tim x biet:
: \(\left(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\right).x=\frac{2012}{51}+\frac{2012}{52}+...+\frac{2012}{99}+\frac{2012}{100}\)
Giải bất phương trình:
\(\left(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\right).\left(x^2-x+1982\right)< 2012.\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)\)
tìm x
(\(\frac{1}{1.2}\)+\(\frac{1}{3.4}\)+...+\(\frac{1}{99.100}\)). x =\(\frac{2012}{51}\)+\(\frac{2012}{52}\)+....+\(\frac{2012}{99}\)+\(\frac{2012}{100}\)
Giải Pt :
a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+........+\frac{1}{x\left(x+1\right)}=\frac{\sqrt{2012-x}+2012}{\sqrt{2012-x}+2013}\)
b) \(\sqrt{2x+3}+\sqrt{x+1}=3x+2\sqrt{2x^2+5x+3}-16\)
b) \(\left(\sqrt{2x+3}-3\right)+\left(\sqrt{x+1}-2\right)+5=3x+2\left(\sqrt{2x^2+5x+3}-6\right)+12-16\)
\(\Leftrightarrow\left(\sqrt{2x+3}-3\right)+\left(\sqrt{x+1}-2\right)=3\left(x-3\right)+2\left(\sqrt{2x^2+5x+3}-6\right)\)
\(\Leftrightarrow\frac{2\left(x-3\right)}{\sqrt{2x+3}+3}+\frac{x-3}{\sqrt{x+1}+2}-3\left(x-3\right)-\frac{2\left(x-3\right)\left(2x+11\right)}{\sqrt{2x^2+5x+3}+6}=0\Leftrightarrow x-3=0\Leftrightarrow x=3.\)
Giải phương trình: \(\left(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)2013x=\dfrac{2012}{51}+\dfrac{2012}{52}+\dfrac{2012}{99}+\dfrac{2012}{100}\)
\(\frac{1}{1.2}+\frac{1}{3.4}+....+\frac{1}{99.100}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{100}-2.\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{100}\right)\)
\(=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{100}-1-\frac{1}{2}-\frac{1}{3}-....-\frac{1}{50}=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
=> \(2013x.\left(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)=2013x.\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)\)
=> \(2013x.\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)=2012.\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)\Rightarrow2013x=2012\Rightarrow x=\frac{2012}{2013}\)
Vậy \(x=\frac{2012}{2013}\)
p/s: --trình bày sai sót mong bỏ qua
Giải phương trình: \(\left(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)2013x=\dfrac{2012}{51}+\dfrac{2012}{52}+\dfrac{2012}{99}+\dfrac{2012}{100}\)
\(\frac{\left(1+\frac{2012}{1}\right)\left(1+\frac{2012}{2}\right).......\left(1+\frac{2012}{100}\right)}{\left(1+\frac{1000}{1}\right)\left(1+\frac{1000}{2}\right).....\left(1+\frac{1000}{2012}\right)}\)
($\frac{1}{1.2}$ + $\frac{1}{2.3}$ + $\frac{1}{3.4}$ + ... + $\frac{1}{2011. 2012}$ ) x = 2011
\(\Leftrightarrow x\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2011}-\dfrac{1}{2012}\right)=2011\)
\(\Leftrightarrow x\cdot\dfrac{2011}{2012}=2011\)
hay x=2012
\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2011.2012}\right)x=2011\)
\(\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2011}-\dfrac{1}{2012}\right)x=2011\)
\(\left(\dfrac{1}{1}-\dfrac{1}{2012}\right)x=2011\)
\(\dfrac{2011}{2012}x=2011\)
\(x=2012\)
`(1/[1.2]+1/[2.3]+1/[3.4]+....+1/[2011.2012])x=2011`
`(1-1/2+1/2-1/3+1/3-1/4+.....+1/2011-1/2012)x=2011`
`(1-1/2012)x=2011`
`2011/2012x=2011`
`x=2011:2011/2012`
`x=2012`
Tính giá trị của biểu thức:
a) A= (153 + 5. 152 - 53) : ( 183 + 6. 182 - 63)
b) \(B=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{2012}-1\right)\)
c) \(C=\frac{-1}{1.2}.\frac{-2^2}{2.3}.\frac{-3^2}{3.4}...\frac{-99^2}{99.100}.\frac{-100^2}{100.101}\)
=>\(-B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2012}\right)\)
=\(\frac{1}{2}.\frac{2}{3}...\frac{2011}{2012}=\frac{1}{2012}\)
Tìm x, biết: (1/1.2+1/3.4+...+1/99.100).x=2012/51+2012/52+...+2012/99+2012/100
giúp mình nhaaaaaaaaaaaaaaaaaaa ^^ (cho bạn luông 1 0 tick í í í ^.*)
Ta có:
\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\)
=> \(\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right).x=\frac{2012}{51}+\frac{2012}{52}+...+\frac{2012}{99}+\frac{2012}{100}\)
=> \(\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right).x=2012.\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\)
=> x = 2012