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22 tháng 4 2019
a)\(\left(7\frac{4}{9}+4\frac{7}{11}\right)-3\frac{4}{9}\)=\(7\frac{4}{9}+4\frac{7}{11}-3\frac{4}{9}\)=\(\left(7\frac{4}{9}-3\frac{4}{9}\right)+4\frac{7}{11}\)= 4+\(4\frac{7}{11}\)=\(8\frac{7}{11}\)
b)\(\frac{-7}{9}.\frac{4}{11}+\frac{-7}{9}.\frac{7}{11}+5\frac{7}{9}\)=\(\frac{-7}{9}.\left(\frac{4}{11}+\frac{7}{11}\right)+5+\frac{7}{9}\)=\(\frac{-7}{9}.1+5+\frac{7}{9}\)=\(\frac{-7}{9}+\frac{7}{9}+5\)=\(\left(\frac{-7}{9}+\frac{7}{9}\right)+5\)= 0+5=5
c)\(50\%.1\frac{1}{3}.10\frac{7}{35}.0,75\)= \(\frac{1}{2}.\frac{4}{3}.10\frac{1}{5}.\frac{3}{4}\)=\(\frac{1}{2}.\frac{4}{3}.\frac{51}{5}.\frac{3}{4}\)=\(\frac{1.4.51.3}{2.3.5.4}\)=\(\frac{51}{2.5}\)=\(\frac{51}{10}\)
d)\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}\)=\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)=\(\frac{43}{43}-\frac{1}{43}\)=\(\frac{42}{43}\)
22 tháng 4 2019
a) \(\left(7\frac{4}{9}+4\frac{7}{11}\right)-3\frac{4}{9}\)
\(=\left(\frac{67}{9}+\frac{51}{11}\right)-\frac{31}{9}\)
\(=\left(\frac{67}{9}-\frac{31}{9}\right)+\frac{51}{11}\)
\(=\frac{36}{9}+\frac{51}{11}\)
\(=\frac{95}{11}=8\frac{7}{11}\)
b) \(-\frac{7}{9}.\frac{4}{11}+-\frac{7}{9}.\frac{7}{11}+5\frac{7}{9}\)
\(=-\frac{7}{9}.\frac{4}{11}+-\frac{7}{9}.\frac{7}{11}+\frac{52}{9}\)
\(=-\frac{7}{9}.\left(\frac{4}{11}+\frac{7}{11}\right)+\frac{52}{9}\)
\(=-\frac{7}{9}.1+\frac{52}{9}\)
\(=-\frac{7}{9}+\frac{52}{9}\)
= 5
d) \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}\)
\(=1.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\right)\)
\(=1.\left(1-\frac{1}{43}\right)\)
\(=1.\frac{42}{43}\)
\(=\frac{42}{43}\)
NV
0
1 tháng 8 2020
a) Ta có: \(15\frac{3}{13}-\left(3\frac{4}{7}+8\frac{3}{13}\right)\)
\(=15+\frac{3}{13}-3-\frac{4}{7}-8-\frac{3}{13}\)
\(=4-\frac{4}{7}=\frac{24}{7}\)
b) Ta có: \(\left(7\frac{4}{9}+4\frac{7}{11}\right)-3\frac{4}{9}\)
\(=7+\frac{4}{9}+4+\frac{7}{11}-3-\frac{4}{9}\)
\(=8+\frac{7}{11}=\frac{95}{11}\)
c) Ta có: \(\frac{-7}{9}\cdot\frac{4}{11}+\frac{-7}{9}\cdot\frac{7}{11}+5\frac{7}{9}\)
\(=\frac{-7}{9}\cdot\frac{4}{11}+\frac{-7}{9}\cdot\frac{7}{11}+\frac{-7}{9}\cdot\frac{-52}{7}\)
\(=\frac{-7}{9}\cdot\left(\frac{4}{11}+\frac{7}{11}-\frac{52}{7}\right)\)
\(=\frac{-7}{9}\cdot\frac{45}{-7}=5\)
d) Ta có: \(50\%\cdot1\frac{1}{3}\cdot10\cdot\frac{7}{35}\cdot0.75\)
\(=\frac{1}{2}\cdot\frac{4}{3}\cdot10\cdot\frac{7}{35}\cdot\frac{3}{4}\)
\(=5\cdot\frac{7}{35}=1\)
e) Ta có: \(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{40\cdot43}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)
\(=1-\frac{1}{43}=\frac{43}{43}-\frac{1}{43}\)
\(=\frac{42}{43}\)
MV
27 tháng 4 2017
1)
a)
\(\dfrac{-5}{11}\cdot\dfrac{4}{7}+\dfrac{-5}{11}\cdot\dfrac{3}{7}-\dfrac{8}{11}\\ =\dfrac{-5}{11}\cdot\left(\dfrac{4}{7}+\dfrac{3}{7}\right)-\dfrac{8}{11}\\ =\dfrac{-5}{11}\cdot1-\dfrac{8}{11}\\ =\dfrac{-5}{11}-\dfrac{8}{11}\\ =\dfrac{-5}{11}+\dfrac{-8}{11}\\ =\dfrac{-13}{11}\)
b)
