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bài 1 a)
\(\frac{15}{7}\times\frac43+\frac63\)
=\(\frac{60}{21}+\frac63\)
=\(\frac{60}{21}+\frac{42}{21}\)
=\(\frac{102}{21}\)
=\(\frac{34}{7}\)
b)\(\frac{-13}{41}\times\frac{17}{11}\times\frac{11}{17}\)
=\(\frac{-13}{41}\times\left(\frac{17}{11}\times\frac{11}{17}\right)\)
=\(\frac{-13}{41}\times1\)
=\(\frac{-13}{41}\)
a) (\(6\frac{2}{7}.x+\frac{3}{7}\))=-1.\(\frac{11}{5}+\frac{3}{7}\)
(\(6\frac{2}{7}.x+\frac{3}{7}\))=\(\frac{-62}{35}\)
\(\frac{44}{7}.x\)=\(\frac{-62}{35}-\frac{3}{7}\)
\(\frac{44}{7}.x=\frac{-77}{35}\)
x=\(\frac{-77}{35}:\frac{44}{7}\)=\(\frac{539}{1540}\)
(6\(\frac27x\) + \(\frac37\)) : 2\(\frac15\) - \(\frac37\) = - 1
(\(\frac{44}{7}x\) + \(\frac37\)) : \(\frac{11}{5}\) = - 1 + \(\frac37\)
(\(\frac{44}{7}x+\) \(\frac37\)) : \(\frac{11}{5}\) = - \(\frac77+\frac37\)
\(\frac{44x+3}{7}\) : \(\frac{11}{5}\) = - \(\frac47\)
(44\(x\) + 3) : \(\frac{11}{5}\) = - 4
44\(x\) + 3 = - 4 x \(\frac{11}{5}\)
44\(x\) + 3 = - \(\frac{44}{5}\)
44\(x\) = - \(\frac{44}{5}\) - 3
44\(x\) = - \(\frac{59}{5}\)
\(x=-\frac{59}{5}:44\)
\(x\) = - \(\frac{59}{250}\)
Vậy \(x\) = - \(\frac{59}{250}\)
j) \(\frac{8}{3}.\frac{2}{5}.\frac{3}{8}.10.\frac{19}{92}=\left(\frac{8}{3}.\frac{3}{8}\right).\left(\frac{2}{5}.10\right).\frac{19}{92}=1.4.\frac{19}{92}\)
\(=\frac{19}{23}\)
k)\(\frac{-5}{7}.\frac{2}{11}.\frac{-13}{17}.\frac{19}{12}.\frac{17}{13}=\left(\frac{2}{11}.\frac{19}{12}\right).\left(\frac{-13}{17}.\frac{17}{13}\right).\frac{-5}{7}\)
\(=\frac{-19}{66}.\frac{-5}{7}=\frac{95}{462}\)
l)\(\frac{12}{19}.\frac{7}{15}.\frac{-13}{17}.\frac{19}{12}.\frac{17}{13}=\left(\frac{12}{19}.\frac{19}{12}\right).\left(\frac{-13}{17}.\frac{17}{13}\right).\frac{7}{15}\)
\(=\frac{-7}{15}\)
cậu tham khảo trên này ạ, nếu đúng cho mk 1 t.i.c.k ạ, thank nhiều
a: \(\Leftrightarrow-\dfrac{9}{46}+\dfrac{108}{46}-\dfrac{93}{23}:\left(\dfrac{13}{4}-\dfrac{5}{3}x\right)=1\)
\(\Leftrightarrow\dfrac{93}{23}:\left(\dfrac{13}{4}-\dfrac{5}{3}x\right)=\dfrac{53}{46}\)
\(\Leftrightarrow-\dfrac{5}{3}x+\dfrac{13}{4}=\dfrac{186}{53}\)
=>-5/3x=55/212
hay x=-33/212
