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S
3 tháng 2
câu d)
\(\dfrac{6}{5}+\left(3+\dfrac{-1}{5}\right)\\ =\dfrac{6}{5}+3+\dfrac{-1}{5}\\ =1+3=4\)
câu e)
\(-\dfrac{3}{5}+\left(-\dfrac{2}{5}+2\right)\\ =\dfrac{-3}{5}+\dfrac{-2}{5}+2\\ =-1+2=1\)
câu f)
\(\dfrac{8}{3}-\left(-\dfrac{4}{3}+3\right)\\ =\dfrac{8}{3}+\dfrac{4}{3}-3\\ =4-3=1\)
AH
Akai Haruma
Giáo viên
3 tháng 2
Lời giải:
g.
$=\frac{-5}{11}+\frac{-6}{11}+1=(\frac{-5}{11}+\frac{-6}{11})+1$
$=\frac{-11}{11}+1=(-1)+1=0$
h.
$=\frac{-17}{13}+\frac{4}{13}+\frac{25}{101}$
$=\frac{-13}{13}+\frac{25}{101}=(-1)+\frac{25}{101}=-(1-\frac{25}{101})$
$=\frac{-76}{101}$
i.
$=\frac{6}{7}+\frac{1}{8}-\frac{3}{16}.4=\frac{6}{7}+\frac{1}{8}-\frac{3}{4}$
$=\frac{6}{7}+\frac{1}{8}-\frac{6}{8}=\frac{6}{7}+\frac{-5}{8}$
$=\frac{13}{56}$
S
3 tháng 2
CÂU A)
\(\dfrac{5}{3}+\left(7+\dfrac{-5}{3}\right)\\ =\dfrac{5}{3}+7+\dfrac{-5}{3}=7\)
CÂI B)
\(-\dfrac{7}{31}+\left(\dfrac{24}{17}+\dfrac{7}{31}\right)\\ =-\dfrac{7}{31}+\dfrac{24}{17}+\dfrac{7}{31}=\dfrac{24}{17}\)
CÂU C)
\(\dfrac{3}{7}+\left(-\dfrac{1}{5}+\dfrac{-3}{7}\right)\\ =\dfrac{3}{7}-\dfrac{1}{5}-\dfrac{3}{7}=-\dfrac{1}{5}\)
AH
Akai Haruma
Giáo viên
3 tháng 2
k.
$=\frac{-5}{7}(\frac{2}{11}+\frac{9}{11})+\frac{12}{7}$
$=\frac{-5}{7}.\frac{11}{11}+\frac{12}{7}$
$=\frac{-5}{7}+\frac{12}{7}=\frac{7}{7}=1$
l.
$=(\frac{1}{5}+\frac{4}{5})+(\frac{-2}{9}+\frac{-7}{9})+\frac{16}{17}$
$=\frac{5}{5}+\frac{-9}{9}+\frac{16}{17}$
$=1+(-1)+\frac{16}{17}=\frac{16}{17}$
AH
Akai Haruma
Giáo viên
3 tháng 2
a.
$=(\frac{-17}{13}+\frac{4}{13})+(\frac{11}{31}+\frac{20}{31})+\frac{2}{135}$
$=\frac{-13}{13}+\frac{31}{31}+\frac{2}{135}=(-1)+1+\frac{2}{135}=\frac{2}{135}$
b.
$=(\frac{3}{17}+\frac{20}{-17})+(\frac{4}{18}+\frac{-2}{9})+(\frac{5}{8}+\frac{21}{56})$
$=(\frac{3}{17}+\frac{-20}{17})+(\frac{2}{9}+\frac{-2}{9})+(\frac{5}{8}+\frac{3}{8})$
$=\frac{-17}{17}+\frac{0}{9}+\frac{8}{8}$
$=(-1)+0+1=0$
AH
Akai Haruma
Giáo viên
3 tháng 2
c.