\(\left(\dfrac{2}{9}:\dfrac{5}{3}+\dfrac{1}{3}:\dfrac{5}{3}\right)^2-\left(\dfrac{1}{3}-\dfrac{5}{8}\right)\\ =\left(\dfrac{2}{9}\cdot\dfrac{3}{5}+\dfrac{1}{3}\cdot\dfrac{3}{5}\right)^2-\left(\dfrac{-7}{24}\right)\\ =\left[\dfrac{3}{5}\cdot\left(\dfrac{2}{9}+\dfrac{1}{3}\right)\right]^2+\dfrac{7}{24}\\ =\left[\dfrac{3}{5}\cdot\dfrac{5}{9}\right]^2+\dfrac{7}{24}\\ =\left[\dfrac{1}{3}\right]^2+\dfrac{7}{24}\\ =\dfrac{1}{9}+\dfrac{7}{24}\\ =\dfrac{29}{72}\)
c) \(14-\left|\dfrac{-3}{4}\right|-\left(\dfrac{1}{3}-\dfrac{5}{8}\right)\\ =14-\dfrac{3}{4}-\left(\dfrac{-7}{24}\right)\\ =14+\dfrac{-3}{4}+\dfrac{7}{24}\\ =13\dfrac{13}{24}\)
11 tháng 8 2016
a)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{3}{4}+...+\frac{1}{9}-\frac{1}{10}\)
= \(1+\left(\frac{-1}{2}+\frac{1}{2}\right)+\left(\frac{-1}{3}+\frac{1}{3}\right)+...+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{10}\)
= \(1-\frac{1}{10}\)
=\(\frac{9}{10}\)
b)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
=\(1+\left(\frac{-1}{3}+\frac{1}{3}\right)+\left(\frac{-1}{5}+\frac{1}{5}\right)+\left(\frac{-1}{7}+\frac{1}{7}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{11}\)
=\(1-\frac{1}{11}\)
= \(\frac{10}{11}\)
c) đặt A=\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}\)
\(\frac{1}{3}A\)=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(\frac{2}{3}A\)=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
\(\frac{2}{3}A\)=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(\frac{2}{3}A\)=\(1+\left(\frac{-1}{3}+\frac{1}{3}\right)+\left(\frac{-1}{5}+\frac{1}{5}\right)+\left(\frac{-1}{7}+\frac{1}{7}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{11}\)
\(\frac{2}{3}A\)=\(\frac{10}{11}\)
A= \(\frac{10}{11}:\frac{2}{3}\)
A= \(\frac{10}{11}.\frac{3}{2}\)=\(\frac{15}{11}\)
d) giả tương tự câu c kết quả \(\frac{25}{11}\)
11 tháng 8 2016
tổng đặc biệt đó bạn
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{9\times10}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(1-\frac{1}{10}=\frac{9}{10}\)
những câu sau cũng áp dụng như vậy nhé
WR
1 tháng 4 2015
a=113/13-(24/7+53/13)=113/13-24/7-53/13=60/13-24/7=396/16
b=(64/9+37/11)-44/9=64/9+37/11-44/9=20/9+37/11=181/33
c=-5/7(2/11+9/11)+15/7=-5/7+15/7=10/7
d=0.7.22/3.20.0.375.5/28=0
e=(6,17+35/9-236/97)(1/3-0.25-1/12)=A.(1/3-1/4-1/12)=A.(1/12-1/12)=A.0=0
1 tháng 4 2015
e=(6,17 + 3 5/9 - 2 36/97 ) . ( 1/3 - 0,25 - 1/12)=(6,17 + 3 5/9 - 2 36/97 ) .(1/3-1/12-0,25)
=(6,17 + 3 5/9 - 2 36/97 ) .(4/12-1/12-0,25)=(6,17 + 3 5/9 - 2 36/97 ) .(3/12-0,25)
=(6,17 + 3 5/9 - 2 36/97 ) .(1/4-0.25)=(6,17 + 3 5/9 - 2 36/97 ) .(0.25-0.25)
=(6,17 + 3 5/9 - 2 36/97 ) .0=0
Vậy E=0
NT
2
PT
0
giúp mình với
\(\left(0,4-\dfrac{2}{9}+\dfrac{2}{11}\right):\left(1,4-\dfrac{7}{9}+\dfrac{7}{11}\right)\\ =\dfrac{0,4-\dfrac{2}{9}+\dfrac{2}{11}}{1,4-\dfrac{7}{9}+\dfrac{7}{11}}\\ =\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}\\ =\dfrac{2\cdot\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7\cdot\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}\\ =\dfrac{2}{7}\)