tính G = \(\dfrac{\left(3\dfrac{2}{15}+\dfrac{1}{5}\right):2\dfrac{1}{2}}{\left(5\dfrac{3}{7}-2\dfrac{1}{4}\right):4\dfrac{43}{56}}-\dfrac{1,2:\left(1\dfrac{1}{5}.1\dfrac{1}{4}\right)}{0,32+\dfrac{2}{25}}\)
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Những câu hỏi liên quan
10 tháng 7 2022
\(A=\dfrac{\left(3+\dfrac{2}{15}+\dfrac{1}{5}\right):\dfrac{5}{2}}{\left(5+\dfrac{3}{7}-2-\dfrac{1}{4}\right):\left(4+\dfrac{43}{56}\right)}\)
\(=\dfrac{\dfrac{10}{3}\cdot\dfrac{2}{5}}{\dfrac{89}{28}:\dfrac{267}{56}}=\dfrac{4}{3}:\dfrac{2}{3}=2\)
\(B=\dfrac{\dfrac{6}{5}:\left(\dfrac{6}{5}\cdot\dfrac{5}{4}\right)}{\dfrac{8}{25}+\dfrac{2}{25}}=\dfrac{\dfrac{6}{5}:\dfrac{3}{2}}{\dfrac{2}{5}}=2\)
Do đó: A=B
VH
3 tháng 8 2018
A = \(\dfrac{\left(\dfrac{47}{15}+\dfrac{3}{15}\right):\dfrac{5}{2}}{\left(\dfrac{38}{7}-\dfrac{9}{4}\right):\dfrac{267}{56}}=\dfrac{\dfrac{10}{3}.\dfrac{2}{5}}{\dfrac{89}{28}.\dfrac{56}{267}}=2\)
B= \(\dfrac{1,2:\left(\dfrac{6}{5}.\dfrac{5}{4}\right)}{0,32+\dfrac{2}{25}}=\dfrac{\dfrac{6}{5}:\dfrac{3}{2}}{\dfrac{8}{25}+\dfrac{2}{25}}=\dfrac{4}{\dfrac{5}{\dfrac{2}{5}}}=2\)
=> A = B
7 tháng 10 2017
c)
Ta có :\(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)
\(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{3}{2}}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{2}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{\dfrac{8}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{3}{8}}\) \(=2+\dfrac{1}{\dfrac{11}{8}}\) \(=2+\dfrac{8}{11}\) \(=\dfrac{30}{11}\)
7 tháng 10 2017
d) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\dfrac{1}{4}:2\)
\(=3-1+\dfrac{1}{8}\)
\(=\dfrac{17}{8}\)
O
18 tháng 3 2022
a) \(\left(\dfrac{2}{3}-\dfrac{1}{2}-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}-\dfrac{1}{7}\right)\)
\(=-\dfrac{1}{6}\cdot\dfrac{17}{28}\)
\(=-\dfrac{17}{168}\)
b) \(\left(\dfrac{15}{21}\div\dfrac{5}{7}\right)\div\left(\dfrac{6}{5}\div2\right)\)
\(=1\div\dfrac{3}{5}\)
\(=\dfrac{5}{3}\)
22 tháng 3 2017
11) \(\dfrac{5}{7}\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{4}{7}\right)+\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{4}{7}\right):\dfrac{7}{5}\)
= \(\dfrac{5}{7}\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{4}{7}\right)+\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{4}{7}\right)\cdot\dfrac{5}{7}\)
= \(\dfrac{5}{7}\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{4}{7}+\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{4}{7}\right)\)
= \(\dfrac{5}{7}\cdot0\)
=0
12) \(\dfrac{43}{5}\left(\dfrac{17}{3}-\dfrac{16}{9}+2\right)-\dfrac{43}{5}\left(\dfrac{17}{3}-\dfrac{16}{9}\right)\)
= \(\dfrac{43}{5}\left(\dfrac{17}{3}-\dfrac{16}{9}+2-\dfrac{17}{3}+\dfrac{16}{9}\right)\)
= \(\dfrac{43}{5}\cdot2=\dfrac{43}{10}\)
25 tháng 7 2017
11, 5/7( 1/2-1/3+1/4)+ (1/3-1/2-1/4):7/5
= 5/7.(1/2 - 1/3 + 1/4 )+( 1/3 - 1/2 - 1/4). 5/7
= 5/7.(1/2 - 2/3 + 1/4 + 1/3 - 1/2 - 1/4)
= 5/7 . -1/3
= -5/21
12, 43/5.(17/3 - 16/9 + 2)- 43/5. (17/3 - 16/9)
= 43/5.( 17/3 - 16/9 + 2 - 17/3 + 16/9)
= 43/5 . 2
= 86/5
30 tháng 11 2021
2: \(=\dfrac{203}{60}\cdot\dfrac{81}{1225}=\dfrac{783}{3500}\)
AH
Akai Haruma
Giáo viên
30 tháng 4 2023
Bài 1:
$(y+\frac{1}{3})+(y+\frac{1}{9})+(y+\frac{1}{27})+(y+\frac{1}{81})=\frac{56}{81}$
$(y+y+y+y)+(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81})=\frac{56}{81}$
$4\times y+\frac{40}{81}=\frac{56}{81}$
$4\times y=\frac{56}{81}-\frac{40}{81}=\frac{16}{81}$
$y=\frac{16}{81}:4=\frac{4}{81}$
AH
Akai Haruma
Giáo viên
30 tháng 4 2023
Bài 2:
$18: \frac{x\times 0,4+0,32}{x}+5=14$
$18: \frac{x\times 0,4+0,32}{x}=14-5=9$
$\frac{x\times 0,4+0,32}{x}=18:9=2$
$x\times 0,4+0,32=2\times x$
$2\times x-x\times 0,4=0,32$
$x\times (2-0,4)=0,32$
$x\times 1,6=0,32$
$x=0,32:1,6=0,2$
\(G=\dfrac{\left(3\dfrac{2}{15}+\dfrac{1}{5}\right):2\dfrac{1}{2}}{\left(5\dfrac{3}{7}-2\dfrac{1}{4}\right)}-\dfrac{1,2:\left(1\dfrac{1}{5}.1\dfrac{1}{4}\right)}{0,32+\dfrac{2}{25}}\)
\(G=\dfrac{\left(\dfrac{47}{15}+\dfrac{1}{5}\right):\dfrac{5}{2}}{\left(\dfrac{38}{7}-\dfrac{9}{4}\right)}-\dfrac{\dfrac{6}{5}:\left(\dfrac{6}{5}.\dfrac{5}{4}\right)}{\dfrac{8}{25}+\dfrac{2}{25}}\)
\(G=\dfrac{\dfrac{4}{3}}{\dfrac{89}{28}}-\dfrac{2}{\dfrac{2}{5}}\)
\(G=\dfrac{89}{21}-\dfrac{1}{5}\)
\(G=\dfrac{424}{105}.\)
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