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a) (417+64):(416+16)
= (417+43):(416+42)
=420:418
=42=16
b)83:24+513:511.30
=(23)3:24+52.1
=26:24+25
=22+25
=4+25=29
c)33.48+34.42+81.42
=33.48+33.3.42+33.3.42
=33.48+33.126+33.126
=33.(48+126+126)
=27.300
=8100
1.
\(B=20182018.2017-20172017.2018\)
\(B=2018.10001.2017-2017.10001.2018\)
\(B=0\)
1 B=20182018.2017-20172017.2018
B=2018.10001.2017-2017.10001.2018
B=0
2 C=12+22+32+...+1002
C=1(1+0)+2(1+1)+3(1+2)+...+100(1+99)
C=1+2+1.2+3+2.3+...+100+99.100
C=(1+2+3+...+100)+(1.2+2.3+...+99.100)
C=[(1+100).100:2]+[(99.100.101):3]
C=5050+333300
C=338350
-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
a, \(5-\left(\frac{a}{b}+\frac{1}{2}\right)=2\frac{1}{3}\) => \(\frac{a}{b}+\frac{1}{2}=5-2\frac{1}{3}\) => \(\frac{a}{b}+\frac{1}{2}=\frac{8}{3}\) => \(\frac{a}{b}=\frac{8}{3}-\frac{1}{2}\) => \(\frac{a}{b}=\frac{13}{6}\)
b, \((\frac{3}{4}+2\frac{1}{2}):\frac{3}{5-3}=\left(\frac{3}{4}+\frac{5}{4}\right):\frac{3}{5}-1=\frac{9}{4}:\frac{-2}{5}=\frac{-45}{8}\)
a, 5-(\(\frac{a}{b}\)+\(\frac{1}{2}\))=2\(\frac{1}{3}\)
<=>5-\(\frac{a}{b}-\frac{1}{2}\)=\(\frac{7}{3}\)
<=>\(\frac{a}{b}=5-\frac{1}{2}-\frac{7}{3}\)
<=>\(\frac{a}{b}=\frac{13}{6}\)
b,(\(\frac{3}{4}\)+2\(\frac{1}{2}\)):\(\frac{3}{5}\)-3
=(\(\frac{3}{4}\)+\(\frac{5}{2}\)).\(\frac{5}{3}\)-3
=\(\frac{23}{4}\).\(\frac{5}{3}\)-3
=\(\frac{115}{12}\)-3
=\(\frac{115-36}{12}\)
=\(\frac{79}{12}\)
\(-23\cdot63+23\cdot21-58\cdot23\)
\(=23\left(-63+21-58\right)\)
\(=23\cdot\left(-100\right)\)
\(=-2300\)