Tìm y:
\(1\frac{1}{3}+1\frac{1}{5}\cdot y-\frac{4}{5}=2\frac{4}{5}\)
Chý ý: dấu chấm là nhân
\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{5}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{2017}\right)\)
giấu chấm ở trên là dấu nhân còn mấy dấu chấm bên dưới là 3 chấm
GIẢI CHI TIẾT GIÚP MÌNH NHÉ
tìm y : a, 2 + \(\frac{3}{4-\frac{1}{2-y}}\)= 7
b, 16 : {\(\frac{\frac{3}{5}\cdot y+8}{21}+9\)} = \(\frac{7}{10}\)
DẤU Ở GIỮA 3/5 VÀ Y LÀ DẤU NHÂN
không nói linh tinh nha thánh troll trả lời thì trả lời đi bị trừ điểm đó
\(5\cdot\left(\frac{1}{5}+\frac{1}{17}\right)-\left(\frac{2}{5}+\frac{2}{17}+\frac{9}{15}+\frac{12}{68}\right)\)
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 ^^
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
\(\frac{19}{37}+\left(1-\frac{19}{37}\right)\)
\(=\frac{19}{37}+1-\frac{19}{37}\)
\(=\left(\frac{19}{37}-\frac{19}{37}\right)+1\)
\(=0+1=1\)
\(1\frac{4}{5}.2\frac{3}{6}.5\frac{1}{4}.\frac{12}{45}.\frac{20}{15}\)
Giải chi tiết nhé!
p/s: dấu chấm là dấu nhân nhé.
a)\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{2017\cdot2018}\) b)\(\left[x\cdot\frac{5}{3}-1\right]:9=3\frac{1}{2}:2,25\)
dấu chấm là dấu nhân nha :3
\(a,\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2017\cdot2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=1-\frac{1}{2018}\)
\(=\frac{2017}{2018}.\)
\(b,\left[x\cdot\frac{5}{3}-1\right]:9=3\frac{1}{2}:2,25\)
\(\Leftrightarrow\left[x\cdot\frac{5}{3}-1\right]:9=\frac{7}{2}:\frac{9}{4}\)
\(\Leftrightarrow\left[x\cdot\frac{5}{3}-1\right]:9=\frac{7}{2}\cdot\frac{4}{9}\)
\(\Leftrightarrow\left[x\cdot\frac{5}{3}-1\right]:9=\frac{14}{9}\)
\(\Leftrightarrow x\cdot\frac{5}{3}-1=\frac{14}{9}\cdot9\)
\(\Leftrightarrow x\cdot\frac{5}{3}-1=14\)
\(\Leftrightarrow x\cdot\frac{5}{3}=14+1\)
\(\Leftrightarrow x\cdot\frac{5}{3}=15\)
\(\Leftrightarrow x=15:\frac{5}{3}\)
\(\Leftrightarrow x=15\cdot\frac{3}{5}\)
\(\Leftrightarrow x=9.\)
a)\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2017.2018}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=\frac{1}{1}-\frac{1}{2018}\)
\(=\frac{2017}{2018}\)
b)\(\left[x.\frac{5}{3}-1\right]:9=3\frac{1}{2}:2,25\)
\(\Leftrightarrow\left[x.\frac{5}{3}-1\right]:9=3\frac{1}{2}:\frac{9}{4}=1\frac{5}{9}\)
\(\Rightarrow x.\frac{5}{3}-1=1\frac{5}{9}.9=14\)
\(\Rightarrow x.\frac{5}{3}=14+1=15\)
\(\Rightarrow x=15:\frac{5}{3}=9\)
a) \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2017\cdot2018}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=\frac{1}{1}-\frac{1}{2018}\)
\(=\frac{2018}{2018}-\frac{1}{2018}\)
\(=\frac{2017}{2018}\)
b) \(\left(x\cdot\frac{5}{3}-1\right):9=3\frac{1}{2}:2,25\)
\(\left(x\cdot\frac{5}{3}-1\right):9=\frac{7}{2}:\frac{9}{4}\)
\(\left(x\cdot\frac{5}{3}-1\right):9=\frac{14}{9}\)
\(x\cdot\frac{5}{3}-1=\frac{14}{9}\cdot9\)
\(x\cdot\frac{5}{3}-1=14\)
\(x\cdot\frac{5}{3}=14+1\)
\(x\cdot\frac{5}{3}=15\)
\(x=15:\frac{5}{3}\)
\(x=9\)
Vậy \(x=9\)
bài 1 : Tìm x và tính nhanh
\(\frac{3}{5}\cdot x-\frac{3}{2}=\frac{5}{6}\)
\(\frac{12}{15}+\frac{37}{7}+\frac{16}{5}+\frac{5}{7}\)
Lưu ý:Dấu . là dấu nhân
Giải rõ ràng cho mình
\(HELP\) \(ME\)\(!\)
1.
