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a) (x-2)(x+3) <0 => x-2 và x+3 phải trái dấu
=> x-2<0 và x+3>0
hoặc x-2>0 và x+3<0
=> x<2 và x>-3 => -3<x<2
hoặc x>2 và x<-3 ( vô lý ) ( loại )
=> x \(\in\) { -2;-1;0;1 }
Đúng 100%, tích nha, please!!
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{63}+...+\frac{1}{\left(x+1\right)\left(x+4\right)}=\frac{199}{400}\)(ĐK \(x\ne-1;-4\))
=> \(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{\left(x+1\right)\left(x+4\right)}=\frac{199}{400}\)
=> \(\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{\left(x+1\right)\left(x+4\right)}\right)=\frac{199}{400}\)
=> \(\frac{1}{2}\left(1-\frac{1}{x+4}\right)=\frac{199}{400}\)
=> \(1-\frac{1}{x+4}=\frac{199}{400}:\frac{1}{2}=\frac{199}{200}\)
=> \(\frac{1}{x+4}=1-\frac{199}{200}=\frac{1}{200}\)
=> x + 4 = 200 => x = 196(tm)
Ta có
1/3 = 1/1 x 3
1/15 = 1/3 x 5
1/35 = 1/5 x 7
.....
1/(x + 1 ) x ( x + 4 )
\(\Rightarrow\)1 - 1/3 + 1/3 - 1/5 +1/5 - 1/7 +............+ 1/( x + 1 ) - 1/( x + 4) = 199/400
\(\Rightarrow\)1 - 1/( x + 4 ) = 199/400
\(\Rightarrow\)1/(x + 4 ) = 1 - 199/400
\(\Rightarrow\)1/(x + 4 ) = 201/400
còn lại bạn tự làm nha
x−42021+x−32020=x−22019+x−12018x−42021+x−32020=x−22019+x−12018
⇔ x−42021+x−32020−x−22019−x−12018=0x−42021+x−32020−x−22019−x−12018=0
⇔ (1+x−42021)+(1+x−32020)−(1+x−22019)−(1+x−12018)=0(1+x−42021)+(1+x−32020)−(1+x−22019)−(1+x−12018)=0⇔ x+20172021+x+20172020−x+20172019−x+20172018=0x+20172021+x+20172020−x+20172019−x+20172018=0
⇔ (x+2017)(12021+12020−12019−12018)=0(x+2017)(12021+12020−12019−12018)=0
⇔ x + 2017 = 0
⇔ x = -2017
\(\frac{x-1}{2020}+\frac{x-2}{2021}=\frac{x+1}{2018}+\frac{x+2}{2017}\)
\(\Leftrightarrow\frac{x-1}{2020}+1+\frac{x-2}{2021}-1=\frac{x+1}{2018}+1+\frac{x+2}{2017}+1\)
\(\Leftrightarrow\frac{x+2019}{2020}+\frac{x+2019}{2021}=\frac{x+2019}{2018}+\frac{x+2019}{2017}\)
\(\Leftrightarrow\left(x+2019\right)\left(\frac{1}{2020}+\frac{1}{2021}-\frac{1}{2018}-\frac{1}{2017}\right)=0\)
mà \(\frac{1}{2020}+\frac{1}{2021}-\frac{1}{2018}-\frac{1}{2017}\ne0\)
\(\Leftrightarrow x+2019=0\)
\(\Leftrightarrow x=-2019\)
B=5+2(x-2019)2020
Vì (x-2019)2020 ≥0
=>5+(x-2019)2020 ≥5
Để B đạt Min
=>x-2019=0
=>x=2019
Vậy MinB=5 <=>x=2019
a) \(A=\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.10}+\dfrac{1}{143}\)
\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)+\dfrac{1}{143}\)
\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{100}\right)+\dfrac{1}{143}=\dfrac{1}{2}.\dfrac{99}{100}+\dfrac{1}{143}=\dfrac{99}{200}+\dfrac{1}{143}=\dfrac{99.143+200.1}{200.143}=\dfrac{14157+200}{28600}=\dfrac{14357}{28600}\)
b) \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+99\right)=14950\)
\(\Rightarrow x+x+...+x+\left(1+2+...+99\right)=14950\)
\(\Rightarrow100x+\left(\left(99+1\right):2\right).99:2=14950\)
\(\Rightarrow100x+2475=14950\Rightarrow100x=12475\Rightarrow x=\dfrac{12475}{100}=\dfrac{499}{4}\)
Ta có : \(A=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\)
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(2A=1-\frac{1}{11}\)
\(2A=\frac{10}{11}\)
\(A=\frac{10}{11}.\frac{1}{2}=\frac{5}{11}\)
Vì x2 luôn luôn lơn hơn hoặc bằng 0
Nên x2 + 2015 luôn luôn lơn hơn hoặc bằng 2015
Nên x - 2016 = 0
=> x = 2016 (t/m)
1)Đặt A = \(\frac{1}{7.8}+\frac{1}{14.10}+\frac{1}{20.13}+...+\frac{1}{38.22}\)
\(\frac{1}{2}A=\frac{1}{8.14}+\frac{1}{14.20}+...+\frac{1}{38.44}\)
\(\frac{1}{2}A=6\left(\frac{1}{8}-\frac{1}{14}+\frac{1}{14}-\frac{1}{20}+...+\frac{1}{38}-\frac{1}{44}\right)\)
\(\frac{1}{2}A=6\left(\frac{1}{8}-\frac{1}{44}\right)\)
\(\frac{1}{2}A=6.\frac{9}{88}\)
\(\frac{1}{2}A=\frac{27}{44}\)
\(A=\frac{1}{44}:\frac{1}{2}=\frac{1}{22}\)
2) a) \(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{x\left(x+2\right)}=\frac{2018}{2020}\)
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{x\left(x+2\right)}=\frac{2018}{2020}\)
\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{x\left(x+2\right)}=\frac{2018}{2020}\)
\(\frac{1}{3}-\frac{1}{9}-\frac{1}{x\left(x+2\right)}=\frac{2018}{2020}\)
\(\frac{2}{9}-\frac{1}{x\left(x+2\right)}=\frac{2018}{2020}\)
\(\frac{1}{x\left(x+2\right)}=\frac{2}{9}-\frac{2018}{2020}\)
Hình như đề sai . Hoặc là mình sai >: