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\(2A=2+\dfrac{1}{2}.6+\dfrac{1}{3}.12+\dfrac{1}{4}.20+...+\dfrac{1}{200}.40200=\)
\(=2+\dfrac{1}{2}.2.3+\dfrac{1}{3}.3.4+\dfrac{1}{4}.4.5+...+\dfrac{1}{200}.200.201=\)
\(=2+3+4+5+...+201=\dfrac{200\left(2+201\right)}{2}\)
\(=20300\Rightarrow A=\dfrac{20300}{2}=10150\)
A=1+\(\dfrac{1+2}{2}+\dfrac{1+2+3}{3}+........+\dfrac{1+2+.......+200}{200}\)
A=1+\(\dfrac{\dfrac{\left(1+2\right).2}{2}}{2}+\dfrac{\dfrac{\left(1+3\right).3}{2}}{3}+.......+\dfrac{\dfrac{\left(1+200\right).200}{2}}{200}\)
A=\(\dfrac{2}{2}\)+\(\dfrac{3}{2}\)+......+\(\dfrac{200}{2}\)=\(\dfrac{2+3+.......+200}{2}\)=\(\dfrac{\dfrac{\left(2+200\right).\text{[}\left(200-2\right):1+1\text{]}}{2}}{2}\)=\(\dfrac{19701}{2}\)
`1//([-1]/2)^2 . |+8|-(-1/2)^3:|-1/16|=1/4 .8+1/8 .16=2+2=4`
`2//|-0,25|-(-3/2)^2:1/4+3/4 .2017^0=0,25-2,25.4+0,75.1=0,25-9+0,75=-8,75+0,75-8`
`3//|2/3-5/6|.(3,6:2 2/5)^3=|-1/6|.(3/2)^3=1/6 . 27/8=9/16`
`4//|(-0,5)^2+7/2|.10-(29/30-7/15):(-2017/2018)^0=|1/4+7/2|.10-1/2:1=|15/4|.10-1/2=15/4 .10-1/2=75/2-1/2=37`
`5// 8/3+(3-1/2)^2-|[-7]/3|=8/3+(5/2)^2-7/3=8/3+25/4-7/3=107/12-7/3=79/12`
1) Ta có
\(C=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2022}\right)\)
\(C=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2021}{2022}\)
\(C=\dfrac{1}{2022}\)
2) \(A=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\)
\(\Rightarrow3A=1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\)
\(\Rightarrow4A=A+3A\) \(=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)
\(\Rightarrow12A=3.4A=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\)
\(\Rightarrow16A=12A+4A=\left(3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\right)+\left(1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\right)\)
\(=3-\dfrac{101}{3^{99}}-\dfrac{100}{3^{100}}\) \(< 3\). Từ đó suy ra \(A< \dfrac{3}{16}\)
Ta có: \(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)\cdot...\cdot\left(1-\dfrac{1}{10^2}\right)\)
\(=\dfrac{-3}{4}\cdot\dfrac{-8}{9}\cdot\dfrac{-15}{16}\cdot...\cdot\dfrac{-99}{100}\)
\(=-\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{99}{100}\)
\(=-\dfrac{10+1}{2\cdot10}=\dfrac{-11}{20}\)
Phải thế này nha bạn!
\(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{10^2}\right)\)
\(=\dfrac{2^2-1^2}{2^2}.\dfrac{3^2-1^2}{3^2}.\dfrac{4^2-1^2}{4^2}...\dfrac{10^2-1^2}{10^2}\)
\(=\dfrac{\left(2+1\right)\left(2-1\right)}{2.2}.\dfrac{\left(3+1\right)\left(3-1\right)}{3.3}.\dfrac{\left(4+1\right)\left(4-1\right)}{4.4}...\dfrac{\left(10+1\right)\left(10-1\right)}{10.10}\)
\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}...\dfrac{\left(10+1\right)\left(10-1\right)}{10.10}\)
\(=\dfrac{\left[1.2.3...\left(10+1\right)\right]\left[3.4.5...\left(10-1\right)\right]}{\left(2.3.4...10\right)\left(2.3.4...10\right)}\)
\(=\left(10+1\right).\dfrac{1}{2.10}\)
\(=\dfrac{11}{20}\)
Theo mình nghĩ phải như thế này.
Vì lớp 6 chưa học hàng đẳng thức nên phải làm thêm bước này:
ta có : a2 - b2 = a2 - ab + ab - b2 = a(a-b) + b( a-b) = (a-b)(a+b)
⇒ a2 - b2 = (a-b)(a+b)
Áp dụng vào biểu thức A ta có :
A=(1−122).(1−132).(1−142)....(1−1302)
A = ( 1 - \(\dfrac{1}{2}\))(1+ \(\dfrac{1}{2}\))(1 - \(\dfrac{1}{3}\)).(1+\(\dfrac{1}{3}\)).....(1-\(\dfrac{1}{30}\))(1+\(\dfrac{1}{30}\))
A = {(1-\(\dfrac{1}{2}\))(1-\(\dfrac{1}{3}\)).(1-\(\dfrac{1}{4}\))........(1-\(\dfrac{1}{30}\))}{(1+\(\dfrac{1}{2}\))(1+\(\dfrac{1}{3}\)).......(1+\(\dfrac{1}{30}\))}
A =( \(\dfrac{1}{2}\).\(\dfrac{2}{3}\).\(\dfrac{3}{4}\).....\(\dfrac{29}{30}\))( \(\dfrac{3}{2}\).\(\dfrac{4}{3}\).\(\dfrac{5}{4}\).........\(\dfrac{31}{30}\))
A = \(\dfrac{2.3.4.5.6......29}{2.3.4.5.6.....29}\) \(\times\) \(\dfrac{1}{30}\) x \(\dfrac{3.4.5.......30}{3.4.5.......30}\) \(\times\) \(\dfrac{31}{2}\)
A =1 \(\times\) \(\dfrac{1}{30}\) \(\times\) \(\dfrac{31}{2}\)
A = \(\dfrac{31}{60}\)
\(A=-\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{2014^2}\right)\)
\(A=\dfrac{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(2012\cdot2014\right)\left(2013\cdot2015\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(2013\cdot2013\right)\left(2014\cdot2014\right)}\)
\(A=\dfrac{\left(1\cdot2\cdot3\cdot...\cdot2012\cdot2013\right)\left(3\cdot4\cdot5\cdot...\cdot2014\cdot2015\right)}{\left(2\cdot3\cdot4\cdot...\cdot2013\cdot2014\right)\left(2\cdot3\cdot4\cdot...\cdot2013\cdot2014\right)}\)
\(A=\dfrac{1\cdot2015}{2014\cdot2}=\dfrac{2015}{4028}\)
Vì \(\dfrac{2015}{4028}>-\dfrac{1}{2}\) nên A > B
e: \(=\left(\dfrac{18}{37}+\dfrac{19}{37}\right)+\left(\dfrac{8}{24}+\dfrac{2}{3}\right)-\dfrac{47}{24}=2-\dfrac{47}{24}=\dfrac{1}{24}\)
f: \(=-8\cdot\dfrac{1}{2}:\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=-4:\dfrac{13}{12}=\dfrac{-48}{13}\)
g: \(=\dfrac{4}{25}+\dfrac{11}{2}\cdot\dfrac{5}{2}-\dfrac{8}{4}=\dfrac{4}{25}+\dfrac{55}{4}-2=\dfrac{1191}{100}\)