(2000×7,5+2012÷3)×(21-3,5×0,25)×(11,2-10-1,2)×(3,75-0,75)/2011+3,14
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Ta có : \(\frac{\left(11,2-10-1,2\right)x\left(3,75-0,75\right)}{2011+3014}=\frac{0x3}{2011+3014}=\frac{0}{2011+3014}=0\)
\(\Rightarrow\left(2000\times7,5+2012:3\right)\times\left(21-3,5\times0,25\right)\times\frac{\left(11,2-10-1,2\right)\times\left(3,75-0,75\right)}{2011+3014}\)
\(=\left(2000\times7,5+2012:3\right)\times\left(21-3,5\times0,25\right)\times0\)
\(=0\)
\(1+\left(1+2\right)+\left(1+2+3\right)+\left(1+2+3+4\right)+...+\left(1+2+3+...+100\right)\)
\(=\left(1+1+1+...+1\right)+\left(2+2+...+2\right)+\left(3+..+3\right)+...+\left(99+99\right)+100\)
( biểu thức trên có 100 số 1, 99 số 2, 98 số 3,...., 2 số 9, 1 số 100)
\(=100\times1+99\times2+98\times3+...+2\times99+1\times100\)
suy ra \(\frac{1+\left(1+2\right)+\left(1+2+3\right)+\left(1+2+3+4\right)+...+\left(1+2+3+...+100\right)}{100\times1+99\times2+98\times3+...+2\times99+1\times100}=1\)
0,75 + 1,5 \(\times\) 97,8 \(\times\) 0,5 + 0,25 \(\times\) 3 \(\times\) 1,2
= 0,75 \(\times\) 1 + (1,5 \(\times\) 0,5)\(\times\) 97,8 + (0,25 \(\times\)3) \(\times\) 1,2
= 0,75 \(\times\) 1 + 0,75 \(\times\) 97,8 + 0,75 \(\times\) 1,2
= 0,75 \(\times\) ( 1 + 97,8 + 1,2)
= 0,75 \(\times\) [1 + ( 97,8 + 1,2)]
= 0,75 \(\times\) [1 + 99]
= 0,75 \(\times\) 100
= 75
1/ (69.210+1210)+(219.273+15.49.94) = 29.39.210+310.220+219.39+5.3.218.38 = 219.39+310.220+219.39+5.218.39
= 218.39(2+3.22+5)=19.218.39
a; \(\dfrac{93}{17}\): \(x\) + (- \(\dfrac{21}{17}\)) : \(x\) + \(\dfrac{22}{7}\): \(\dfrac{22}{3}\) = \(\dfrac{5}{14}\)
\(\dfrac{94}{17}\) \(\times\) \(\dfrac{1}{x}\) - \(\dfrac{21}{17}\) \(\times\) \(\dfrac{1}{x}\) + \(\dfrac{3}{7}\) = \(\dfrac{5}{14}\)
\(\dfrac{72}{17}\) \(\times\) \(\dfrac{1}{x}\) + \(\dfrac{3}{7}\) = \(\dfrac{5}{14}\)
\(\dfrac{72}{17x}\) = \(\dfrac{5}{14}\) - \(\dfrac{3}{7}\)
\(\dfrac{72}{17x}\) = - \(\dfrac{1}{14}\)
17\(x\) = 72.(-14)
17\(x\) = - 1008
\(x\) = - 1008 : 17
\(x\) = - \(\dfrac{1008}{17}\)
Vậy \(x\) \(=-\dfrac{1008}{17}\)
b; - \(\dfrac{32}{27}\) - (3\(x\) - \(\dfrac{7}{9}\))3 = - \(\dfrac{24}{27}\)
- \(\dfrac{32}{27}\) + \(\dfrac{24}{27}\) = (3\(x\) - \(\dfrac{7}{9}\))3
(3\(x-\dfrac{7}{9}\))3 = - \(\dfrac{8}{27}\)
(3\(x-\dfrac{7}{9}\))3 = (- \(\dfrac{2}{3}\))3
3\(x-\dfrac{7}{9}\) = - \(\dfrac{2}{3}\)
3\(x\) = - \(\dfrac{2}{3}\) + \(\dfrac{7}{9}\)
3\(x\) = \(\dfrac{1}{9}\)
\(x\) = \(\dfrac{1}{9}\) : 3
\(x\) = \(\dfrac{1}{27}\)
Vậy \(x=\dfrac{1}{27}\)
1,2 x ( 2,5:P)=3,6
2,5:P = 3,6:1,2
2,5:P = 3
P = 2,5:3
P = \(\dfrac{5}{6}\)
7,5 * x + x * 3,5 = 39,93
x * (7,5 + 3,5) = 39,93
x * 11 = 39,93
x = 39,93 : 11
x = 3,63
x : 4 + x = 3,75
x : (4 + 1) = 3,75
x : 5 = 3,75
x = 3,75 * 5
x = 18,75
(7,5+3,5)*x=39,93 x:4+x=3,75
11*x=39,93 x:4+x:1=3,75
x=39,93:11 x:(4+1)=3,75
x=3,63 x:5=3,75 x=3,75*5 x= 18,75
(2000×7,5+2012÷3)×(21-3,5×0,25)×(11,2-10-1,2)×(3,75-0,75)/2011+3,14=15670.6666667x20.125x0x3:2014.14=0
học tót
(2000×7,5+2012÷3)×(21-3,5×0,25)×(11,2-10-1,2)×(3,75-0,75)/2011+3,14
=(2000×7,5+2012÷3)×(21-3,5×0,25)×(1,2-1,2)×(3,75-0,75)/2011+3,14
=(2000×7,5+2012÷3)×(21-3,5×0,25)×0×(3,75-0,75)/2011+3,14
=0