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=-14.(-13).(-12).....83.84.85
= 14.13.12.11.10.9.8.7.6.5.4.3.2.1.0.1.2.3.4....85
=0
a) \(53.\left(-15\right)+\left(-15\right).47=\left(-15\right).\left(53+47\right)=\left(-15\right).100=-1500\)
b) \(\left(-43\right).92-46.27+46.41=\left(-86\right).46-46.27+46.41\)
\(=46\left[\left(-86\right)-27+41\right]=46.\left(-72\right)=-3312\)
c) \(\left(-72\right)\left(15-49\right)+15\left(-56+72\right)=\left(-72\right).15-72.49+15.\left(-56\right)+15.72\)
\(=\left[\left(-72\right).15+15.72\right]-72.49-15.56=0-3528-840=4368\)
d) \(\left(-2^4\right).17.\left(-3\right)^0.\left(-5\right)^6\left(-1^{2n}\right)=16.17.1.12625.\left(-1^{2n}\right)\)
a,\(53.\left(-15\right)+\left(-15\right).47\)
\(=-15.\left(53+47\right)=-15.100=-1500\)
\(b,-43.92-46.27+46.41\)
\(=-43.92-\left[46.\left(27-41\right)\right]=-43.92-\left[46.\left(-14\right)\right]\)
\(=-3956+644=-3312\)
\(c,-72\left(15-49\right)+15\left(-56+72\right)\)
\(=-15.72+49.72+15.\left(-56\right)+15.72\)
\(=72.\left(-15+49+15\right)-15.56=3528-840=2688\)
Thực hiện phép tính
-a) -5/7.7/10-3/11:6/5
b) 5/14.1/3.-4/15:5/7
c) 2 13/15.11/15-13/15+2 13/15 . 4/15
a) -5/7.7/10 - 3/11 : 6/5
= -1/2 - 5/22
= -8/11
b) 5/14.1/3.-4/15:5/7
= -2/63:5/7 = -2/45
c) 2 13/15.11/15 - 13/15 + 2 13/15.4/15
= 2 13/15.(11/15 + 4/15) - 13/15
= 2 13/15.1 - 13/15
= 2 13/15 - 13/15
= 2
a) 4+(12-15) và 4+12-15
4+(-3) và 16-15
1 và 1
=> Bằng nhau
b) 4-(12-15) và 4-12+15
4-(-3) và -8+15
7 và 7
=> bằng nhau
a) 53.(-15)+(-15).47
=53.[(-15)+(-15)].47
=53.(-30).47
=-1590.47
=-74730
a) \(x+\left(-81\right)-\left(11-81\right)=x-81-11+81=x-11\)
b) \(\left(-1-x+2\right)+1=\left(-1\right)-x+2+1=-x+2=2-x\)
c) \(15-\left(11-x\right)+\left(11-15\right)=15-11-x+11-15\)
\(=\left(15-15\right)+\left(11-11\right)+x=0+0+x=x\)
d) \(15-\left(15-x+202\right)=15-15+x-202=x-202\)
a)x-81-11+81=x-11
b)(-1-x+2)+1=-1-x+2+1=2-x
c)15-(11-x)+(11-15)=15-11+x+11-15=x
d)15-(15-x+202)+202=15-15+x-202+202=x
Ta có: \(\frac{15}{90.94}+\frac{15}{94.98}+\frac{15}{98.102}+...+\frac{15}{146.150}\)
= \(15.\left(\frac{1}{90.94}+\frac{1}{94.98}+\frac{1}{98.102}+...+\frac{1}{146+150}\right)\)
= \(15.\left[\frac{1}{4}.\left(\frac{4}{90.94}+\frac{4}{94.98}+\frac{4}{98.102}+...+\frac{4}{146+150}\right)\right]\)
= \(15.\left[\frac{1}{4}.\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+\frac{1}{98}-\frac{1}{102}+...+\frac{1}{146}-\frac{1}{150}\right)\right]\)
= \(15.\left[\frac{1}{4}.\left(\frac{1}{90}-\frac{1}{150}\right)\right]\)
= \(15.\left(\frac{1}{4}.\frac{1}{225}\right)\)
= \(=\frac{1}{60}\)
Bài làm
\(\frac{15}{90.94}+\frac{15}{94.98}+\frac{15}{98.102}+...+\frac{15}{146.150}\)
= \(15.\frac{1}{90.94}+15.\frac{1}{94.98}+15.\frac{1}{98.102}+...+15.\frac{1}{146.150}\)
= \(15.\left(\frac{1}{90.94}+\frac{1}{94.98}+\frac{1}{98.102}+...+\frac{1}{146.150}\right)\)
= \(15.\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+\frac{1}{98}-\frac{1}{102}+...+\frac{1}{146}-\frac{1}{150}\right)\)
= \(15.\left(\frac{1}{90}-\frac{1}{150}\right)\)
= \(15.\left(\frac{5}{450}-\frac{3}{450}\right)\)
= \(15.\frac{2}{450}\)
= \(\frac{2}{30}\)
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=1 nha bạn
bằng 46 nha bạn