\(\frac{3}{4}\)*\(\frac{8}{9}\)*\(\frac{15}{16}\cdot...\cdot\frac{9999}{10000}\)
dấu sao với dấu . là nhân nha các bạn
các bạn giải nhanh và đầy đủ dùm mik nha. mik đag cần gấp. cảm ơn trước nha :)
\(5\cdot\left(\frac{1}{5}+\frac{1}{17}\right)-\left(\frac{2}{5}+\frac{2}{17}+\frac{9}{15}+\frac{12}{68}\right)\)
Dấu chấm là dấu nhân nha các bạn giải đầy dủ giùm mình
5.(1/5+1/17)-(2/5+2/17+9/15+12/68)
=5.22/85-22/17
=22/17-22/17
=0
Ta có : \(5\cdot\left(\frac{1}{5}+\frac{1}{17}\right)-\left(\frac{2}{5}+\frac{2}{17}+\frac{9}{15}+\frac{12}{68}\right)\)
\(=\) \(5\cdot\frac{1}{5}+5\cdot\frac{1}{17}-\left(\frac{2}{5}+\frac{2}{17}+\frac{3}{5}+\frac{3}{17}\right)\)
\(=\) \(1+\frac{5}{17}-\left[\left(\frac{2}{5}+\frac{3}{5}\right)+\left(\frac{2}{17}+\frac{3}{17}\right)\right]\)
\(=\) \(1+\frac{5}{17}-\left(1+\frac{5}{17}\right)\)
\(=\) \(1+\frac{5}{17}-1-\frac{5}{17}\)
\(=\)\(0\)
Vậy ...
Tk ủng hộ mk nha các bn ❣❣ C.ơn nhiều ^^
\(\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot\cdot\cdot\cdot\frac{9999}{10000}\)=?
\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{9999}{10000}=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}....\frac{99.101}{100.100}\)
=\(\frac{1.3.2.4.3.5....999.101}{2.2.3.3.4.4....100.100}=\frac{1.101}{2.100}=\frac{101}{200}\)
\(\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}..........\frac{9999}{10000}\)
tính nhanh
\(\frac{19}{37}+\left(1-\frac{19}{37}\right)\)
\(\frac{7}{13}\cdot\frac{5}{14}\cdot\frac{39}{15}\)
\(2\frac{3}{7}\cdot\frac{1}{2}-\frac{1}{2}\cdot\frac{3}{7}+\frac{1}{3}\)
\(\frac{9}{5}:\frac{17}{15}+\frac{8}{5}:\frac{17}{15}\)
\(\frac{2017}{2018}\cdot\frac{1}{2019}+\frac{2017}{2018}:\frac{2019}{2018}+\frac{1}{2018}\)
\(\frac{637\cdot527-189}{526\cdot637+448}\)
\(\frac{4}{5\cdot7}+\frac{4}{7\cdot9}+\frac{4}{9\cdot11}+...+\frac{4}{23\cdot25}\)
dấu . là dấu nhân nha mọi người
\(\frac{19}{37}+\left(1-\frac{19}{37}\right)\)
\(=\frac{19}{37}+1-\frac{19}{37}\)
\(=\left(\frac{19}{37}-\frac{19}{37}\right)+1\)
\(=0+1=1\)
tìm x
\(\frac{2017}{2017\cdot\left(x\cdot1\right)}\)=21
dấu chấm là dấu nhân nha ghi đầy đủ cách giải
ta có x.1=21(vì 2017:2017=1 nên triệt tiêu)
x =21:1=21
\(\frac{2017}{2017\cdot\left(x\cdot1\right)}=21\)
\(\Leftrightarrow\frac{1}{x}=21\)
\(\Leftrightarrow x=\frac{1}{21}\)
\(\frac{2017}{2017.\left(x.1\right)}=21\)
\(\frac{1}{x}=21\)
\(x=\frac{1}{21}\)
\(\frac{7^2.3}{2.3.5^2}\)dấu . là dấu nhân nha
giải nhanh hộ mik với cảm ơn nhìu please
\(\frac{7^2.3}{2.3.5^2}\)
\(=\frac{7^2.1}{2.1.5^2}\)
\(=\frac{7^2}{2.5^2}\)
\(=\frac{49}{2.25}\)
\(=\frac{49}{50}\)
