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a: \(=\left(\dfrac{17}{10}+\dfrac{70}{10}-\dfrac{87}{10}\right):\left(\dfrac{23}{4}-\dfrac{11}{4}+\dfrac{9}{25}\right)\cdot\left(12,98\cdot0,25\right)+12,5\)
\(=0:\left(3+\dfrac{9}{25}\right)\cdot\left(12,98+0,25\right)+12,5\)
=12,5
b: \(=\dfrac{13}{12}\cdot\dfrac{27}{5}\cdot2\cdot\dfrac{34}{9}\cdot2\cdot\dfrac{2}{17}\)
\(=\dfrac{13}{12}\cdot2\cdot\dfrac{27}{5}\cdot\dfrac{34}{9}\cdot\dfrac{4}{17}\)
\(=\dfrac{13}{6}\cdot\dfrac{27}{5}\cdot\dfrac{8}{9}=\dfrac{8}{6}\cdot3\cdot\dfrac{13}{5}=4\cdot\dfrac{13}{5}=\dfrac{52}{5}\)
a) \(\left(-16\right).12=\left(-192\right)\)
b) \(22.\left(-5\right)=\left(-110\right)\)
c) \(\left(-2500\right).\left(-100\right)=\left(+250000\right)\)
d) \(\left(-11\right)^2=\left(-11\right).\left(-11\right)=\left(+121\right)\)
a, ( - 16 ) . 12 b, 22 . ( - 5 ) c, ( - 2500 ) . ( - 100 )
= - ( 16.12 ) = - ( 22 . 5 ) = 2500 . 100
= - 192 = - 110 = 250000
d, ( -11 )2 = [ ( -11 ) . ( -11 ) ]
= 121
a)\(VT=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(=\frac{1}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right]\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\)
\(=\frac{1}{2}-\frac{1}{3n+2}=\frac{3n+2}{2\cdot\left(3n+2\right)}-\frac{2}{2\cdot\left(3n+2\right)}\)
\(=\frac{3n+2-2}{6n+4}=\frac{3n}{6n+4}=VP\)
\(a,\left(19x+2.5^2\right):14=\left(13-8\right)^2-4^2\)
\(\Leftrightarrow\left(19x+50\right):14=5^2-4^2\)
\(\Leftrightarrow\left(19x+50\right):14=9\)
\(\Leftrightarrow19x+50=126\)
\(\Leftrightarrow19x=76\Leftrightarrow x=4\)
b) x + ( x + 1 ) + ( x + 2 ) + ... + ( x + 30 ) = 1240
x + x + 1 + x + 2 + ... + x + 30 = 1240
( x + x + ... + x ) + ( 1 + 2 + ... + 30 ) = 1240
Số số hạng là : ( 30 - 1 ) : 1 + 1 = 30 ( số )
Tổng là : ( 30 + 1 ) . 30 : 2 = 465
=> 31x + 465 = 1240
=> 31x = 775
=> x = 25
Vậy........
c
\(=160:\left\{17+\left[9\cdot5-\left(14+2^3\right)\right]\right\}\)
\(=160:\left\{17+\left[45-\left(14+8\right)\right]\right\}\)
\(=160:\left\{17+\left[45-22\right]\right\}\)
\(=160:\left(17+23\right)\)
\(=160:40\)
\(=4\)
d
\(=798+100:\left[16-2\cdot\left(25-22\right)\right]\)
\(=798+100:\left(16-2\cdot3\right)\)
\(=798+100:\left(16-6\right)\)
\(=798+100:10\)
\(=798+10\)
\(=808\)