$=\frac{-5}{12}+\frac{6}{11}+\frac{7}{17}+\frac{5}{11}+\frac{5}{12}$
$=(\frac{-5}{12}+\frac{5}{12})+(\frac{6}{11}+\frac{5}{11})+\frac{7}{17}$
$=\frac{0}{12}+\frac{11}{11}+\frac{7}{17}$
$=0+1+\frac{7}{17}=\frac{24}{17}$
d.
$=\frac{9}{16}+\frac{-8}{27}+1+\frac{7}{16}+\frac{-19}{27}$
$=(\frac{9}{16}+\frac{7}{16})+1+(\frac{-8}{27}+\frac{-19}{27})$
$=\frac{16}{16}+1+\frac{-27}{27}$
$=1+1+(-1)=1$
AH
Akai Haruma
Giáo viên
3 tháng 2
Lời giải:
e.
$=(\frac{-2}{11}+\frac{-9}{11})+(\frac{6}{7}+\frac{1}{7})+\frac{1}{2}$
$=\frac{-11}{11}+\frac{7}{7}+\frac{1}{2}$
$=(-1)+1+\frac{1}{2}=\frac{1}{2}$
f.
$=(\frac{8}{19}+\frac{-27}{19})+(\frac{4}{21}+\frac{17}{21})+\frac{2}{5}$
$=\frac{-19}{19}+\frac{21}{21}+\frac{2}{5}$
$=(-1)+1+\frac{2}{5}=\frac{2}{5}$
g.
$=\frac{(-2)+(-8)+17}{15}=\frac{7}{15}$
h.
$=\frac{-10}{30}+\frac{12}{30}+\frac{-75}{30}$
$=\frac{(-10)+12+(-75)}{30}=\frac{-73}{30}$
3 tháng 2
i) \(\dfrac{1}{2}+\dfrac{-3}{8}+\dfrac{5}{9}\)
\(=\dfrac{1}{8}+\dfrac{5}{9}\)
\(=\dfrac{49}{72}\)
k)\(\dfrac{13}{-30}+\dfrac{17}{45}+\dfrac{-7}{18}\)
\(=\dfrac{-1}{18}+\dfrac{-7}{18}\)
\(=\dfrac{-4}{9}\)
m) \(\dfrac{-5}{12}+\dfrac{4}{9}+\dfrac{-11}{6}\)
\(=\dfrac{1}{36}+\dfrac{-11}{6}\)
\(=\dfrac{-65}{36}\)
n) \(\dfrac{-7}{31}+\dfrac{24}{17}+\dfrac{7}{31}\)
\(=\left(\dfrac{-7}{31}+\dfrac{7}{31}\right)+\dfrac{24}{17}\)
\(=0+\dfrac{24}{17}\)
\(\dfrac{24}{17}\)
AH
Akai Haruma
Giáo viên
3 tháng 2
Lời giải:
o.
$=\frac{17}{21}+\frac{5}{9}+\frac{-17}{21}+\frac{2}{3}+\frac{4}{9}$
$=(\frac{17}{21}+\frac{-17}{21})+(\frac{5}{9}+\frac{4}{9})+\frac{2}{3}$
$=0+1+\frac{2}{3}=\frac{5}{3}$
p.
$=\frac{13}{5}+\frac{7}{16}+\frac{-15}{16}+\frac{6}{15}$
$=(\frac{13}{5}+\frac{6}{15})+(\frac{7}{16}+\frac{-15}{16})$
$=\frac{45}{15}+\frac{-8}{16}$
$=3-\frac{1}{2}=\frac{5}{2}$
AH
Akai Haruma
Giáo viên
3 tháng 2
r.
\(=(\frac{-5}{9}+\frac{4}{-9})+(\frac{8}{15}+\frac{7}{15})+\frac{-2}{11}\\ =(\frac{-5}{9}+\frac{-4}{9})+\frac{15}{15}+\frac{-2}{11}\\ =\frac{-9}{9}+\frac{15}{15}+\frac{-2}{11}=-1+1+\frac{-2}{11}=\frac{-2}{11}\)
s.
