a, 544544- 444444 c , 165 : 11 . 999 + 85 .999
b, 31313- 10101 - 20202 d, 25.4 + 25 : 5 - 4 .( 30- 5 ) - 5
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a) 544544 – 444444 = 544.1001 – 444.1001 = 1001.(544 – 444) = 1001.100 = 100100
b) 131313 – 10101 – 20202 = 10101.13 – 10101 – 10101.2
= 10101.(13 – 1 – 2) = 10101.10 = 101010
\(a,165:11\cdot999+85\cdot999=15\cdot999+85\cdot999=999\left(15+85\right)=999\cdot100=99900\)
\(b,25\cdot4+25:5-4\left(30-5\right)-5=25\cdot4+5-4\cdot25-5=\left(25\cdot4-4\cdot25\right)+\left(5-5\right)=0+0=0\)
a, 165:11.999 + 85.999 = (15.11):11.999 + 85.999 = [(15.11):11].999 + 85.999
= 15.(11:11).999 + 85.999 = 15.999 + 85.999 = (15+85).999 = 99900
b, 25.4 + 25:5 – 4.(30 – 5) – 5 = 25.4 + 25:5 – 4.25 – 5 = (25.4 – 4.25) + (25:5 – 5) = 0 + 0 = 0
Ta có: \(B=\frac{3.5.7.11.13.17-10101}{1313130+20202}\)
\(\Rightarrow B=\frac{255255-10101}{1313130+20202}\)
\(\Rightarrow B=\frac{1001.255-10101}{10101.13.10+10101.2}\)
\(\Rightarrow B=\frac{1001.255-10101}{10101.\left(13.10+2\right)}\)
\(\Rightarrow B=\frac{1001.255-10101}{10101.132}\)
\(\Rightarrow B=\frac{10101.255-19201}{10101.132}\)
\(10101.\left(\frac{5}{10101}-\frac{5}{20202}-\frac{5}{30303}-\frac{5}{40404}\right)=\frac{10101.5}{10101}-\frac{10101.5}{10101.2}-\frac{10101.5}{10101.3}-\frac{10101.5}{10101.4}\)\(=5-\frac{5}{2}-\frac{5}{3}-\frac{5}{4}=-\frac{5}{12}\)(nhưng lớp 4 chưa học phân số âm)
=10101x5/10101-10101x5/20202+10101x5/30303-10101x5/40404
=5-5/2+5/3-5/4
=35/12
cái kia nó bị lỗi bn nha
a) 544544 – 444444
= 544.1001 – 444.1001
= 1001.(544 – 444)
= 1001.100
= 100100
b) 131313 – 10101 – 20202
= 10101.13 – 10101 – 10101.2
= 10101.(13 – 1 – 2)
= 10101.10
= 101010
a, 100100
b, 1010
c, 99900
d, 0