Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\left(-\frac{2}{3}+\frac{3}{7}\right):\frac{4}{5}+\left(-\frac{1}{3}+\frac{4}{7}\right):\frac{4}{5}\)
\(=\left[\left(-\frac{2}{3}+\frac{3}{7}\right)+\left(-\frac{1}{3}+\frac{4}{7}\right)\right]:\frac{4}{5}\)
\(=\left[\left(-\frac{2}{3}-\frac{1}{3}\right)+\left(\frac{3}{7}+\frac{4}{7}\right)\right]:\frac{4}{5}\)
\(=0:\frac{4}{5}\)
\(=0\)
b) \(\frac{5}{9}\left(\frac{1}{11}-\frac{5}{22}\right)+\frac{5}{9}\left(\frac{1}{15}-\frac{2}{3}\right)\)
\(=\frac{5}{9}\left[\left(\frac{2}{22}-\frac{5}{22}\right)+\left(\frac{1}{15}-\frac{10}{15}\right)\right]\)
\(=\frac{5}{9}\left[-\frac{3}{22}+\left(-\frac{3}{5}\right)\right]\)
\(=\frac{5}{9}\left(-\frac{81}{110}\right)\)
\(=-\frac{9}{22}\)
BÀI 1
a, \(5\times\frac{-7}{10}=\frac{-35}{10}=\frac{-7}{2}\)
b, \(\frac{4}{5}\times\frac{-7}{10}=\frac{-28}{50}=\frac{-14}{25}\)
c, \(\frac{4}{9}+\frac{4}{3}\times\frac{16}{4}=\frac{4}{9}+\frac{16}{3}=\frac{52}{9}\)
d, \(\frac{11}{22}-\frac{3}{9}\times\frac{14}{21}=\frac{11}{22}-\frac{2}{9}=\frac{55}{198}=\frac{5}{18}\)
BÀI 2
\(A=\frac{6}{13}\times\frac{5}{7}+\frac{6}{13}\times\frac{2}{7}+\frac{17}{13}\)
\(A=\frac{30}{91}+\frac{12}{91}+\frac{17}{13}\)
\(A=\frac{30}{91}+\frac{12}{91}+\frac{119}{91}\)
\(A=\frac{161}{91}=\frac{23}{13}\)
\(B=\frac{11}{15}\times\frac{4}{11}+\frac{11}{15}\times\frac{5}{11}+\frac{11}{15}\times\frac{2}{11}\)
\(B=\frac{4}{15}+\frac{1}{3}+\frac{2}{15}\)
\(B=\frac{11}{15}\)
\(C=\left(\frac{19}{64}-\frac{33}{22}+\frac{24}{51}\right)\times\left(\frac{1}{5}-\frac{1}{15}-\frac{2}{15}\right)\)
\(C=\frac{-797}{1088}\times0\)
\(C=0\)
\(D=\frac{8}{13}\times\frac{7}{12}+\frac{8}{13}\times\frac{5}{12}-\frac{1}{12}\)
\(D=\frac{14}{39}+\frac{10}{39}-\frac{1}{12}\)
\(D=\frac{83}{156}\)
bạn biết câu náy không (24 + 11) . {546 - [14 . (64 - 2^{3}3) : 2]} =
a) 5/9 + 4/9 . 3/7 + 4/9 . 4/7
= 5/9 + 4/9 . (3/7 + 4/7)
= 5/9 + 4/9 . 1
= 5/9 + 4/9
= 1
a)\(-\dfrac{2}{5}.\dfrac{4}{7}+\dfrac{-3}{5}.\dfrac{2}{7}+\dfrac{-3}{5}\)
=\(-\dfrac{2}{5}.\dfrac{4}{7}+\dfrac{3}{7}.\dfrac{-2}{5}+\dfrac{-3}{5}\)
=\(-\dfrac{2}{5}.\left(\dfrac{4}{7}+\dfrac{3}{7}\right)+\dfrac{-3}{5}\)
=\(\dfrac{-2}{5}.1+\dfrac{-3}{5}\)
=\(-\dfrac{2}{5}+\dfrac{-3}{5}\)
=\(-\dfrac{5}{5}\) = -1
\(\dfrac{5}{9}.\dfrac{14}{17}+\dfrac{1}{17}.\dfrac{5}{9}+\dfrac{2}{9}+\dfrac{5}{12}\)
=\(\dfrac{5}{9}.\left(\dfrac{14}{17}+\dfrac{1}{17}\right)+\dfrac{2}{9}+\dfrac{5}{12}\)
=\(\dfrac{5}{9}.\dfrac{15}{17}+\dfrac{2}{9}+\dfrac{5}{12}\)
=\(\dfrac{25}{51}+\dfrac{2}{9}+\dfrac{5}{12}\)
=\(\dfrac{691}{612}\)
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 \(\frac{25}{11}\)
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
a)\(1-2+3-4+5-6+7-8+8-9+9-10\)
=\(\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+\left(7-8\right)+\left(8-9\right)+\left(9-10\right)\)
\(=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)\)
\(=\left(-1\right).6\)
\(=-6\)
b)\(1-2+3-4+...+99-100\)
\(=\left(1-2\right)+\left(3-4\right)+...+\left(99-100\right)\)}\(\left[\left(100-1\right):1+1\right]:2=50\)(cặp)
\(=\left(-1\right)+\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)} 50 số (-1)
