dfrac{5}{4}+\left(\dfrac{-2}{5}+\dfrac{5}{4}\right)-\dfrac{5}{2}\\ =\dfrac{2}{5}+\dfrac{5}{4}-\dfrac{2}{5}+\dfrac{5}{4}-\dfrac{5}{2}\\ =\left(\dfrac{2}{5}-\dfrac{2}{5}\right)+\left(\dfrac{5}{4}+\dfrac{5}{4}\right)-\dfrac{5}{2}\\ =0+\dfrac{10}{4}-\dfrac{5}{2}\\ =\dfrac{5}{2}-\dfrac{5}{2}\\ =0\)
8 tháng 4 2022
b: =12+5/14-3-5/7-5-5/14
=4-5/7
=28/7-5/7=23/7
c: =(-2/5-11/10)+(7/11-7/11)
=-4/10-11/10=-15/10=-3/2
K
8 tháng 4 2022
\(a,\dfrac{5}{9}\cdot\dfrac{7}{13}+\dfrac{5}{9}\cdot\dfrac{8}{13}-\dfrac{5}{13}\cdot\dfrac{2}{9}\)
\(=\dfrac{5}{9}\cdot\dfrac{7}{13}+\dfrac{5}{9}\cdot\dfrac{8}{13}-\dfrac{2}{13}\cdot\dfrac{5}{9}\)
\(=\dfrac{5}{9}\cdot\left(\dfrac{7}{13}+\dfrac{8}{13}-\dfrac{2}{13}\right)\)
\(=\dfrac{5}{9}\cdot\dfrac{14}{13}\)
\(=\dfrac{70}{117}\)
\(d,\dfrac{1}{2}+\dfrac{-2}{3}+\dfrac{1}{6}+\dfrac{-2}{5}\)
\(=\left(\dfrac{1}{2}+\dfrac{-2}{3}+\dfrac{1}{6}\right)+\dfrac{-2}{5}\)
\(=0+\dfrac{-2}{5}\)
\(=\dfrac{-2}{5}\)
OY
17 tháng 8 2021
a) \(\dfrac{-5}{9}+\dfrac{3}{5}-\dfrac{3}{9}+\dfrac{-2}{5}\)
=\(\left(\dfrac{-5}{9}-\dfrac{3}{9}\right)+\left(\dfrac{3}{5}+\dfrac{-2}{5}\right)\)
=\(\dfrac{-8}{9}+\dfrac{1}{5}=-\dfrac{31}{45}\)
b) \(\dfrac{5}{17}-\dfrac{9}{15}-\dfrac{2}{17}+\dfrac{-2}{5}\)
=\(\left(\dfrac{5}{17}-\dfrac{2}{17}\right)-\left(\dfrac{9}{15}+\dfrac{2}{5}\right)\)
=\(\dfrac{3}{17}-1=\dfrac{-14}{17}\)
1 tháng 7 2021
\(a,5x\dfrac{7}{3}=\dfrac{5}{1}x\dfrac{7}{3}=\dfrac{35}{3};b,\dfrac{13}{4}:7=\dfrac{13}{4} :\dfrac{7}{1}=\dfrac{13}{4}x\dfrac{1}{7}=\dfrac{13}{28}\)
H
1 tháng 7 2021
1. Tính
\(a,5\times\dfrac{7}{3}=\dfrac{35}{3}\)
\(b,\dfrac{13}{4}:7=\dfrac{13}{4}\times\dfrac{1}{7}=\dfrac{13}{28}\)
2. Tính
\(a,\dfrac{3}{7}+\dfrac{2}{5}+\dfrac{3}{4}\)
\(=\dfrac{15}{35}+\dfrac{14}{35}+\dfrac{3}{4}\)
\(=\dfrac{29}{35}+\dfrac{3}{4}\)
\(=\dfrac{116}{140}+\dfrac{105}{140}\)
\(=\dfrac{221}{140}\)
\(b,\dfrac{9}{7}-\dfrac{5}{11}\times\dfrac{11}{7}\)
\(=\dfrac{9}{7}-\dfrac{55}{77}\)
\(=\dfrac{99}{77}-\dfrac{55}{77}\)
\(=\dfrac{44}{77}=\dfrac{4}{7}\)
\(c,\dfrac{3}{5}\times\dfrac{5}{7}+\dfrac{4}{7}\)
\(=\dfrac{3}{5}\times\left(\dfrac{5}{7}+\dfrac{4}{7}\right)\)
\(=\dfrac{3}{5}\times\dfrac{9}{7}\)
\(=\dfrac{27}{35}\)
\(d,\dfrac{7}{9}\times\dfrac{2}{5}:\dfrac{3}{11}\)
\(=\dfrac{14}{45}:\dfrac{3}{11}\)
\(=\dfrac{14}{45}\times\dfrac{11}{3}\)
\(=\dfrac{154}{135}\)
\(e,\dfrac{9}{7}+\dfrac{2}{3}-\dfrac{1}{4}\)
\(=\dfrac{27}{21}+\dfrac{14}{21}-\dfrac{1}{4}\)
\(=\dfrac{41}{21}-\dfrac{1}{4}\)
\(=\dfrac{164}{84}-\dfrac{21}{84}\)
\(=\dfrac{143}{84}\)
\(g,\dfrac{4}{9}:\dfrac{3}{5}\times\dfrac{2}{11}\)
\(=\dfrac{4}{9}\times\dfrac{5}{3}\times\dfrac{2}{11}\)
\(=\dfrac{20}{27}\times\dfrac{2}{11}\)
\(=\dfrac{40}{297}\)
\(h,\dfrac{7}{2}-\dfrac{3}{10}:\dfrac{2}{5}\)
\(=\left(\dfrac{7}{2}-\dfrac{3}{10}\right):\dfrac{2}{5}\)
\(=\left(\dfrac{35}{10}-\dfrac{3}{10}\right):\dfrac{2}{5}\)
\(=\dfrac{32}{10}:\dfrac{2}{5}\)
