Tính:
4/5 + 3
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3: \(=20-12-8+12=20-8=12\)
5: \(=-18-42-21-35=-116\)
3: \(=-15+18-12+8=-27+26=-1\)
2: \(=-12+21-15+10=9-5=4\)
`3/4 xx4/5 +3/4xx8/5-3/4xx7/5`
`=3/4xx(4/5+8/5-7/5)`
`=3/4xx5/5`
`=3/4xx1`
`=3/4`
Ta có: \(\left(7-\dfrac{4}{3}+\dfrac{1}{3}\right)-\left(6+\dfrac{5}{4}-\dfrac{4}{3}\right)-\left(5\cdot\dfrac{7}{4}+\dfrac{5}{3}\right)\)
\(=6-\left(\dfrac{72}{12}+\dfrac{15}{12}-\dfrac{16}{12}\right)-\left(\dfrac{35}{4}+\dfrac{5}{3}\right)\)
\(=6-\dfrac{71}{12}-\dfrac{125}{12}\)
\(=\dfrac{-124}{12}=\dfrac{-62}{6}=\dfrac{-31}{3}\)
5 - 1 = 4 1 + 4 = 5 2 + 3 = 5
5 - 2 = 3 4 + 1 = 5 3 + 2 = 5
5 - 3 = 2 5 - 1 = 4 5 - 2 = 3
5 - 4 = 1 5 - 4 = 1 5 - 3 = 2
5-1=4 1+4=5 2+3=5
5-2=3 4+1=4 3+2=5
5-3=2 5-1=4 5-2=3
5-4=1 5-4=1 5-3=2
Bài 3:
a: a*S=a^2+a^3+...+a^2023
=>(a-1)*S=a^2023-a
=>\(S=\dfrac{a^{2023}-a}{a-1}\)
b: a*B=a^2-a^3+...-a^2023
=>(a+1)B=a-a^2023
=>\(B=\dfrac{a-a^{2023}}{a+1}\)
a: =-1/3+1/3=0
b: \(=\dfrac{4}{11}\left(-\dfrac{2}{7}-\dfrac{4}{7}-\dfrac{1}{7}\right)=\dfrac{4}{11}\cdot\left(-1\right)=-\dfrac{4}{11}\)
c: \(=10+\dfrac{5}{9}-3-\dfrac{5}{7}-4-\dfrac{5}{9}=3-\dfrac{5}{7}=\dfrac{16}{7}\)
d: \(=\dfrac{1}{3}+\dfrac{7}{4}-\dfrac{7}{4}+\dfrac{4}{5}=\dfrac{1}{3}+\dfrac{4}{5}=\dfrac{5+12}{15}=\dfrac{17}{15}\)
a: =-1/3+1/3=0
b: =411(−27−47−17)=411⋅(−1)=−411=411(−27−47−17)=411⋅(−1)=−411
c: =10+59−3−57−4−59=3−57=167=10+59−3−57−4−59=3−57=167
d: =13+74−74+45=13+45=5+1215=1715
`4/5+3`
`=4/5+3/1`
`=4/5+15/5`
`=19/5`
4/5+3
=4/5+3/1=4/5+3/1
=4/5+15/5=4/5+15/5
=19/5=19/5