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\(\left(5\times6+11-41\right)\times\left(1+2+3+\cdot\cdot\cdot+100\right)\)
\(=\left(30+11-41\right)\times\left(1+2+\cdot\cdot\cdot+100\right)\)
\(=\left(41-41\right)\times\left(1+2+\cdot\cdot\cdot+100\right)\)
\(=0\times\left(1+2+\cdot\cdot\cdot+100\right)\)
\(=0\)
\(\left(2+4+\cdot\cdot\cdot+100\right)\times\left(11\times7-78+1\right)\)
\(=\left(2+4+\cdot\cdot\cdot+100\right)\times\left(77+1-78\right)\)
\(=\left(2+4+\cdot\cdot\cdot+100\right)\times0\)
\(=0\)
Giải:
\(B=\left(5+10+15+...+2015\right)\times\left(20142014\times2015-20152015\times2014\right)\)
\(B=\left(5+10+15+...+2015\right)\times\left(10001\times2014\times2015-10001\times2015\times2014\right)\)
\(B=\left(5+10+15+...+2015\right)\times0\)
\(B=0\)
Chúc em học tốt!
B = (5+10+15+.........+2015) x (20142014 x 2015-20152015 x 2014)
B = (5+10+15+.........+2015) x ( 2014 x 1001 x 2015 - 2015 x 1001 x 2014 ] B = (5+10+15+.........+2015) x 0B = 0 nha
(5 + 6 + 11 - 41) x (1 + 2 + 3 + ... + 100)
= [(5 + 6) - (11 - 41)] x (1 + 2 +...+ 100)
= [11 - 30 ] x (1 + 2 + ... + 100)
= - 19 x (1 + 2 +...+ 100)
Lớp 5 chưa học số âm.
12 x 85 + 35 - 182 - 35 x 94
= 1020 - 35 x (94 - 1) - 182
= 1020 - 35 x 93 - 182
= 1020 - 3255 - 182
= - 2235 - 182
= - 2417
Lớp 5 chưa học số âm
\(A=\frac{3333}{4545}+\frac{1}{15}+\frac{1111}{5555}\)
\(=\frac{11}{15}+\frac{1}{15}+\frac{1}{5}\)
\(=\frac{12}{15}+\frac{3}{15}\)
\(=1\)
\(A=\frac{3333}{4545}+\frac{1}{15}+\frac{1111}{5555}\)
\(=\frac{11}{15}+\frac{1}{15}+\frac{1}{5}\)
\(=\frac{12}{15}+\frac{3}{15}\)
\(=1\)
a; 5\(\dfrac{3}{4}\) : 3 + 2\(\dfrac{1}{4}\).\(\dfrac{1}{3}\) - \(\dfrac{3}{8}\)
= \(\dfrac{23}{4}\) : 3 + \(\dfrac{9}{4}\).\(\dfrac{1}{3}\) - \(\dfrac{3}{8}\)
= \(\dfrac{23}{4}\) x \(\dfrac{1}{3}\) + \(\dfrac{3}{4}\) - \(\dfrac{3}{8}\)
= \(\dfrac{23}{12}\) + \(\dfrac{3}{4}\) - \(\dfrac{3}{8}\)
= \(\dfrac{46}{24}\) + \(\dfrac{18}{24}\) - \(\dfrac{9}{24}\)
= \(\dfrac{64}{24}\) - \(\dfrac{9}{24}\)
= \(\dfrac{55}{24}\)
\(\frac{20152015}{20162016}+\frac{20162016}{20162016}\)
\(=\frac{2015\times10001}{2016\times10001}+\frac{2016\times10001}{2016\times10001}\)
\(=\frac{2015}{2016}+\frac{2016}{2016}\)
\(=\frac{4031}{2016}\)