c: \(\Leftrightarrow\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{x\left(x+3\right)}=\dfrac{18}{19}\)
\(\Leftrightarrow1-\dfrac{1}{x+3}=\dfrac{18}{19}\)
=>x+3=19
hay x=16
=>(5/17+12/17)+(-20/31-11/31)-4/9<=x/9<=(-3/7-4/7)+(7/15+8/15)+2/3
=>-4/9<=x/9<=6/9
=>-4<=x<=6
hay \(x\in\left\{-4;-3;-2;-1;0;...;6\right\}\)
5\(\frac{8}{17}\) : x - (\(\frac{8}{17}\)) : x + 3\(\frac{1}{17}\): 17\(\frac13\) = \(\frac{4}{17}\)
(5\(\frac{8}{17}\) - \(\frac{8}{17}\)) : x + \(\frac{52}{17}\) : \(\frac{52}{3}\) = \(\frac{4}{17}\)
5:x + \(\frac{52}{17}\times\) \(\frac{3}{52}\) = \(\frac{4}{17}\)
5 : x + \(\frac{3}{17}\) = \(\frac{4}{17}\)
5 : x = \(\frac{4}{17}\) - \(\frac{3}{17}\)
5 : x = \(\frac{1}{17}\)
x = 5 : \(\frac{1}{17}\)
x = 5 x 17
x = 85
Vậy x = 85
1/1.4 + 1/4.7 + ...+1/x(x+3) = 6/19
3/1.4 + 3/4.7 +..+3/x(x+3) = 18/19
1/1 - 1/4+ 1/4 - 1/7 +...+1/x-1/(x+3) = 1 - 1/19
1- 1/(x+3) = 1 - 1/19
1/(x+3) = 1/19
x + 3 = 19
x = 19 - 3
x = 16
Vậy x = 16
\(1\frac57-\frac57\times\frac{2}{11}-\frac57\times\frac{9}{11}\)
=\(\frac57\times\left(1-\frac{2}{11}-\frac{9}{11}\right)\)
=\(\frac57\times\left(\frac{11}{11}-\frac{2}{11}-\frac{9}{11}\right)\)
=\(\frac57\times1\)
=\(\frac57\)
\(\frac{5}{11}\times\frac{18}{29}-\frac{5}{11}\times\frac{8}{29}+\frac{5}{11}\times\frac{19}{26}\)
=\(\frac{5}{11}\times\left(\frac{18}{29}-\frac{8}{29}+\frac{19}{29}\right)\)
=\(\frac{5}{11}\times1\)
=\(\frac{5}{11}\)
Giúp Mik nhanh nha
\(\frac{3}{17}.\frac{29}{17}-\frac{3}{17}.\frac{12}{17}-\frac{3}{17}\)
\(=\frac{3}{17}.\frac{29}{17}-\frac{3}{17}.\frac{12}{17}-\frac{3}{17}.1\)
\(=\frac{3}{17}.\left(\frac{29}{17}-\frac{12}{17}-1\right)\)
\(=\frac{3}{17}.0\)
\(=0\)
\(=\frac{3}{17}.\frac{29}{7}-\frac{3}{17}.\frac{12}{7}-\frac{3}{17}\\ =\frac{3}{17}\left(\frac{29}{7}-\frac{12}{7}-1\right)\\ =\frac{3}{17}.\frac{10}{7}\\ =\frac{30}{119}\)
\(\frac{3}{17}.\frac{29}{17}-\frac{3}{17}.\frac{12}{17}-\frac{3}{17}\)
\(=\frac{3}{17}\left(\frac{29}{17}-\frac{12}{17}-1\right)\)
\(=\frac{3}{17}.0\)
\(=0\)
Study well ! >_<
\(\frac{3}{7}\cdot\left(\frac{29}{17}-\frac{2}{17}-1\right)\)
\(=\frac{3}{7}\cdot\frac{29-2-17}{17}\)
\(=\frac{3}{7}\cdot0=0\)