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Nếu \(a+b+c=0\)
\(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+0.\dfrac{2}{abc}\)
\(=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2\left(a+b+c\right)}{abc}=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}\)
\(=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\)
a. Đề bài sai, chắc chắn thiếu dữ kiện \(a+b+c=0\)
b.
\(\sqrt{1+\dfrac{1}{99^2}+\dfrac{1}{100^2}}=\sqrt{\dfrac{1}{1^2}+\dfrac{1}{99^2}+\dfrac{1}{\left(-100\right)^2}}=\sqrt{\left(1+\dfrac{1}{99}-\dfrac{1}{100}\right)^2}\)
\(=1+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{9901}{9900}\)
a)\(=\left(\dfrac{2}{2}+\dfrac{1}{2}\right)\times\left(\dfrac{3}{3}+\dfrac{1}{3}\right)\times...\times\left(\dfrac{2005}{2005}+\dfrac{1}{2005}\right)\)
\(=\dfrac{3}{2}\times\dfrac{4}{3}\times...\times\dfrac{2006}{2005}=\dfrac{2006}{2}=1003\)
b)\(=\left(\dfrac{2}{3}+\dfrac{1}{3}\right)\times\dfrac{1}{2}=\dfrac{3}{3}\times\dfrac{1}{2}=\dfrac{1}{2}\)
a) \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{100}}\)
\(2A=2\cdot\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right)\)
\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{101}}\)
\(2A-A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}-\dfrac{1}{2}-\dfrac{1}{2^2}-...-\dfrac{1}{2^{100}}\)
\(A=1-\dfrac{1}{2^{100}}\)
b) \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2023\cdot2024}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2023}-\dfrac{1}{2024}\)
\(=1-\dfrac{1}{2024}\)
\(=\dfrac{2024}{2024}-\dfrac{1}{2024}\)
\(=\dfrac{2023}{2024}\)
a) Ta có: \(A=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{2014}\right)\left(1-\dfrac{1}{2015}\right)\left(1-\dfrac{1}{2016}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2013}{2014}\cdot\dfrac{2014}{2015}\cdot\dfrac{2015}{2016}\)
\(=\dfrac{1}{2016}\)
b) Ta có: \(\dfrac{x-2}{12}+\dfrac{x-2}{20}+\dfrac{x-2}{30}+\dfrac{x-2}{42}+\dfrac{x-2}{56}+\dfrac{x-2}{72}=\dfrac{16}{9}\)
\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)=\dfrac{16}{9}\)
\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\right)=\dfrac{16}{9}\)
\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{3}-\dfrac{1}{9}\right)=\dfrac{16}{9}\)
\(\Leftrightarrow\left(x-2\right)\cdot\dfrac{2}{9}=\dfrac{16}{9}\)
\(\Leftrightarrow x-2=\dfrac{16}{9}:\dfrac{2}{9}=\dfrac{16}{9}\cdot\dfrac{9}{2}=8\)
hay x=10
Vậy: x=10
`A=(8 2/7-4 2/7)-3 4/9`
`=8+2/7-4-2/7-3-4/9`
`=4-3-4/9`
`=1-4/9=5/9`
`B=(10 2/9-6 2/9)+2 3/5`
`=10+2/9-6-2/9+2+3/5`
`=4+2+3/5`
`=6+3/5=33/5`
Bài 2:
`a)5 1/2*3 1/4`
`=11/2*13/4`
`=143/8`
`b)6 1/3:4 2/9`
`=19/3:38/9`
`=19/3*9/38=3/2`
`c)4 3/7*2`
`=31/7*2`
`=62/7`
Bài 1:
\(A=\left(8\dfrac{2}{7}-4\dfrac{2}{7}\right)-3\dfrac{4}{9}\)
\(A=\left(\dfrac{58}{7}-\dfrac{30}{7}\right)-\dfrac{31}{9}\)
\(A=4-\dfrac{31}{9}\)
\(A=\dfrac{5}{9}\)
\(B=\left(10\dfrac{2}{9}-6\dfrac{2}{9}\right)+2\dfrac{3}{5}\)
\(B=\left(\dfrac{92}{9}-\dfrac{56}{9}\right)+\dfrac{13}{5}\)
\(B=4+\dfrac{13}{5}\)
\(B=\dfrac{33}{5}\)
\(B=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)
\(B=\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+\left(\dfrac{3}{47}+1\right)+...+\left(\dfrac{48}{2}+1\right)+\dfrac{49}{1}\)
\(B=\left(\dfrac{50}{49}+\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}\right)+1\)
\(B=\dfrac{50}{50}+\dfrac{50}{49}+\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}\)
\(B=50\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+...+\dfrac{1}{2}\right)\)
\(\Rightarrow\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}+\dfrac{1}{50}}{50\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+...+\dfrac{1}{2}\right)}=\dfrac{1}{50}\)
A= 1+(\(\dfrac{1}{2014}\)+1)+(\(\dfrac{2}{2013}\)+1)+...+(\(\dfrac{2013}{2}\)+1)
= \(\dfrac{2015}{2015}\)+(\(\dfrac{1}{2014}\)+1)+(\(\dfrac{2}{2013}\)+1)+...+(\(\dfrac{2013}{2}\)+1)
= 2015.(\(\dfrac{1}{2015}\)+\(\dfrac{1}{2014}\)+\(\dfrac{1}{2013}\)+...+\(\dfrac{1}{2}\))=2015.B
\(\Rightarrow\) \(\dfrac{A}{B}\)=2015
Bài 1
\(1+\frac{1}{a^2}+\frac{1}{(a+1)^2}=(1+\frac{1}{a})^2-\frac{2}{a}+\frac{1}{(a+1)^2}\)
\(=(\frac{a+1}{a})^2-2.\frac{a+1}{a}.\frac{1}{a+1}+(\frac{1}{a+1})^2=(\frac{a+1}{a}-\frac{1}{a+1})^2\)
\(=(1+\frac{1}{a}-\frac{1}{a+1})^2\)
$\Rightarrow A=|1+\frac{1}{a}-\frac{1}{a+1}|=1+\frac{1}{a}-\frac{1}{a+1}$ với $a>0$
Bài 2:
Áp dụng kết quả bài 1 thì:
\(B=1+\frac{1}{1}-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+...+1+\frac{1}{2011}-\frac{1}{2012}\)
\(=2011+(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011})-(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012})\)
\(=2012-\frac{1}{2012}\)
a) \(1-\dfrac{1}{2}=\dfrac{1}{2}\)
\(\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{3-2}{6}=\dfrac{1}{6}\)
\(\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{4-3}{12}=\dfrac{1}{12}\)
\(\dfrac{1}{4}-\dfrac{1}{5}=\dfrac{5-4}{20}=\dfrac{1}{20}\)
\(\dfrac{1}{5}-\dfrac{1}{6}=\dfrac{6-5}{30}=\dfrac{1}{30}\)
b) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}\)
\(=\left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+\left(\dfrac{1}{4}-\dfrac{1}{5}\right)+\left(\dfrac{1}{5}+\dfrac{1}{6}\right)\)
\(=1+\left(-\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+\left(-\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(-\dfrac{1}{5}+\dfrac{1}{5}\right)+-\dfrac{1}{6}\)\(=1+-\dfrac{1}{6}\)
\(=\dfrac{5}{6}\)