Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{21}{36}x\frac{39}{14}x\frac{54}{13}=\frac{21x39x54}{36x14x13}=\frac{3x7x13x3x9x6}{6x6x7x2x13}=\frac{3x9}{2x2}=\frac{27}{4}\)
\(\frac{21}{36}x\frac{39}{14}x\frac{54}{13}=\frac{13}{8}x\frac{54}{13}=\frac{54}{8}=\frac{27}{4}\)
Vì bớt cả trên lẫn dưới đi 13
a)<=>2/7:(13/24+5/24) b)<=>15/24-9/24-4/24 c)=7.5.39/13.14.15
<=>2/7:3/4 <=>2/24=1/12 =3/2.3
<=>2/7x4/3 =1/2
<=>8/21
a. \(\frac{2}{7}:\frac{13}{24}+\frac{2}{7}:\frac{5}{24}\)
= \(\frac{2}{7}:\left(\frac{13}{24}+\frac{5}{24}\right)\)
= \(\frac{2}{7}:\frac{3}{4}\)
= \(\frac{8}{21}\)
b. \(\frac{15}{24}-\frac{3}{8}-\frac{1}{6}\)
= \(\frac{15}{24}-\frac{9}{24}-\frac{4}{24}\)
= \(\frac{2}{24}=\frac{1}{12}\)
c. \(\frac{7}{13}.\frac{5}{14}.\frac{39}{15}\)
= \(\frac{7.5.3.13}{13.2.7.3.5}\)
= \(\frac{1}{2}\)
Dấu \(.\)là dấu nhân
\(\left(7.13+8.13\right):\left(9\frac{2}{3}-y\right)=39\)
\(\Rightarrow13.\left(7+8\right):\left(\frac{29}{3}-y\right)=39\)
\(\Rightarrow13.15:\left(\frac{29}{3}-y\right)=39\)
\(\Rightarrow195:\left(\frac{29}{3}-y\right)=39\)
\(\Rightarrow\frac{29}{3}-y=195:39\)
\(\Rightarrow\frac{29}{3}-y=5\)
\(\Rightarrow y=\frac{29}{3}-5\)
\(\Rightarrow y=\frac{29}{3}-\frac{15}{3}\)
\(\Rightarrow y=\frac{14}{3}\)
Vậy \(y=\frac{14}{3}\)
~ Ủng hộ nhé .
P/s : Đúng nha
(91+104):(29|3-x)=39
195:(29|3-x)=39
29|3-x=195:39=5
x=29|3-5
x=14|3
12+13+14+15-45+25
=25+14+15-45+25
=39+15-45+25
=54-45+25
=9+25
=34
\(\frac{21}{36}\)x \(\frac{39}{14}\)x \(\frac{54}{13}\)= \(\frac{189}{82}\).
Mình sửa lại đề nè:
\(x+\frac{5}{13}+\frac{10}{26}+\frac{15}{39}+\frac{20}{52}+\frac{40}{104}=\frac{25}{13}\)
\(=x+\frac{5}{13}+\frac{5}{13}+\frac{5}{13}+\frac{5}{13}+\frac{5}{13}=\frac{25}{13}\)
\(=x+\frac{5}{13}x5=\frac{25}{13}\)
\(x+\frac{25}{13}=\frac{25}{13}\)
\(x=\frac{25}{13}-\frac{25}{13}\)
\(x=0\)
Chúc bạn học tốt!
a) \(\dfrac{6}{13}:\left(\dfrac{1}{2}-x\right)=\dfrac{15}{39}\)
\(\dfrac{1}{2}-x=\dfrac{6}{13}:\dfrac{15}{39}\)
\(\dfrac{1}{2}-x=\dfrac{6}{5}\)
\(x=\dfrac{1}{2}-\dfrac{6}{5}\)
\(x=-\dfrac{7}{10}\)
b) \(3\times\left(\dfrac{x}{4}+\dfrac{x}{28}+\dfrac{x}{70}+\dfrac{x}{130}\right)=\dfrac{60}{13}\)
\(3\times x\times\left(\dfrac{1}{4}+\dfrac{1}{28}+\dfrac{1}{70}+\dfrac{1}{130}\right)=\dfrac{60}{13}\)
\(x\times\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+\dfrac{3}{7\times13}\right)=\dfrac{60}{13}\)
\(x\times\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}\right)=\dfrac{60}{13}\)
\(x\times\left(1-\dfrac{1}{13}\right)=\dfrac{60}{13}\)
\(x\times\dfrac{12}{13}=\dfrac{60}{13}\)
\(x=\dfrac{60}{13}:\dfrac{12}{13}\)
\(x=5\)
\(\dfrac{70}{39}\times\dfrac{13}{14}-\dfrac{13}{25}:\dfrac{39}{50}=\dfrac{5}{3}-\dfrac{13}{25}\times\dfrac{50}{39}=\dfrac{5}{3}-\dfrac{2}{3}=1\)
\(\dfrac{70}{39}\cdot\dfrac{13}{14}-\dfrac{13}{25}:\dfrac{39}{50}\)
\(=\dfrac{5}{3}-\dfrac{13}{25}\cdot\dfrac{50}{39}\)
\(=\dfrac{5}{3}-\dfrac{2}{3}\)
=1