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3/ Chu vi hình chữ nhật:
\(\left(\dfrac{1}{4}+\dfrac{3}{10}\right)\cdot2=\dfrac{11}{10}\) (chưa biết đơn vị)
Diện tích hình chữ nhật:
\(\dfrac{1}{4}\cdot\dfrac{3}{10}=\dfrac{11}{20}\) (chưa biết đơn vị)
Theo đề bài ta có :
\(A=\frac{n+1}{n-1}=\frac{1}{2}\)
\(\Leftrightarrow2\left(n+1\right)=n-1\)
\(\Leftrightarrow2n+2=n-1\)
\(\Leftrightarrow2n-n=-1-2\)
\(\Rightarrow n=-3\)
Vậy với n = - 3 thì A = \(\frac{1}{2}\)
Để mình sửa lại cái đề bạn chút nghen !!
\(x_1+x_2=x_3+x_4=....=x_{49}+x_{50}=x_{51}+x_1\\ \Rightarrow x_1+x_2+x_3+x_4+....+x_{49}+x_{50}+x_{51}+x_1=50\\ \)
Mà : \(x_1+x_2+...+x_{51}=0\\ \Rightarrow x_1=49\\ M\text{à};x_{51}+x_1=1\\ \Rightarrow x_{51}=-50\\ \Rightarrow x_{50}=51\)
Chúc bạn học tốt !!!
\(=\dfrac{2}{2}\).(\(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{10}\)+...+\(\dfrac{2}{x.\left(x+1\right)}\))
=2.(\(\dfrac{1}{6}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{20}\)+...+\(\dfrac{2}{x.\left(x+1\right)}\))
=2.(\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+\(\dfrac{1}{4.5}\)+...+\(\dfrac{1}{x.\left(x+1\right)}\))
=2.[(\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\))+(\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\))+(\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\))+...+(\(\dfrac{1}{x}\)-\(\dfrac{1}{x+1}\))
=2.[\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+...+\(\dfrac{1}{x}\)-\(\dfrac{1}{x+1}\)]
2.[(\(\dfrac{1}{3}\)-\(\dfrac{1}{3}\))+(\(\dfrac{1}{4}\)-\(\dfrac{1}{4}\))+...+(\(\dfrac{1}{x}\)-\(\dfrac{1}{x}\))+(\(\dfrac{1}{2}\)-\(\dfrac{1}{x+1}\))]
=2.[0+0+...+0+(\(\dfrac{1}{2}\)-\(\dfrac{1}{x+1}\))]
=2.(\(\dfrac{1}{2}\)-\(\dfrac{1}{x+1}\))
=2.(\(\dfrac{1.x+1-1.2}{2.x+1}\))
=2.(\(\dfrac{x+1-2}{2x}\))=2.\(\dfrac{x-1}{2x}\)=\(\dfrac{2.\left(x-1\right)}{2x}\)=\(\dfrac{2x-2}{2x}\)
\(\dfrac{2x-2}{2x}\)=\(\dfrac{2014}{2016}\)\(\Rightarrow\)(2x-2).2016=2014.2x=4032x-4032=4028x
\(\Rightarrow\)4032x-4028x=4x=4032\(\Rightarrow\)x=4032:4=1008
Đặt A=\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x.\left(x+1\right)}\)
\(A=\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+1\right)}\)
\(A=\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{x.\left(x+1\right)}\)
1,=0 . [2017/2018+2018/2019]
=>0
2,TH1 x-3=0=>x=3
TH2 y-4=0=>y=4
3, -2/4 = -x/10 = 16/y
=>-1/2 = -x/10 = 16/y
=>-1/2 = -x/10 => -5/10 = -x/10 => x=5
-1/2 = 16/y => 16/-32 = 16/y => y = -32
\(\left(x-2\right)\left(x-4\right)< 0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2< 0\\x-4>0\end{matrix}\right.=>4< x< 2\left(1\right)\\\left\{{}\begin{matrix}x-2>0\\x-4< 0\end{matrix}\right.=>2< x< 4\left(2\right)}\end{matrix}\right.\)(1 ) vô lý=> loại
=> (x-2).(x-4)<0 <=> 2<x<4
b. ta có\(x^2+1>0\forall x\)
=>(x2 -1).(x2+1)<0 <=> (x2 -1)<0 <=> x2<1
<=> -1<x<1
câu c bạn làm tương tự
Vì : \(x^2-3>x^2-10\)
\(\Rightarrow\left\{\begin{matrix}x^2-3>0\\x^2-10< 0\end{matrix}\right.\)\(\Leftrightarrow\left\{\begin{matrix}x^2>3\\x^2< 10\end{matrix}\right.\)\(\Leftrightarrow3< x^2< 10\)
\(\Rightarrow x^2\in\left\{4;5;6;7;8\right\}\)
Mà : \(x\in Z\Rightarrow x^2\) là số chính phương
\(\Rightarrow x^2=4=2^2\Rightarrow x=2\)
Vậy x = 2