\(\frac{3}{4}x\frac{8}{5}:1\frac{1}{6}=\)\(2\frac{1}{3}x1\frac{1}{4}-\frac{7}{5}=\)\(4\frac{2}{3}+1\frac{1}{4}+2\frac{1}{3}+2\frac{3}{7}=\)2.
\(x.\frac{1}{2}=\frac{3}{4}+\frac{1}{5}\)\(x-\frac{1}{5}=\frac{2}{3}.\frac{9}{4}\)\(x.\frac{1}{5}+\frac{2}{3}=\frac{5}{4}\)\(\frac{13}{5}:x-\frac{1}{2}=\frac{4}{5}.\frac{15}{16}\)\(x:\frac{4}{9}+\frac{1}{3}=\frac{2}{5}.1\frac{13}{2}\)Chú ý: Dấu chấm là dấu nhân\(\dfrac{3}{4}\times\dfrac{8}{5}:1\dfrac{1}{6}\)
=\(\dfrac{6}{5}:\) \(\dfrac{7}{6}\)
=\(\dfrac{6}{5}\times\dfrac{6}{7}=\dfrac{36}{35}\)
2\(\dfrac{1}{3}\) x 1\(\dfrac{1}{4}\) -\(\dfrac{7}{5}\)
\(\dfrac{7}{3}\times\dfrac{5}{4}-\) \(\dfrac{7}{5}\)
\(\dfrac{35}{12}-\dfrac{7}{5}\)
\(\dfrac{175}{60}-\dfrac{84}{60}=\dfrac{91}{60}\)
4\(\dfrac{2}{3}+1\dfrac{1}{4} +2\dfrac{1}{3}+2\dfrac{3}{7}\)
(4 +2) + \(\left(\dfrac{2}{3}+\dfrac{1}{3}\right)\) +1\(\dfrac{1}{4}\) + \(2\dfrac{3}{7}\)
6 + 1 + \(\dfrac{5}{4}\) + \(\dfrac{17}{7}\)
7 + \(\dfrac{103}{28}\)
\(\dfrac{299}{28}\)
x. \(\dfrac{1}{2}\) =\(\dfrac{3}{4}+\dfrac{1}{5}\)
x . \(\dfrac{1}{2}=\dfrac{15}{20}+\dfrac{4}{20}\)
x.\(\dfrac{1}{2}=\dfrac{19}{20}\)
x = \(\dfrac{19}{20}:\dfrac{1}{2}\)
x = 19/10
\(x-\dfrac{1}{5}=\dfrac{2}{3}\) . \(\dfrac{9}{4}\)
\(x-\dfrac{1}{5}=\) \(\dfrac{3}{2}\)
x = \(\dfrac{3}{2}+\dfrac{1}{5}\)
x =\(\dfrac{15}{10}\) + \(\dfrac{3}{10}\)
x=\(\dfrac{18}{10}\) =\(\dfrac{9}{5}\)