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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 \(x\) \(\in\) - \(\dfrac{29}{15}\)
A = SCSH: ( 102 - 1 ) : 1 + 1 = 102
A = Tổng: ( 102 + 1 ) . 102 : 2 = 5253
Vậy KQ là: 5253
B = SCSH: ( 2998 - 1 ) : 3 + 1 = 1000
B = Tổng: ( 2998 + 1 ) . 1000 : 2 = 1499500
Vậy KQ là 1499500
\(\text{C=1+2-3-4+5+6-7-8+9+...+2002-2003-2004+2005+2006}\)
\(\text{C=1+(2-3-4+5)+(6-7-8+9)+...+(2002-2003-2004+2005)+2006}\)
\(\text{C=1+0+0+...+0+2006}\)
\(\text{C=1+2006}\)
\(C=2007\)
HỌC TỐT!!!
\(a,[\left(8.x-12\right):4].3^3.3=3^6.6\)
\(\left(8x-12\right):4=54\)
\(8x-12=216\)
\(8x=228\)
\(x=28,5\)
\(b,41-2^{x+1}=9\)
\(2^{x+1}=41-9\)
\(2^{x+1}=32\)
\(2^{x+1}=2^5\)
\(\Rightarrow x+1=5\)
\(\Rightarrow x=4\)
\(A=\frac{1\cdot2+2\cdot4+3\cdot6+4\cdot8+5\cdot10+6\cdot12}{3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20+18\cdot24}\)
\(A=\frac{2\cdot3\left[1\cdot2\right]+2\cdot3\left[2\cdot4\right]+2\cdot3\left[3\cdot6\right]+2\cdot3\left[4\cdot8\right]+2\cdot3\left[5\cdot10\right]}{3\cdot4\left[3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20\right]}\)
\(A=\frac{\left[3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20\right]}{2\cdot3\left[3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20\right]}=\frac{1}{2\cdot3}=\frac{1}{6}\)
(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