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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
3 tháng 11 2023
\(\dfrac{8}{9}\) : ( 2 - 3 \(\times\) y) = \(\dfrac{5}{3}\)
2 - 3 \(\times\) y = \(\dfrac{8}{9}\) : \(\dfrac{5}{3}\)
2 - 3 \(\times\) y = \(\dfrac{8}{15}\)
3 \(\times\) y = 2 - \(\dfrac{8}{15}\)
3 \(\times\) y = \(\dfrac{22}{15}\)
y = \(\dfrac{22}{15}\) : 3
y = \(\dfrac{22}{45}\)
BN
5
PN
10 tháng 8 2020
\(\frac{1}{3}-\frac{3}{5}+\frac{5}{7}-\frac{7}{9}+\frac{9}{11}-\frac{11}{13}-\frac{9}{11}+\frac{7}{9}-\frac{5}{7}+\frac{3}{5}-\frac{1}{3}\)
\(=\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{3}{5}-\frac{3}{5}\right)+\left(\frac{5}{7}-\frac{5}{7}\right)+\left(\frac{7}{9}-\frac{7}{9}\right)+\left(\frac{9}{11}-\frac{9}{11}\right)-\frac{11}{13}\)
\(=0+0+0+0+0-\frac{11}{13}=-\frac{11}{13}\)
VC
Vũ Cao Minh⁀ᶦᵈᵒᶫ ( ✎﹏IDΣΛ亗 )
CTV
VIP
10 tháng 8 2020
\(\frac{1}{3}-\frac{3}{5}+\frac{5}{7}-\frac{7}{9}+\frac{9}{11}+\frac{7}{9}-\frac{5}{7}+\frac{3}{5}-\frac{1}{3}\)
\(=\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{3}{5}-\frac{3}{5}\right)+\left(\frac{5}{7}-\frac{5}{7}\right)+\left(\frac{7}{9}-\frac{7}{9}\right)-\frac{11}{13}\)
\(=0+0+0+0+0-\frac{11}{13}=-\frac{11}{13}\)
NL
1
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
NQ
1
N
6 tháng 7 2018
A=3.(1/1.2+1/2.3+1/3.4+.....+1/399.400)
A=3.(1/1-1/2+1/2-1/3+......+1/399-1/400)
A=3.(1-1/400)
A=3.399/400
A=1197/400
BT
6 tháng 7 2018
A=3.(1/1.2+1/2.3+1/3.4+.....+1/399.400)
A=3.(1/1-1/2+1/2-1/3+......+1/399-1/400)
A=3.(1-1/400)
A=3.399/400
A=1197/400
MP
5 tháng 7 2018
Bài 1: Tính nhanh:
A = 3/1*2 + 3/2*3 + 3/3*4 + ... + 3/399*400
=>3A = 1/1*2 + 1/2*3 + 1/3*4 + ... + 1/399*400
3A = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/399 - 1/400
3A = 1 - 1/400
3A = 400/400 - 1/400
3A = 399/400
A = 399/400 : 3
A = 399/400 . 1/3
A = 133/400.
Có gì ko hiểu bn ib mk nha.^^
5 tháng 7 2018
\(A=\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{399.400}\)
\(A=3.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{399.400}\right)\)
\(A=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{399}-\frac{1}{400}\right)\)
\(A=3.\left(1-\frac{1}{400}\right)\)
\(A=3.\frac{399}{400}\)
\(A=\frac{1197}{400}\)
\(B=\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+...+\frac{5}{399.400}\)
\(B=5.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{399.400}\right)\)
\(B=5.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{399}-\frac{1}{400}\right)\)
\(B=5.\left(1-\frac{1}{400}\right)\)
\(B=5.\frac{399}{400}\)
\(B=\frac{399}{80}\)
\(C=\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{149.151}\)
\(C=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{149}-\frac{1}{151}\)
\(C=\frac{1}{5}-\frac{1}{151}\)
\(C=\frac{146}{755}\)
\(D=\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}+...+\frac{3}{149.151}\)
\(D=\frac{3}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{149.151}\right)\)
\(D=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{149}-\frac{1}{151}\right)\)
\(D=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{151}\right)\)
\(D=\frac{3}{2}.\frac{146}{755}\)
\(D=\frac{219}{755}\)
\(E=\frac{11}{1.3}+\frac{11}{3.5}+\frac{11}{5.7}+...+\frac{11}{99.101}\)
\(E=\frac{11}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(E=\frac{11}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(E=\frac{11}{2}.\left(1-\frac{1}{101}\right)\)
\(E=\frac{11}{2}.\frac{100}{101}\)
\(E=\frac{550}{101}\)
_Chúc bạn học tốt_
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
-6 , tick mk nha bạn thân !!!
1+(-3)+5+(-7)+9+(-11)
=[1+5+9]+[(-3)+(-7)+(-11)]
= 15 + (-21)
= (-6)