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H9
HT.Phong (9A5)
CTVHS
17 tháng 1
a) \(\left(\dfrac{3}{29}-\dfrac{1}{5}\right)\cdot\dfrac{29}{3}\)
\(=\dfrac{3}{29}\cdot\dfrac{29}{3}-\dfrac{1}{5}\cdot\dfrac{29}{3}\)
\(=1-\dfrac{29}{15}\)
\(=\dfrac{15-29}{15}\)
\(=-\dfrac{14}{15}\)
b) \(\dfrac{3}{4}\cdot\dfrac{7}{9}+\dfrac{1}{4}\cdot\dfrac{7}{9}\)
\(=\dfrac{7}{9}\cdot\left(\dfrac{1}{4}+\dfrac{3}{4}\right)\)
\(=\dfrac{7}{9}\cdot1\)
\(=\dfrac{7}{9}\)
c) \(\dfrac{1}{7}\cdot\dfrac{5}{9}+\dfrac{5}{9}\cdot\dfrac{1}{7}+\dfrac{5}{9}\cdot\dfrac{3}{7}\)
\(=\dfrac{5}{9}\cdot\left(\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{3}{7}\right)\)
\(=\dfrac{5}{9}\cdot\dfrac{5}{7}\)
\(=\dfrac{25}{63}\)
d) \(4\cdot11\cdot\dfrac{3}{4}\cdot\dfrac{9}{121}\)
\(=\left(4\cdot\dfrac{3}{4}\right)\cdot\left(11\cdot\dfrac{9}{121}\right)\)
\(=3\cdot\dfrac{9}{11}\)
\(=\dfrac{27}{11}\)
29 tháng 5 2018
\(A=\frac{1}{1\cdot2}+\frac{2}{2\cdot4}+\frac{3}{4\cdot7}+\frac{4}{7\cdot11}+...+\frac{10}{46\cdot56}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{46}-\frac{1}{56}\)
\(A=1-\frac{1}{56}\)
\(A=\frac{55}{56}\)
\(B=\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{23\cdot27}\)
\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{23}-\frac{1}{27}\)
\(B=\frac{1}{3}-\frac{1}{27}\)
\(B=\frac{8}{27}\)
\(C=\frac{4}{3\cdot6}+\frac{4}{6\cdot9}+\frac{4}{9\cdot12}+...+\frac{4}{99\cdot102}\)
\(C=\frac{4}{3}\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+\frac{3}{9\cdot12}+...+\frac{3}{99\cdot102}\right)\)
\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{99}-\frac{1}{102}\right)\)
\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{102}\right)\)
\(C=\frac{4}{3}\cdot\frac{33}{102}\)
\(C=\frac{22}{51}\)
LX
1
H
10 tháng 1 2023
\(a,\left(-\dfrac{4}{9}\right)\times\dfrac{3}{8}+\dfrac{1}{18}=-\dfrac{1}{6}+\dfrac{1}{18}=\dfrac{-1.3}{6.3}+\dfrac{1}{18}=\dfrac{-3+1}{18}=-\dfrac{2}{18}=-\dfrac{1}{9}\)
xem lại đề câu b
\(c,\left(\dfrac{1}{2}-\dfrac{2}{3}\right)\times\dfrac{4}{4}=\dfrac{1}{2}-\dfrac{2}{3}=\dfrac{3-4}{6}=-\dfrac{1}{6}\\ d,\dfrac{9}{7}\times\left(\dfrac{3}{7}-\dfrac{1}{2}\right)=\dfrac{9}{7}\times\left(\dfrac{3.2-7}{14}\right)=\dfrac{9}{7}\times\dfrac{-1}{14}=-\dfrac{9}{98}\)
PK
2
19 tháng 12 2015
( 1+2+3+4-1-2-3-4) * 1*2*...........*8*9
= 0 * 1*2*...........*8*9
= 0
AH
Akai Haruma
Giáo viên
15 tháng 1
a/
$\frac{7}{4}+\frac{9}{16}=\frac{28}{16}+\frac{9}{16}=\frac{28+9}{16}=\frac{37}{16}$
b/
$\frac{-70}{27}+\frac{35}{9}=\frac{-70}{27}+\frac{105}{27}=\frac{35}{27}$
c/
$\frac{-4}{-5}+\frac{-2}{3}=\frac{4}{5}-\frac{2}{3}=\frac{12}{15}-\frac{10}{15}=\frac{2}{15}$
d/ Đề thiếu. Bạn xem lại.
i/
$\frac{-7}{2}+(\frac{-5}{-4}-\frac{4}{-9}$
$=\frac{-7}{2}+\frac{5}{4}+\frac{4}{9}$
$=\frac{-14}{4}+\frac{5}{4}+\frac{4}{9}$
$=\frac{-9}{4}+\frac{4}{9}=\frac{-81}{36}+\frac{16}{36}=\frac{-65}{36}$
AH
Akai Haruma
Giáo viên
15 tháng 1
k/
$\frac{1}{4}+\frac{1}{3}-\frac{1}{2}$
$=\frac{3}{12}+\frac{4}{12}-\frac{6}{12}=\frac{1}{12}$
m/
$\frac{-1}{9}-\frac{1}{7}-\frac{3}{4}$
$=-(\frac{1}{9}+\frac{1}{7}+\frac{3}{4})$
$=-(\frac{28}{252}+\frac{36}{252}+\frac{189}{252})$
$=\frac{-253}{252}$
n/
$\frac{11}{12}-(\frac{-3}{18}+\frac{5}{6}$
$=\frac{11}{12}+\frac{1}{6}+\frac{5}{6}$
$=\frac{11}{12}+1=\frac{23}{12}$
o/
$\frac{20}{17}+\frac{-11}{2}-\frac{30}{4}$
$=\frac{20}{17}-\frac{11}{2}-\frac{15}{2}$
$=\frac{20}{7}-13=\frac{-71}{7}$
v/
$\frac{1}{2}-\frac{3}{4}+\frac{5}{6}$
$=\frac{6}{12}-\frac{9}{12}+\frac{10}{12}$
$=\frac{6-9+10}{12}=\frac{7}{12}$
\(-\dfrac{3}{4}+\dfrac{3}{7}-\dfrac{1}{4}+\dfrac{4}{9}+\dfrac{4}{9}\\ =\left(-\dfrac{3}{4}-\dfrac{1}{4}\right)+\left(\dfrac{4}{9}+\dfrac{4}{9}\right)+\dfrac{3}{7}\\ =-1+\dfrac{8}{9}+\dfrac{3}{7}\\ =\dfrac{20}{63}.\)