\(\frac{7^2.3}{2.3.5^2}=\frac{7^2.3:3}{2.5^2.3:3}=\frac{7^2}{2.5^2}=\frac{49}{50}\)
Tk nha
\(\frac{1}{3}+x\cdot\frac{2}{7}=\frac{11}{12}\)
dấu chấm đằng sau x là nhân nha các bạn
à, bài này là tìm x nha các bạn
\(\frac{1}{3}+x\times\frac{2}{7}=\frac{11}{12}\)
\(x\times\frac{2}{7}=\frac{11}{12}-\frac{1}{3}\)
\(x\times\frac{2}{7}=\frac{7}{12}\)
\(x=\frac{7}{12}:\frac{2}{7}\)
\(x=\frac{49}{24}\)
~Moon~
\(\frac{1}{3}+x.\frac{2}{7}=\frac{11}{12}\)
\(x.\frac{2}{7}=\frac{11}{12}-\frac{1}{3}\)
\(x.\frac{2}{7}=\frac{7}{12}\)
\(x=\frac{7}{12}\div\frac{2}{7}\)
\(x=\frac{49}{24}\)
bạn làm thế này nhé
1/3 + x . 2/7 = 11/12
<=> 2x/7=11/12 - 1/3
<=> 2x/7= 7/12
<=> 2x.12=7.7
<=> 24x = 49
<=> x = 49/24
Vậy x = 49/24
Bài 1:Tính bằng cách hợp lý:
a)\(\frac{8.4.125.25+96524+3476}{1+5+9+...+61+62,5-26}\)
b)\(\frac{5+55+555+5555}{9+99+999+9999}\)
c)39,2.27+39,2.43+78,4.15
d)\(\frac{4}{17}.\frac{3}{11}+\frac{8}{11}.\frac{4}{17}-\frac{4}{17}\)
e)(812,1.10,5-91.10,5).(41.11-4100.0,1-41)
Bài 2:Tính nhanh các tổng sau:
a)\(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{57.59}\)
b)\(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right)-\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)\)
c)\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).\left(1-\frac{1}{5}\right)\cdot...\cdot\left(1-\frac{1}{2012}\right)\)
GẤP!GẤP!GẤP!MAI MIK NỘP RỒI!BẠN NÀO ĐẦU TIÊN MIK TIK CHO NHA!
*Lưu ý:Dấu chấm là dấu nhân đó
Bài 1.
b) \(\frac{5+55+555+5555}{9+99+999+9999}\)
= \(\frac{5\left(1+11+111+1111\right)}{9\left(1+11+111+1111\right)}=\frac{5}{9}\)
c) \(39,2\cdot27+39,2\cdot43+78,4\cdot15\)
= \(39,2\cdot27+39,2\cdot43+39,2\cdot2\cdot15\)
= \(39,2\left(27+43+30\right)=39,2\cdot100=3920\)
d) \(\frac{4}{17}\cdot\frac{3}{11}+\frac{8}{11}\cdot\frac{4}{17}-\frac{4}{17}\)
= \(\frac{4}{17}\left(\frac{3}{11}+\frac{8}{11}-1\right)=\frac{4}{17}\cdot0=0\)
Bài 2.
a) \(\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+...+\frac{1}{57\cdot59}\)
= \(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{57}-\frac{1}{59}\)
= \(\frac{1}{5}-\frac{1}{59}=\frac{54}{295}\)
b) \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right)-\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)\)
= \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}-\frac{1}{6}-\frac{1}{7}-\frac{1}{8}\)
= \(\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)
c) \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)...\left(1-\frac{1}{2012}\right)\)
= \(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot...\cdot\frac{2011}{2012}=\frac{1}{2012}\)
\(\frac{3}{4\cdot}\frac{8}{9\cdot}\frac{15}{16}.........\cdot\frac{9999}{10.000}\)=?