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\(=\dfrac{20}{21}x\dfrac{21}{22}x\dfrac{22}{23}x...x\dfrac{1999}{2000}\)
\(=\dfrac{20}{2000}=\dfrac{1}{100}\)
=20/21x21/22x22/23x..............x1998/1999x1999/2000
=20x21x22x23x.....................x1998x1999/21x22x23x24x...............x1999x2000
=20/2000
1/100
`(1-1/4) \times (1-1/9) \times (1-1/16) \times ... \times (1-1/10000)`
`= 3/4 \times 8/9 \times 15/16 \times ... \times 9999/10000`
`= (3 \times 8 \times 15 \times ... \times 9999)/(4 \times 9 \times 16 \times ... \times 10000)`
`=`\(\dfrac{\left(1\times3\right)\times\left(2\times4\right)\times\left(3\times5\right)\times...\times\left(99\times101\right)}{\left(2\times2\right)\times\left(3\times3\right)\times\left(4\times4\right)\times...\times\left(100\times100\right)}\)
`=` \(\dfrac{\left(1\times2\times3\times...\times99\right)\times\left(3\times4\times5\times...\times101\right)}{\left(2\times3\times4\times...\times100\right)\times\left(2\times3\times4\times...\times100\right)}\)
`=` \(\dfrac{1\times101}{2\times100}\)
`= 101/200.`
(1-1/4) x ( 1-1/9) x ( 1- 1/16) x....x (1 - 1/10000)
= 3/4 x 8/9 x 15/16 x ... x 9999/10000
= (1x3/2x2) x ( 2x4/3x3 ) x ( 3 x 5 / 4x4 ) x ... x ( 99 x 101 / 100x 100 )
= \(\dfrac{\left(1.2.3...99\right).\left(3.4.5.101\right)}{\left(2.3.4...100\right).\left(2.3.4...100\right)}\)
= 1 x 101 / 100 x 2
= 101/200
0,25 x 3 + 1 : 4 x 7
= 0,25 (3+7)
= 0,25 x 10
= 2,5
X x 1.2 + X x 1.8 = 45
X [1(2+8)] = 45
X [1 x 10] = 45
X x 10 = 45
X = 45 : 10
X = 4,5
Lời giải:
$x-\frac{x}{3}\times \frac{3}{2}=2-\frac{1}{2}$
$x-x\times \frac{1}{2}=\frac{3}{2}$
$x\times (1-\frac{1}{2})=\frac{3}{2}$
$x\times \frac{1}{2}=\frac{3}{2}$
$x=\frac{3}{2}: \frac{1}{2}=3$
\(\left(1-\frac{1}{4}\right)x\left(1-\frac{1}{9}\right)x\left(1-\frac{1}{16}\right)x\left(1-\frac{1}{25}\right)=\frac{3}{4}x\frac{8}{9}x\frac{15}{16}x\frac{24}{25}\)
\(=\frac{3x8x15x24}{4x9x16x25}\)
\(=\frac{3x2x4x3x5x4x2x3}{2x2x3x3x4x4x5x5}\)(chỗ này rút gọn đi)
\(=\frac{3}{5}\)
a: \(x+\dfrac{3}{9}=\dfrac{7}{6}\cdot\dfrac{2}{3}\)
=>\(x+\dfrac{1}{3}=\dfrac{14}{18}=\dfrac{7}{9}\)
=>\(x=\dfrac{7}{9}-\dfrac{1}{3}=\dfrac{7}{9}-\dfrac{3}{9}=\dfrac{4}{9}\)
b: \(x-\dfrac{2}{3}=\dfrac{1}{8}:\dfrac{5}{4}\)
=>\(x-\dfrac{2}{3}=\dfrac{1}{8}\cdot\dfrac{4}{5}=\dfrac{1}{10}\)
=>\(x=\dfrac{1}{10}+\dfrac{2}{3}=\dfrac{3+20}{30}=\dfrac{23}{30}\)
\(A=\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\cdot...\cdot\left(1-\dfrac{1}{9801}\right)\)
\(=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)\cdot...\left(1-\dfrac{1}{99}\right)\left(1+\dfrac{1}{99}\right)\)
\(=\left(\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{98}{99}\right)\cdot\left(\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{100}{99}\right)\)
\(=\dfrac{1}{99}\cdot\dfrac{100}{2}=\dfrac{50}{99}\)