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5:
a: Số đoạn thẳng tạo thành là: 100*99/2=4950(đoạn)
b: \(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{2022}\right)\left(1+\dfrac{1}{2}\right)\cdot\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1+\dfrac{1}{2022}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{2021}{2022}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{2023}{2022}\)
=1/2022*2023/2
=2023/4044
`(15-x)+(x-12)=7-(-5+x)`
`=>15-x+x-12=7+5-x`
`=>3=12-x`
`=>x=12-3`
`=>x=9`
Vậy `x=9`
a)
\(\dfrac{6}{13}\cdot\dfrac{8}{7}\cdot\dfrac{-26}{3}\cdot\dfrac{-7}{8}\)
\(=\dfrac{6}{13}\cdot\dfrac{-26}{3}\cdot\dfrac{8}{7}\cdot\dfrac{-7}{8}\)
\(=-4\cdot\left(-1\right)\\ =4\)
b)
\(\dfrac{6}{11}+\dfrac{11}{3}\cdot\dfrac{3}{22}\)
\(=\dfrac{6}{11}+\dfrac{1}{2}\\ =\dfrac{12}{22}+\dfrac{11}{22}\\ =\dfrac{23}{22}\)
\(\frac{2}{2.4}+\frac{2}{4.6}+....+\frac{2}{x\left(x+2\right)}=\frac{4}{9}\)
<=> \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{x}-\frac{1}{x+2}=\frac{4}{9}\)
<=> \(\frac{1}{2}-\frac{1}{x+2}=\frac{4}{9}\)
<=> \(\frac{1}{x+2}=\frac{1}{18}\)
=> \(x+2=18\)
<=> \(x=16\)
Vậy...
( 2 + x ) + ( 4 + x ) + ( 6 + x ) + ... + ( 52 + x ) = 780
( x + x + x + ... + x ) + ( 2 + 4 + 6 + ... + 52 ) = 780
26x = 780 - 702
26x = 78
x = 78 : 26
x = 3
a) \(\frac{28\times7-45\times7+7\times18}{45\times14}\)
\(=\frac{7\left(28-45+7\right)}{45\times14}\)
\(=\frac{7\times\left(-10\right)}{45\times14}=\frac{-1}{9}\)
b) \(\frac{12.3-2.6}{4.5.6}\)
\(=\frac{2.6.3-2.6}{4.5.6}\)
\(=\frac{2.6\left(3-1\right)}{2.2.5.6}\)
\(=\frac{2.6.2}{2.2.5.6}\)\(=\frac{1}{5}\)
ta phân tích thành
\(\frac{1}{1\cdot2}\)x\(\frac{2\cdot2}{2\cdot3}\)x\(\frac{3\cdot3}{3\cdot4}\)x......x\(\frac{5\cdot5}{5\cdot6}\)x\(\frac{6\cdot6}{6\cdot7}\)
Tử nhân tử mẫu nhân mẫu ta có
1x2x2x3x3x.........x5x5x6x6
1x2x2x3x3x4x....x5x6x6x7
Rút gọn ta có
\(\frac{1}{7}\)
Vậy A=\(\frac{1}{7}\)
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