22.(x-3)-5=15
K
Khách
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26 tháng 7 2023
a: \(=\dfrac{3}{5}:\dfrac{7}{5}=\dfrac{3}{5}\cdot\dfrac{5}{7}=\dfrac{3}{7}\)
b: \(=\dfrac{9}{17}\left(\dfrac{8}{5}-\dfrac{3}{5}\right)+\dfrac{8}{17}\)
=9/17+8/17=1
c: =>x-3/10=7/15*1/5=7/75
=>x=7/75+3/10=59/150
2 tháng 7 2021
1) Ta có: \(\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{7}{25}\cdot\dfrac{5}{7}\right)\)
\(=\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{1}{5}\right)\)
=0
2) Ta có: \(\dfrac{8}{17}\cdot\dfrac{4}{15}+\dfrac{8}{17}\cdot\dfrac{22}{15}-\dfrac{8}{15}\cdot\dfrac{9}{17}\)
\(=\dfrac{8}{17}\left(\dfrac{4}{15}+\dfrac{22}{15}-\dfrac{9}{15}\right)\)
\(=\dfrac{8}{17}\cdot\dfrac{15}{15}=\dfrac{8}{17}\)
3) Ta có: \(\dfrac{2021}{2}\cdot\dfrac{1}{3}+\dfrac{4042}{4}\cdot\dfrac{1}{5}+\dfrac{6063}{3}\cdot\dfrac{22}{15}\)
\(=\dfrac{2021}{2}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)+2021\cdot\dfrac{22}{15}\)
\(=\dfrac{2021}{2}\cdot\dfrac{8}{15}+\dfrac{2021}{2}\cdot\dfrac{44}{15}\)
\(=\dfrac{2021}{2}\cdot\dfrac{52}{15}\)
\(=\dfrac{52546}{15}\)
4) Ta có: \(\dfrac{4}{7}\cdot\dfrac{2}{13}+\dfrac{8}{13}:\dfrac{7}{4}+\dfrac{4}{7}:\dfrac{13}{2}+\dfrac{4}{7}\cdot\dfrac{1}{13}\)
\(=\dfrac{4}{7}\left(\dfrac{2}{13}+\dfrac{8}{13}+\dfrac{2}{13}+\dfrac{1}{13}\right)\)
\(=\dfrac{4}{7}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
24 tháng 3 2023
a, \(\dfrac{5\times11\times15\times18}{6\times15\times22\times10}\)
= \(\dfrac{5\times11\times15\times6\times3}{6\times15\times11\times2\times2\times5}\)
= \(\dfrac{5\times11\times15\times6}{5\times11\times15\times6}\) \(\times\) \(\dfrac{3}{2\times2}\)
= 1 \(\times\) \(\dfrac{3}{4}\)
= \(\dfrac{3}{4}\)
b, \(\dfrac{7}{12}\) \(\times\) \(\dfrac{3}{5}\) + \(\dfrac{3}{5}\) \(\times\) \(\dfrac{5}{12}\)
= \(\dfrac{3}{5}\) \(\times\) ( \(\dfrac{7}{12}+\dfrac{5}{12}\))
= \(\dfrac{3}{5}\) \(\times\) \(\dfrac{12}{12}\)
= \(\dfrac{3}{5}\) \(\times\) 1
= \(\dfrac{3}{5}\)
NT
0
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
15 tháng 12 2023
a, (\(\dfrac{9}{10}\) - \(\dfrac{15}{16}\)) \(\times\) ( \(\dfrac{5}{12}\) - \(\dfrac{11}{15}\) - \(\dfrac{7}{20}\))
= (\(\dfrac{72}{80}\) - \(\dfrac{75}{80}\)) \(\times\) (\(\)\(\dfrac{25}{60}\) - \(\dfrac{44}{60}\) - \(\dfrac{21}{60}\))
= - \(\dfrac{3}{80}\) \(\times\) (- \(\dfrac{2}{3}\))
= \(\dfrac{1}{40}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
15 tháng 12 2023
b, (-1)3 + (- \(\dfrac{2}{3}\))2 : 2\(\dfrac{2}{3}\) + \(\dfrac{5}{6}\)
= -13 + \(\dfrac{4}{9}\) : \(\dfrac{8}{3}\) + \(\dfrac{5}{6}\)
= -1 + \(\dfrac{4}{9}\) \(\times\) \(\dfrac{3}{8}\) + \(\dfrac{5}{6}\)
= -1 + \(\dfrac{1}{6}\) + \(\dfrac{5}{6}\)
= -1 + 1
= 0
AH
Akai Haruma
Giáo viên
12 tháng 4 2023
Lời giải:
$\frac{22}{15}:(11-x)+\frac{2}{3}=\frac{7}{5}$
$\frac{22}{15}:(11-x)=\frac{7}{5}-\frac{2}{3}=\frac{11}{15}$
$11-x=\frac{22}{15}: \frac{11}{15}=2$
$x=11-2=9$
KC
1
BE
15 tháng 5 2018
\(\frac{15}{8}\)+\(\frac{7}{12}\)-\(\frac{13}{3}\)x\(\frac{3}{8}\)
=\(\frac{45}{24}\)+\(\frac{14}{24}\)-\(\frac{13}{8}\)
=\(\frac{59}{24}\)-\(\frac{39}{24}\)
=\(\frac{20}{24}\)=\(\frac{5}{6}\)
\(\frac{11}{5}\)+\(\frac{5}{22}\):2+\(\frac{5}{6}\)
=\(\frac{11}{5}\)+\(\frac{5}{44}\)+\(\frac{5}{6}\)
=\(\frac{484}{220}\)+\(\frac{25}{220}\)+\(\frac{5}{6}\)
=\(\frac{509}{220}\)+\(\frac{5}{6}\)
=\(\frac{1527}{660}\)+\(\frac{550}{660}\)
=\(\frac{2077}{660}\)
\(2^2.\left(x-3\right)-5=15\)
\(4.\left(x-3\right)=15+5\)
\(4.\left(x-3\right)=20\)
\(x-3=20:4\)
\(x-3=5\)
\(\Rightarrow x=8\)