2/3:(x-1/3)^3 -9=23/3
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NC
0
HN
0
NT
1
15 tháng 5 2017
\(\dfrac{-9}{46}-4\dfrac{1}{23}:\left(3\dfrac{1}{4}-x:\dfrac{3}{5}\right)+2\dfrac{8}{23}=1\)
\(\dfrac{-9}{46}+2\dfrac{8}{23}-4\dfrac{1}{23}:\left(3\dfrac{1}{4}-x:\dfrac{3}{5}\right)=1\)
\(\dfrac{-9}{46}+\dfrac{54}{23}-\dfrac{93}{23}:\left(\dfrac{13}{4}-x:\dfrac{3}{5}\right)=1\)
\(\dfrac{-9}{46}+\dfrac{108}{46}-\dfrac{93}{23}:\left(\dfrac{13}{4}-x:\dfrac{3}{5}\right)=1\)
\(\dfrac{99}{46}-\dfrac{93}{23}:\left(\dfrac{13}{4}-x:\dfrac{3}{5}\right)=1\)
\(\dfrac{93}{23}:\left(\dfrac{13}{4}-x:\dfrac{3}{5}\right)=\dfrac{99}{46}-1\)
\(\dfrac{93}{23}:\left(\dfrac{13}{4}-x:\dfrac{3}{5}\right)=\dfrac{99}{46}-\dfrac{46}{46}\)
\(\dfrac{93}{23}:\left(\dfrac{13}{4}-x:\dfrac{3}{5}\right)=\dfrac{53}{46}\)
\(\dfrac{13}{4}-x:\dfrac{3}{5}=\dfrac{93}{23}:\dfrac{53}{46}\)
\(\dfrac{13}{4}-x:\dfrac{3}{5}=\dfrac{93}{23}\cdot\dfrac{46}{53}\)
\(\dfrac{13}{4}-x:\dfrac{3}{5}=\dfrac{186}{53}\)
\(x:\dfrac{3}{5}=\dfrac{13}{4}-\dfrac{186}{53}\)
\(x:\dfrac{3}{5}=\dfrac{689}{212}-\dfrac{744}{212}\)
\(x:\dfrac{3}{5}=\dfrac{-55}{212}\)
\(x=\dfrac{-55}{212}\cdot\dfrac{3}{5}\)
\(x=\dfrac{-33}{212}\)
Vậy \(x=\dfrac{-33}{212}\).
7 tháng 8 2019
\(1,\frac{2}{3}+\frac{4}{9}+\frac{1}{5}+\frac{2}{15}+\frac{3}{2}-\frac{17}{18}\)
\(< =>\frac{4}{9}+\frac{3}{2}+\left(\frac{2}{3}+\frac{1}{5}+\frac{2}{15}\right)-\frac{17}{18}\)
\(< =>\frac{8}{18}+\frac{27}{18}+\left(\frac{10}{15}+\frac{3}{15}+\frac{2}{15}\right)-\frac{17}{18}\)
\(< =>\frac{35}{18}+1-\frac{17}{18}\)
\(< =>\frac{53}{18}-\frac{17}{18}\)
\(< =>2\)
\(2,\frac{13}{28}\cdot\frac{5}{12}-\frac{5}{28}\cdot\frac{1}{12}\)
\(< =>\left(\frac{13}{28}-\frac{5}{28}\right)\cdot\left(\frac{5}{12}-\frac{1}{12}\right)\)
\(< =>\frac{2}{7}\cdot\frac{1}{3}\)
\(< =>\frac{2}{21}\)
\(3,\frac{19}{4}\cdot\frac{15}{23}-\frac{15}{4}\cdot\frac{7}{23}+\frac{15}{4}\cdot\frac{11}{23}\)
\(< =>\frac{285}{92}-\frac{105}{92}+\frac{165}{92}\)
\(< =>\frac{15}{4}\)
TG
1
CN
16 tháng 4 2016
a)<=>10+15(3x-3,7)=-5,3
<=>10+45x-55,5=-5,3
<=>45x-45,5=-5,3
<=>45x=40,2
<=>x=\(\frac{40,2}{45}\)
b)
T
19 tháng 6 2019
#)Giải ;
b) Đặt \(N=1+2+2^2+2^3+...+2^{2012}\)
\(\Rightarrow2N=2+2^2+2^3+2^4+...+2^{2013}\)
\(\Rightarrow2N-N=N=\left(2+2^2+2^3+2^4+...+2^{2013}\right)-\left(1+2+2^2+2^3+...+2^{2012}\right)\)
\(\Rightarrow N=2^{2013}-1\)
Thay N vào M, ta có :
\(M=\frac{2^{2013}-1}{2^{2014}-2}\)
19 tháng 6 2019
Thêm Cho pen
\(M=\frac{2^{2013}-1}{2^{2014}-2}=\frac{2^{2013}-1}{2.\left(2^{2013}-1\right)}=\frac{1}{2}\)
Phải tính hết nhé
NT
2
LH
21 tháng 9 2018
a) 15-3(2x-9)=6
=)3(2x-9)=15-6=9
=)6x-27=9 =)6x=9+27=36
=)x=36:6=6
21 tháng 9 2018
a, 15-3 (2 x -9) = 6
3 (2 x -9) = 9
2 x - 9 = 3
2 x = 12
x = 6
b, 8 (x-3)-2(x-3) = 24
(8-2) (x-3) = 24
6 (x - 3) = 24
x - 3 = 4
x = 7
còn câu c mik ko bik lm thông cảm
\(\dfrac{2}{3}:\left(x-\dfrac{1}{3}\right)^3-9=\dfrac{23}{3}\)
=>\(\dfrac{2}{3}:\left(x-\dfrac{1}{3}\right)^3=\dfrac{23}{3}+9=\dfrac{50}{3}\)
=>\(\left(x-\dfrac{1}{3}\right)^3=\dfrac{2}{3}:\dfrac{50}{3}=\dfrac{1}{25}\)
=>\(x-\dfrac{1}{3}=\dfrac{\sqrt[3]{5}}{5}\)
=>\(x=\dfrac{\sqrt[3]{5}}{5}+\dfrac{1}{3}=\dfrac{3\sqrt[3]{5}+5}{15}\)