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a/ =60+4-6^2
=64-36
=28
b/ = (9+8.5):7
=(9+40):7
= 49:7
= 7
c/ = 11^2 -243:9 - 60
= 121-27-60
= 34
1−2−3+4+5−6−7+8+...+21−22−23+24+25
= (1 - 2 - 3 + 4) + (5 - 6 - 7 + 8) + ... + (21 - 22 - 23 + 24) + 25=(1−2−3+4)+(5−6−7+8)+...+(21−22−23+24)+25
= 0 + 0 + ... + 0 + 25=0+0+...+0+25
= 25
Câu 17: =8/5-1=8/5-5/5=3/5
Câu18: =8/5+1/20=32/20+1/20=33/20
-3/11.(-22)/66.121/15
=(-3).(-22).121
11.66.15
=11
15
3/7.2/5.7/3.20.19/72
=3.2.7.20.19
7.5.3.72
=76
16
6/7.8/13+6/13.9/7-3/13.6/7
=6/7.8/13+6/7.9/13-3/13.6/7
=6/7.(8/13+9/13-3/13)
=6/7.14/13
=12/13
-1/4.152/11+68/4.(-1)/11
=152/4.(-1)/11+68/4.(-1)/11
=(-1)/11.(152/4+68/4)
=(-1)/11.220/4
=-110/22
-5/7.2/11+(-5)/7.9/11+12/7
=-5/7.2/11+-5/7.9/11+12/7
=-5/7.(2/11+9/11)+12/7
=-5/7.1+12/7
=(-5)/7+12/7
=7/7
=1
146/13-(18/7+68/13)
=146/13-18/7-68/13
=(146/13-68/13)-18/7
=78/13-18/7
=6-18/7
=42/7-18/7
=24/7
c: Ta có: \(\dfrac{5}{3}+\dfrac{5}{3\cdot5}+\dfrac{5}{5\cdot7}+...+\dfrac{5}{101\cdot103}\)
\(=\dfrac{5}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{101\cdot103}\right)\)
\(=\dfrac{5}{2}\left(1-\dfrac{1}{103}\right)\)
\(=\dfrac{5}{2}\cdot\dfrac{102}{103}\)
\(=\dfrac{255}{103}\)
TL:
5 - 128 : [76 : 75 + (23 - 18)2]
= 5 - 128 : [71 + 25]
= 5 - 128 : 32
= 5 - 4
= 1
HT
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