bài 1: tính nhanh:
A = 4/1*3*5+4/3*5*7+4/5*7*9+4/7*9*11+4/9*11*13
giúp mik nha !! Mik cảm ơn nhiều
K
Khách
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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) \(4\cdot11\cdot\dfrac{3}{4}\cdot\dfrac{9}{121}\)
\(=\left(4\cdot\dfrac{3}{4}\right)\cdot\left(11\cdot\dfrac{9}{121}\right)\)
\(=3\cdot\dfrac{9}{11}\)
\(=\dfrac{27}{11}\)
7 tháng 3 2022
A) 3\7 + 5\9 +4\7+ 4\9=(3\7+4\7) + ( 5\9 +4\9) = 7\7+9\9=63\63+63\63=126\63=42\21
b) 1\5+4\11+4\5+ 7\11=(1\5+4\5)+ (4\11+7\11)=5\5+11\11=55\55+55\55=110\55=22\11
27 tháng 2 2022
\(a,15\dfrac{3}{13}-\left(3\dfrac{4}{7}+8\dfrac{3}{13}\right)=15\dfrac{3}{13}-3\dfrac{4}{7}-8\dfrac{3}{13}=\left(15\dfrac{3}{13}-8\dfrac{3}{13}\right)-\dfrac{25}{7}=7-\dfrac{25}{7}=\dfrac{49}{7}-\dfrac{25}{7}=\dfrac{24}{7}\)
\(b,\left(7\dfrac{4}{9}+4\dfrac{7}{11}\right)-3\dfrac{4}{9}=\left(7\dfrac{4}{9}-3\dfrac{4}{9}\right)+4\dfrac{4}{9}=4+\dfrac{40}{9}=\dfrac{36}{9}+\dfrac{40}{9}=\dfrac{76}{9}\)
\(c,\dfrac{-7}{9}.\dfrac{4}{11}+\dfrac{-7}{9}.\dfrac{7}{11}+5\dfrac{7}{9}=\dfrac{-7}{9}\left(\dfrac{4}{11}+\dfrac{7}{11}\right)+\dfrac{52}{9}=\dfrac{-7}{9}.1+\dfrac{52}{9}=\dfrac{-7}{9}+\dfrac{52}{9}=\dfrac{45}{9}=5\)
\(d,50\%.1\dfrac{1}{3}.10.\dfrac{7}{35}.0,75=\dfrac{1}{2}.\dfrac{4}{3}.10.\dfrac{1}{5}.\dfrac{3}{4}=\left(\dfrac{1}{2}.\dfrac{1}{5}.10\right).\left(\dfrac{4}{3}.\dfrac{3}{4}\right)=1.1=1\)
\(e,\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{40.43}=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{40}-\dfrac{1}{43}=1-\dfrac{1}{43}=\dfrac{42}{43}\)
KN
5
3 tháng 2 2021
nhóm các số có 2 chữ số với số đuôi tương tự.VD:(11-1)+(12-2)...
Chúc bạn làm đc
TL
3 tháng 2 2021
Bài làm:
1-2+3-4+5-6+7-8+9-10+11-12+13-14+15-16
= ( 1-2 ) + ( 3-4 ) + ( 5 -6 ) + ( 7- 8 ) + ( 9 - 10 ) + ( 11 - 12 ) + ( 13 -14 ) + ( 15-16 ) { có 8 nhóm }
= ( -1 ) + ( -1 ) + ( -1 ) + ( -1 ) + ( -1 ) + ( -1 ) + ( -1 )+ ( -1 )
= ( -1 ) . 8
= 1 ( vì các số âm là số chẵn )
Đây là bài làm của mình, mình tính theo cách tính nhanh nha
16 tháng 4 2022
Bài 2:
a: 5/7-3/7x=1
=>3/7x=-2/7
hay x=-2/3
b: \(x+\dfrac{2}{3}=\dfrac{-1}{12}\cdot\dfrac{-4}{5}=\dfrac{1}{15}\)
=>x=1/15-2/3=-9/15=-3/5
c: \(x-4=\dfrac{-14}{35}:\dfrac{7}{5}=\dfrac{-14}{35}\cdot\dfrac{5}{7}=\dfrac{-70}{245}=\dfrac{-2}{7}\)
=>x=4-2/7=26/7
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_
\(A=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)
\(\Leftrightarrow A=\frac{1}{1.3}-\frac{1}{11.13}=\frac{1}{3}-\frac{1}{143}=\frac{140}{429}\)
\(A=\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+\frac{4}{7.9.11}+\frac{4}{9.11.13}\)
\(\Rightarrow A=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)
\(\Rightarrow A=\frac{1}{1.3}-\frac{1}{11.13}=\frac{1}{3}-\frac{1}{143}=\frac{140}{429}\)