\(=(\frac{5}{13}+\frac{8}{13})+(\frac{-20}{41}+\frac{-21}{41})+\frac{-5}{17}\\ =\frac{13}{13}+\frac{-41}{41}+\frac{-5}{17}\\ =1+(-1)+\frac{-5}{17}=\frac{-5}{17}\)
3 tháng 2
a: \(-\dfrac{3}{4}\cdot31\dfrac{11}{23}-0,75\cdot8\dfrac{12}{23}\)
\(=-\dfrac{3}{4}\left(31+\dfrac{11}{23}+8+\dfrac{12}{23}\right)\)
\(=-\dfrac{3}{4}\cdot\left(39+1\right)=-\dfrac{3}{4}\cdot40=-30\)
b: \(\left(2\dfrac{1}{3}+3\dfrac{1}{2}\right):\left(-4\dfrac{1}{6}+3\dfrac{1}{7}\right)+7\dfrac{1}{2}\)
\(=\left(\dfrac{7}{3}+\dfrac{7}{2}\right):\left(-\dfrac{25}{6}+\dfrac{22}{7}\right)+7\dfrac{1}{2}\)
\(=\dfrac{35}{6}:\dfrac{-25\cdot7+22\cdot6}{42}+\dfrac{15}{2}\)
\(=\dfrac{35}{6}\cdot\dfrac{42}{-175+132}+\dfrac{15}{2}\)
\(=\dfrac{35\cdot7}{-43}+\dfrac{15}{2}=\dfrac{155}{86}\)
c: \(4\dfrac{5}{9}:\dfrac{-5}{7}+\left(5\dfrac{4}{9}\right):\dfrac{-5}{7}\)
\(=\dfrac{41}{9}\cdot\dfrac{-7}{5}+\dfrac{49}{9}\cdot\dfrac{-7}{5}\)
\(=-\dfrac{7}{5}\left(\dfrac{41}{9}+\dfrac{49}{9}\right)\)
\(=-\dfrac{7}{5}\cdot10=-14\)
d: \(\dfrac{\left(13\dfrac{1}{4}-2\dfrac{5}{27}-10\dfrac{5}{6}\right)\cdot230\dfrac{1}{25}+46\dfrac{3}{4}}{\left(1\dfrac{3}{7}+\dfrac{10}{3}\right):\left(12\dfrac{1}{3}-14\dfrac{2}{7}\right)}\)
\(=\dfrac{\left(13+\dfrac{1}{4}-2-\dfrac{5}{27}-10-\dfrac{5}{6}\right)\cdot230,04+46,75}{\left(\dfrac{10}{7}+\dfrac{10}{3}\right):\left(12+\dfrac{1}{3}-14-\dfrac{2}{7}\right)}\)
\(=\dfrac{\left(1-\dfrac{83}{108}\right)\cdot230,04+46,75}{\dfrac{100}{21}:\left(-2+\dfrac{7-6}{21}\right)}\)
\(=\dfrac{\dfrac{25}{108}\cdot230,04+46,75}{\dfrac{100}{21}:\dfrac{-42+1}{21}}\)
\(=\dfrac{100}{\dfrac{100}{21}\cdot\dfrac{21}{-41}}=100:\dfrac{100}{-41}=-41\)
31: (-29)+8-9
\(=\left(-29\right)+\left(8-9\right)\)
\(=-29-1=-30\)
32: \(\left(-14\right)-6-10\)
\(=\left(-14-6\right)-10\)
=-20-10
=-30
33: \(\left(-42\right)+10+2\)
\(=\left(-42\right)+\left(10+2\right)\)
=-42+12
=-30
34: \(\left(-45\right)+25-20\)
\(=\left(-45\right)+\left(25-20\right)\)
=-45+5
=-40
35: \(\left(-60\right)+17+3\)
\(=\left(-60\right)+\left(17+3\right)\)
=-60+20
=-40
36: \(\left(-48\right)+18-10\)
\(=\left(-48+18\right)-10\)
=-30-10
=-40