giúp mik nhé:
5/6-1/9+3/4=................
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\(a,\Leftrightarrow x^3=\dfrac{20}{3}\Leftrightarrow x=\sqrt[3]{\dfrac{20}{3}}\\ b,\Leftrightarrow x-1=9\Leftrightarrow x=10\\ c,\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ d,\Leftrightarrow2x+1=5\Leftrightarrow x=2\\ e,\Leftrightarrow2x-4=4\Leftrightarrow x=4\)
Câu a) xem lại đề giùm nhé em
b) \(\left(x-1\right)^3=9^3\)
\(x-1=9\)
\(x=10\)
Vậy \(x=10\)
c) \(\left(x-1\right)^2=25\)
\(x-1=5\) hoặc \(x-1=-5\)
* \(x-1=5\)
\(x=6\)
* \(x-1=-5\)
\(x=-4\)
Vậy \(x=-4\); \(x=6\)
d) \(\left(2x+1\right)^3=125\)
\(\left(2x+1\right)^3=5^3\)
\(2x+1=5\)
\(2x=4\)
\(x=2\)
Vậy \(x=2\)
e) Sửa đề: \(\left(2x+4\right)^3=64\)
\(\left(2x+4\right)^3=4^3\)
\(2x+4=4\)
\(2x=0\)
\(x=0\)
Vậy \(x=0\)
\(a,\dfrac{6}{11}+6+\dfrac{5}{7}=\dfrac{42+462+55}{77}=\dfrac{559}{77}\)
\(b,\dfrac{9}{8}\times\dfrac{3}{12}:\dfrac{5}{9}=\dfrac{9}{8}\times\dfrac{3}{12}\times\dfrac{9}{5}=\dfrac{243}{480}=\dfrac{81}{160}\)
\(c,\dfrac{8}{7}:4+2=\dfrac{8}{7}\times\dfrac{1}{4}+2=\dfrac{8}{28}+2=\dfrac{2}{7}+2=\dfrac{16}{7}\)
\(d,\dfrac{3}{5}+4:\dfrac{6}{4}=\dfrac{3}{5}+4\times\dfrac{4}{6}=\dfrac{3}{5}+\dfrac{8}{3}=\dfrac{49}{15}\)
(2 + 4 + 6 + 8 + ... + 2014) - (3 + 5 + 7 + 9 + ... + 2011)
= 2 + 4 + 6 + 8 + ... + 2014 - 3 - 5 - 7 - 9 - ... - 2011
= 2 + (4 - 3) + (6 - 5) + (8 - 7) + ... + (2012 - 2011) + 2014 (có 1005 cặp)
= 2 + 2014 + 1 + 1 + ... + 1 (có 1005 số 1)
= 2016 + 1005
= 3021
(2+4+6+8+...+2014)-(3+5+7+9+...+2011)
= 2+4+6+8+..+2014-3-5-7-9-...-2011
= (2-3)+(4-5)+(6-7)+(8-9)+...+(2010-2011)+2011+2012+2013+2014
= (-1)+(-1)+(-1)+(-1)+...+(-1)+2011+2012+2013+2014 gồm [(2011-2):1+1]:2=1005 số -1
=(-1).1005+2011+2012+2013+2014
=-1005+2011+2012+2013+2014
=7045
a; \(\dfrac{2}{3}\)\(x\) - \(\dfrac{3}{2}\)\(x\) = \(\dfrac{5}{12}\)
(\(\dfrac{2}{3}\) - \(\dfrac{3}{2}\))\(x\) = \(\dfrac{5}{12}\)
- \(\dfrac{5}{6}\)\(x\) = \(\dfrac{5}{12}\)
\(x\) = \(\dfrac{5}{12}\) : (- \(\dfrac{5}{6}\))
\(x=\) - \(\dfrac{1}{2}\)
Vậy \(x=-\dfrac{1}{2}\)
b; \(\dfrac{2}{5}\) + \(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\)
\(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\) - \(\dfrac{2}{5}\)
\(\dfrac{3}{5}\).(3\(x\) - 3,7) = - \(\dfrac{57}{10}\)
3\(x\) - 3,7 = - \(\dfrac{57}{10}\) : \(\dfrac{3}{5}\)
3\(x\) - 3,7 = - \(\dfrac{19}{2}\)
3\(x\) = - \(\dfrac{19}{2}\) + 3,7
3\(x\) = - \(\dfrac{29}{5}\)
\(x\) = - \(\dfrac{29}{5}\) : 3
\(x\) = - \(\dfrac{29}{15}\)
Vậy \(x\) \(\in\) - \(\dfrac{29}{15}\)
-1+2-3+4-5+6.....-99+100
=(2+4+6+....+98+100)-(1+3+5+...+99)
=(100+2).50:2-(99+1).50:2
=102.50:2-100.50:2
=2550-2500
=50
= 51/20
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= 3/14
------------------------------------
= 4/7 - 2/7
= 2/7
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= 17/45
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= 2/3
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= 2 x 1/4
= 1/2
\(\dfrac{5}{6}-\dfrac{1}{9}+\dfrac{3}{4}\text{=}\dfrac{30}{36}-\dfrac{4}{36}+\dfrac{27}{36}\text{=}\dfrac{53}{36}\)
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