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\(\dfrac{2022\times2023-1}{2023\times2021+2022}\)
= \(\dfrac{\left(2021+1\right)\times2023-1}{2023\times2021+2022}\)
= \(\dfrac{2023\times2021+2023-1}{2023\times2021+2022}\)
= \(\dfrac{2023\times2021+2022}{2023\times2021+2022}\)
= 1
\(A=\dfrac{2019\times2021-1}{2019\times2021}=\dfrac{2019\times2021}{2019\times2021}-\dfrac{1}{2019\times2021}=1-\dfrac{1}{2019\times2021}\)
\(B=\dfrac{2021\times2023-1}{2021\times2023}=\dfrac{2021\times2023}{2021\times2023}-\dfrac{1}{2021\times2023}=1-\dfrac{1}{2021\times2023}\)
\(...=1+1+...+1+1\)
Số số 1 là :
\(\left(2022-2\right):2+1+1=1012\left(số\right)\)
Vậy kết quả là \(1x1012=1012\)
Bài 1:
a: x+1/2=5/6
nên x=5/6-1/2=1/3
b: x+1/4=3/4
nên x=3/4-1/4=2/4=1/2
c: x+3/10=1/2
nên x=1/2-3/10=5/10-3/10=1/5
d: x+1/4=3/8
nên x=3/8-1/4=3/8-2/8=1/8
\(A=\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{2021}\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{2022}{2021}\)
\(=\frac{2022}{2}\)
\(=1011\)
A= 1* (1/2+1/3+1/4+...+2021)
A= 1/2+1/3+1/4+...+2021
Mik sẽ ko tính giúp bạn hết toàn bộ để bạn có thể tự làm được!
`a, 3/4 + 1/2 xx 7/2`
`= 3/4 + 7/4`
`=10/4`
`=5/2`
`b, 6/15 - 1/3 : 5/3`
`= 6/15 - 1/3 xx 3/5`
`= 6/15 - 3/15`
`= 3/15`
`=1/5`
`c, x-4/9 = 3/7 : 9/4`
`=> x-4/9= 3/7 xx 4/9`
`=> x-4/9= 12/63`
`=> x-4/9=4/21`
`=> x= 4/21 +4/9`
`=>x= 40/63`
`d, 7/9 xx 3/5 -1/2=1/5`
`->` sao lại bằng có `x` ko vậy ạ?
`a,`
`3/4+1/2 \times 7/2=3/4+7/4=10/4=5/2`
`b,`
`6/15 - 1/3 \div 5/3=6/15-1/5=1/5`
`c,` Tìm x?
`x-4/9=3/7 \div 9/4`
`x-4/9=4/21`
`x=4/21+4/9`
`x=40/63`
`d, 7/9x \times 3/5-1/2=1/5`
`7/9x \times 3/5=1/5+1/2`
`7/9x \times 3/5=7/10`
`7/9x=7/10 \div 3/5`
`7/9x=7/6`
`x=7/6 \div 7/9=3/2`
a, \(\dfrac{1}{2}.\dfrac{1}{4}.\dfrac{1}{6}=\dfrac{1}{48}\)
b, \(\dfrac{1}{2}.\dfrac{1}{4}:\dfrac{1}{6}=\dfrac{1}{8}.6=\dfrac{6}{8}=\dfrac{3}{4}\)
\(a,\dfrac{1}{2}\times\dfrac{1}{4}\times\dfrac{1}{6}=\dfrac{1}{8}\times\dfrac{1}{6}=\dfrac{1}{48}\)
\(b,\dfrac{1}{2}\times\dfrac{1}{4}:\dfrac{1}{6}=\dfrac{1}{8}:\dfrac{1}{6}=\dfrac{6}{8}=\dfrac{3}{4}\)
\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times...\times\left(1-\dfrac{1}{2023}\right)\\ =\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{2022}{2023}\\ =\dfrac{1}{2023}\)
đáp án B bạn nha