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a.17⋅(29−(−111)]+29⋅(−17)17⋅(29−(−111))+29⋅(−17)
=17⋅(29+111)−29⋅17=17⋅(29+111)−29⋅17
=17⋅(29+111−29)=17⋅(29+111−29)
=17⋅(29−29+111)=17⋅(29−29+111)
=17⋅111=17⋅111
=2997=2997
b.19⋅43+(−20)⋅43−(−40)19⋅43+(−20)⋅43−(−40)
=19⋅43−20⋅43+40=19⋅43−20⋅43+40
=43(19−20)+40
a) 2575 + 37 - 2576 - 39
= 2576 + 36 - 2576 - 29
= 36 - 29
= 7
b) 34 + 35 + 36 + 37 - 14 - 15 - 16 - 17
= (34 - 14) + (35 - 15) + (36 - 16) + (37 - 17)
= 20 + 20 + 20 + 20
= 20 . 4
= 80
a) Ta có: \(17\cdot\left[29-\left(-111\right)\right]+29\cdot\left(-17\right)\)
\(=17\cdot\left[29+111-29\right]\)
\(=17\cdot111=1887\)
b) Ta có: \(19\cdot43+\left(-20\right)\cdot43-\left(-40\right)\)
\(=43\cdot\left(19-20\right)+40\)
\(=-43+40=-3\)
\(\frac{3}{29}-\frac{1}{5}.\frac{29}{3}=\frac{3}{29}-\frac{29}{15}=\frac{-362}{145}\)
3/29 - 1/5 . 29/3
= 3/29 . 29/3 - 1/5
= 1 - 1/5
= 5/5 - 1/5
= 4/5
\(\frac{1}{2}B=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{15}-\frac{1}{16}=\frac{1}{4}-\frac{1}{16}=\frac{3}{16}\)
=>\(B=\frac{3}{16}:\frac{1}{2}=\frac{3}{8}\)
\(C=\left(\frac{3}{29}-\frac{1}{5}\right)\cdot\frac{29}{3}=1-\frac{1}{5}\cdot\frac{29}{3}=1-\frac{29}{15}=-\frac{14}{15}\)
17.(29-(-111))+29.(-17)
=17⋅(29+111)+(−17)⋅29
=17⋅140−17⋅29
=17(140-29)
=17.111
=1887
17.[29 - (-111)] + 29.(-17)
= 17.29 + 17.111 + 29.(-17)
= (17.29 - 29.17) + 17.111
= 0 + 1887
= 1887
1887