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Bài 1 :
a, Ta có : \(\left(-123\right)+\left|-13\right|+\left(-7\right)\)
= \(\left(-123\right)+13+\left(-7\right)=\left(-117\right)\)
b, Ta có : \(\left|-10\right|+\left|45\right|+\left(-\left|-455\right|\right)+\left|-750\right|\)
= \(10+45-455+750=350\)
c, Ta có : \(-\left|-33\right|+\left(-15\right)+20-\left|45-40\right|-57\)
= \(\left(-33\right)+\left(-15\right)+20-5-57=-90\)
\(\frac{2^{10}\cdot3^{10}-2^{10}\cdot3^9}{2^9\cdot3^{10}}=\frac{2^{10}\cdot3^9\left(3-1\right)}{2^9\cdot3^{10}}=\frac{2^{11}\cdot3^9}{2^9\cdot3^{10}}=\frac{2^2}{3}=\frac{4}{3}\)
a)\(\left(10^2+11^2+12^2\right)\div\left(13^2+14^2\right)\)
\(=\left(100+121+144\right)\div\left(169+196\right)\)
\(=365\div365\)
\(=1\)
b) \(1.2.3...9-1.2.3...8-1.2.3...8^2\)
\(=1.2.3...8\left(9-1-8\right)\)
\(=1.2.3...8.0\)
\(=0\)
d) \(1152-\left(374+1152\right)+\left(-65+374\right)\)
\(=1152-374-1152-65+374\)
\(=\left(1152-1152\right)-65+\left(374-374\right)\)
\(=0-65+0\)
\(=-65\)
e) \(13-12+11+10-9+8-7-6+5-4+3+2-1\)
\(=13-\left(12-11\right)+\left(10-9\right)+\left(8-7\right)-\left(6-5\right)-\left(4-3\right)\)\(+\left(2-1\right)\)
\(=13-1+1+1-1-1+1\)
\(=13+0+0+0\)
\(=13\)