\(=\left(-1\right).50\)
\(=-50\)
c)\(1-3+5-7+9-11+13-15\)
\(=\left(1-3\right)+\left(5-7\right)+\left(9-11\right)+\left(13-15\right)\)
\(=\left(-2\right)+\left(-2\right)+\left(-2\right)+\left(-2\right)\)
\(=\left(-2\right).4\)
\(=-8\)
d)\(1-3+5-7+...-99+101\) (Đối với bài này, có vẻ đề sai, mình đã sửa lại rồi
\(=\left(1-3\right)+\left(5-7\right)+...+\left(97-99\right)+101\) } \(\left[\left(99-1\right):2+1\right]:2=25\)(cặp)
\(=\left(-2\right)+\left(-2\right)+\left(-2\right)+...+\left(-2\right)\) } 25 số (-2)
\(=\left(-2\right).25\)
\(=-50\)
e)\(-1-2-3-4-...-99-100\)
\(=\left(-1\right)+\left(-2\right)+\left(-3\right)+...+\left(-99\right)+\left(-100\right)\)
\(=\left[\left(-1\right)+\left(-100\right)\right]+\left[\left(-2\right)+\left(-99\right)\right]+...+\left[\left(-51\right)+\left(-50\right)\right]\) } \(\left[\left(100-1\right):1+1\right]:2=50\)(cặp) (phần này của đề bài, không thay được như (-100) hoặc (-1))
\(=\left(-100\right)+\left(-100\right)+\left(-100\right)+...+\left(-100\right)\)} 50 số (-100)
\(=\left(-100\right).50\)
\(=-5000\)
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
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
a: \(\left(-\dfrac{2}{3}-\dfrac{3}{7}\right):\dfrac{4}{5}+\left(-\dfrac{1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)
\(=\left(-\dfrac{2}{3}-\dfrac{3}{7}-\dfrac{1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)
\(=\left(-1+\dfrac{1}{7}\right)\cdot\dfrac{5}{4}=\dfrac{-6}{7}\cdot\dfrac{5}{4}=\dfrac{-30}{28}=-\dfrac{15}{14}\)
b: \(\dfrac{5}{9}:\left(\dfrac{1}{11}-\dfrac{5}{22}\right)+\dfrac{5}{9}:\left(\dfrac{1}{15}-\dfrac{2}{3}\right)\)
\(=\dfrac{5}{9}:\dfrac{-3}{22}+\dfrac{5}{9}:\left(\dfrac{1}{15}-\dfrac{10}{15}\right)\)
\(=\dfrac{5}{9}\cdot\dfrac{-22}{3}+\dfrac{5}{9}:\dfrac{-9}{15}\)
\(=\dfrac{5}{9}\cdot\dfrac{-22}{3}+\dfrac{5}{9}\cdot\dfrac{15}{-9}\)
\(=\dfrac{5}{9}\left(-\dfrac{22}{3}-\dfrac{5}{3}\right)=\dfrac{5}{9}\cdot\dfrac{-27}{3}=\dfrac{5}{9}\cdot\left(-9\right)=-5\)
c: \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right)\cdot\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\)
\(=\left(\dfrac{12}{12}+\dfrac{8}{12}-\dfrac{3}{12}\right)\cdot\left(\dfrac{16}{20}-\dfrac{15}{20}\right)^2\)
\(=\dfrac{17}{12}\cdot\left(\dfrac{1}{20}\right)^2=\dfrac{17}{12}\cdot\dfrac{1}{400}=\dfrac{17}{4800}\)
d: \(2:\left(\dfrac{1}{2}-\dfrac{2}{3}\right)^2=2:\left(\dfrac{3}{6}-\dfrac{4}{6}\right)^2\)
\(=2:\left(-\dfrac{1}{6}\right)^2=2:\dfrac{1}{36}=72\)
a; (- \(\dfrac{2}{3}\) - \(\dfrac{3}{7}\)): \(\dfrac{4}{5}\) + (- \(\dfrac{1}{3}\) + \(\dfrac{4}{7}\)): \(\dfrac{4}{5}\)
= (- \(\dfrac{2}{3}\) - \(\dfrac{3}{7}\)) x \(\dfrac{5}{4}\) + (- \(\dfrac{1}{3}\) + \(\dfrac{4}{7}\)) x \(\dfrac{5}{4}\)
= (- \(\dfrac{2}{3}\) - \(\dfrac{3}{7}\) - \(\dfrac{1}{3}\) + \(\dfrac{4}{7}\)) x \(\dfrac{5}{4}\)
= [ (- \(\dfrac{2}{3}\) - \(\dfrac{1}{3}\)) - (\(\dfrac{3}{7}\) - \(\dfrac{4}{7}\))] x \(\dfrac{5}{4}\)
= [ - 1 + \(\dfrac{1}{7}\)] x \(\dfrac{5}{4}\)
= [- \(\dfrac{7}{7}\) + \(\dfrac{1}{7}\)] x \(\dfrac{5}{4}\)
= - \(\dfrac{6}{7}\) x \(\dfrac{5}{4}\)
= - \(\dfrac{15}{14}\)