\(=\dfrac{16}{5}\times\dfrac{5}{2}\)
\(=\dfrac{80}{10}=8\)
BD
18 tháng 8 2023
\(\dfrac{1}{2}-\dfrac{2}{3}+\dfrac{3}{4}-\dfrac{4}{5}+\dfrac{5}{6}-\dfrac{6}{7}-\dfrac{6}{5}+\dfrac{4}{5}-\dfrac{3}{4}+\dfrac{2}{3}-\dfrac{1}{2}\)
\(=\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(-\dfrac{2}{3}+\dfrac{2}{3}\right)+\left(\dfrac{3}{4}-\dfrac{3}{4}\right)+\left(-\dfrac{4}{5}+\dfrac{4}{5}\right)+\left(\dfrac{5}{6}-\dfrac{6}{7}-\dfrac{6}{5}\right)\)
\(=0+0+0+0-\dfrac{257}{210}\)
\(=\dfrac{257}{210}\)
H9
HT.Phong (9A5)
CTVHS
22 tháng 7 2023
a) \(\dfrac{4}{5}\cdot\dfrac{5}{8}:\dfrac{4}{5}=\left(\dfrac{4}{5}:\dfrac{4}{5}\right)\cdot\dfrac{5}{8}=1\cdot\dfrac{5}{8}=\dfrac{5}{8}\)
b) \(\dfrac{5}{6}+\left(\dfrac{1}{2}:\dfrac{3}{2}+\dfrac{4}{5}\right)=\dfrac{5}{6}+\left(\dfrac{1}{2}\cdot\dfrac{2}{3}+\dfrac{4}{5}\right)=\dfrac{5}{6}+\left(\dfrac{1}{3}+\dfrac{4}{5}\right)=\dfrac{17}{15}+\dfrac{5}{6}=\dfrac{59}{30}\)
Y
15 tháng 8 2023
\(a,A=\dfrac{\dfrac{5}{4}+\dfrac{5}{5}+\dfrac{5}{7}-\dfrac{5}{11}}{\dfrac{10}{4}+\dfrac{10}{5}+\dfrac{10}{7}-\dfrac{10}{11}}\\ =\dfrac{5.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{10.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\\ =\dfrac{5}{10}\\ =\dfrac{1}{2}\)
Vậy \(A=\dfrac{1}{2}\)
\(b,B=\dfrac{2+\dfrac{6}{5}-\dfrac{6}{7}-\dfrac{6}{11}}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =\dfrac{3.\left(\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}\right)}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =3\)
Vậy \(B=3\)
AH
Akai Haruma
Giáo viên
17 tháng 9 2023
Lời giải:
1.
$3\frac{1}{4}-2\frac{1}{3}=3+\frac{1}{4}-(2+\frac{1}{3})$
$=3-2+\frac{1}{4}-\frac{1}{3}$
$=1+\frac{1}{4}-\frac{1}{3}=\frac{5}{4}-\frac{1}{3}=\frac{11}{12}$
2.
$6\frac{5}{9}+2\frac{5}{6}=6+\frac{5}{9}+2+\frac{5}{6}=8+\frac{5}{6}+\frac{5}{9}=8+\frac{25}{18}=8+1+\frac{7}{18}=9+\frac{7}{18}=9\frac{7}{18}$
3.
$6\frac{5}{9}-2\frac{5}{6}=6+\frac{5}{9}-(2+\frac{5}{6})$
$=6+\frac{5}{9}-2-\frac{5}{6}$
$=(6-2)+\frac{5}{9}-\frac{5}{6}$
$=4+\frac{5}{9}-\frac{5}{6}=\frac{41}{9}-\frac{5}{6}=\frac{67}{8}$
H9
HT.Phong (9A5)
CTVHS
29 tháng 10 2023
\(\dfrac{1}{3}+\dfrac{2}{9}=\dfrac{3}{3\times3}+\dfrac{2}{9}=\dfrac{3}{9}+\dfrac{2}{9}=\dfrac{5}{9}\)
\(\dfrac{1}{2}+\dfrac{3}{8}=\dfrac{4}{2\times4}+\dfrac{3}{8}=\dfrac{4}{8}+\dfrac{3}{8}=\dfrac{7}{8}\)
\(\dfrac{5}{12}+\dfrac{2}{3}=\dfrac{5}{12}+\dfrac{2\times4}{3\times4}=\dfrac{5}{12}+\dfrac{8}{12}=\dfrac{13}{12}\)
\(\dfrac{5}{16}+\dfrac{3}{8}=\dfrac{5}{16}+\dfrac{3\times2}{8\times2}=\dfrac{5}{16}+\dfrac{6}{16}=\dfrac{11}{16}\)
\(\dfrac{4}{15}+\dfrac{3}{5}=\dfrac{4}{15}+\dfrac{3\times3}{5\times3}=\dfrac{4}{15}+\dfrac{9}{15}=\dfrac{13}{15}\)
\(\dfrac{8}{63}+\dfrac{7}{10}=\dfrac{8\times10}{63\times10}+\dfrac{7\times63}{10\times63}=\dfrac{80}{630}+\dfrac{441}{630}=\dfrac{521}{630}\)
Lời giải:
$-1,221.\frac{2}{-5}=-1,221.(-0,4)=0,4884$