BÀI 1 tính
a) (7 4/9+4 7/11)-3 4/9 b) -7/9.4/11+-7/9.7/11+5 7/9 c) 50% . 1 1/3 . 10 7/35.0,75 đ) 3/1.4+3/4.7+3/7.10+....+3/40.43
GIÚP MK VỚI CÁC BN ƠI!!!
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
7 tháng 6 2018
\(A=\frac{-3}{5}+\left(\frac{-2}{5}+2\right)\)
\(A=\frac{-3}{5}+\frac{8}{5}\)
\(A=1\)
\(B=\frac{3}{7}+\left(\frac{-1}{5}+\frac{-3}{7}\right)\)
\(B=\frac{3}{7}+\frac{-1}{5}+\frac{-3}{7}\)
\(B=\left(\frac{3}{7}+\frac{-3}{7}\right)+\frac{-1}{5}\)
\(B=0+\frac{-1}{5}\)
\(B=\frac{-1}{5}\)
\(C=\frac{-7}{9}.\frac{4}{11}+\frac{-7}{9}.\frac{7}{11}+\frac{52}{9}\)
\(C=\frac{-7}{9}.\left(\frac{4}{11}+\frac{7}{11}\right)+\frac{52}{9}\)
\(C=\frac{-7}{9}.1+\frac{52}{9}\)
\(C=\frac{-7}{9}+\frac{52}{9}\)
\(C=5\)
\(D=50\%.\frac{4}{3}.10.\frac{7}{35}.0,75\)
\(D=\frac{50}{100}.\frac{4}{3}.10.\frac{7}{35}.\frac{75}{100}\)
\(D=\frac{1}{2}.\frac{4}{3}.10.\frac{1}{5}.\frac{3}{4}\)
\(D=\left(\frac{1}{2}.\frac{1}{5}.10\right).\left(\frac{4}{3}.\frac{3}{4}\right)\)
\(D=1.1\)
\(D=1\)
7 tháng 6 2018
A= 1
B= -1/5
C=5
D=1
mik có ý kiến lần sau bạn đặt câu hỏi nhớ ghi phân số trên dưới cho dễ nhìn nha
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: \(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{40\cdot43}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)
\(=1-\frac{1}{43}=\frac{43}{43}-\frac{1}{43}\)
\(=\frac{42}{43}\)
YN
2 tháng 5 2021
a, -3/4 + 3/7 + -1/4 + 4/9 + 4/7
=(−34+−14)+(37+47)+49(−34+−14)+(37+47)+49
=−1+1+49−1+1+49
=49
4 tháng 5 2016
\(a,\frac{-3}{4}+\frac{3}{7}+\frac{-1}{4}+\frac{4}{9}+\frac{4}{7}\)
\(=\frac{-3}{4}+\frac{-1}{4}+\frac{3}{7}+\frac{4}{7}+\frac{4}{9}\)
\(=-1+1+\frac{4}{9}\)
\(=\frac{4}{9}\)
\(b,\frac{-7}{9}\cdot\frac{4}{11}+\frac{-7}{9}\cdot\frac{7}{11}+5\frac{7}{9}\)
\(=\frac{-7}{9}\cdot\left(\frac{4}{11}+\frac{7}{11}\right)+\frac{52}{9}\)
\(=\frac{-7}{9}\cdot1+\frac{52}{9}\)
\(=\frac{-7}{9}+\frac{52}{9}=\frac{45}{9}=5\)
Ý c) bn tự làm nha. Mk ko làm đc. Nhìn nó rắc rối lắm.
Chúc bn hc tốt nha !!! TỨ DIỆP THẢO
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
23 tháng 5
\(\dfrac{1}{2}+\dfrac{3}{4}-\left(\dfrac{3}{4}-\dfrac{4}{5}\right)\)
=\(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{3}{4}+\dfrac{4}{5}\)
=\(\left(\dfrac{1}{2}+\dfrac{4}{5}\right)+\left(\dfrac{3}{4}-\dfrac{3}{4}\right)\)
=\(\left(\dfrac{5}{10}+\dfrac{8}{10}\right)+0\)
=\(\dfrac{13}{10}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
23 tháng 5
\(-\dfrac{7}{25}.\dfrac{11}{13}+\left(-\dfrac{7}{25}\right).\dfrac{2}{13}-\dfrac{18}{25}\)
=\(-\dfrac{7}{25}.\cdot\left(\dfrac{11}{13}+\dfrac{2}{13}\right)-\dfrac{18}{25}\)
=\(-\dfrac{7}{25}.1-\dfrac{18}{25}\)
=\(-\dfrac{7}{25}-\dfrac{18}{25}\)
=\(-\dfrac{25}{25}\) = \(-1\)
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
5 tháng 3 2023
a: \(=\left(15-6-\dfrac{13}{18}\right):\dfrac{298}{27}-\dfrac{17}{8}:\dfrac{51}{40}\)
\(=\dfrac{149}{18}\cdot\dfrac{27}{298}-\dfrac{5}{3}=\dfrac{3}{2}-\dfrac{5}{3}=\dfrac{9-10}{6}=\dfrac{-1}{6}\)
b: \(=\dfrac{-16}{5}\cdot\dfrac{-15}{64}+\dfrac{-22}{15}:\dfrac{11}{2}\)
\(=\dfrac{3}{4}-\dfrac{4}{15}=\dfrac{29}{60}\)
c: \(=\dfrac{-7}{9}\left(\dfrac{4}{11}+\dfrac{7}{11}\right)+5+\dfrac{7}{9}=\dfrac{-7}{9}+\dfrac{7}{9}+5=5\)
d: \(=\dfrac{1}{2}\cdot\dfrac{4}{3}\cdot10\cdot\dfrac{1}{5}\cdot\dfrac{3}{4}=1\)
e: \(=\dfrac{4}{25}+\dfrac{11}{2}\cdot\dfrac{5}{2}+\dfrac{-23}{4}=\dfrac{204}{25}\)
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)\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}\)=\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)=\(\frac{43}{43}-\frac{1}{43}\)=\(\frac{42}{43}\)
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) \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}\)
\(=1.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\right)\)
\(=1.\left(1-\frac{1}{43}\right)\)
\(=1.\frac{42}{43}\)
\(=\frac{42}{43}\)