Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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)
\(4x\cdot\left(x:2\right)-3\left(1-2x\right)=7-2\left(x+1\right)\)
\(\Leftrightarrow4x\cdot\dfrac{x}{2}-3+6x=7-2x-2\)
\(\Leftrightarrow2x\cdot x-3+6x=5-2x\)
\(\Leftrightarrow2x^2-3+6x=5-2x\)
\(\Leftrightarrow2x^2-3+6x-5+2x=0\)
\(\Leftrightarrow2x^2-8+8x=0\)
\(\Leftrightarrow2\left(x^2-4+4x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2+2\sqrt{2}\\x=-2-2\sqrt{2}\end{matrix}\right.\)
Vậy \(x_1=-2-2\sqrt{2};x_2=-2+2\sqrt{2}\)
\(4x\left(x:2\right)-3x\left(1-2x\right)=7-2\left(x+1\right)\)
\(\Leftrightarrow4x.\dfrac{x}{2}-3+6x-7+2x+2=0\Leftrightarrow2x^2+8x-8=0\Leftrightarrow2\left(x^2+4x-4\right)=0\)
\(\Leftrightarrow\left(x^2+4x+4\right)-8=0\)
\(\Leftrightarrow\left(x+2\right)^2=8\Rightarrow\left[{}\begin{matrix}x-2=\sqrt{8}\\x-2=-\sqrt{8}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\sqrt{2}+2\\x=-\sqrt{8}+2\end{matrix}\right.\)
\(=\dfrac{2}{2}\).(\(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{10}\)+...+\(\dfrac{2}{x.\left(x+1\right)}\))
=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=\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x.\left(x+1\right)}\)
\(A=\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+1\right)}\)
\(A=\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{x.\left(x+1\right)}\)
a)
ta có:
\(\left\{{}\begin{matrix}\dfrac{b-a}{b-a}=1..\forall a\ne b\\\dfrac{b-a}{a.b}=\dfrac{1}{a}-\dfrac{1}{b}..\forall a,b\ne0\end{matrix}\right.\)(*)
\(A=\dfrac{1}{2.5}+\dfrac{1}{5.8}+..+\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(\left\{{}\begin{matrix}a=3n-1\\b=3n+2\end{matrix}\right.\)\(\Rightarrow b-a=3..\forall n\)
Thay (*) vào dãy A
\(A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-....+\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\right)\)
\(A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{3n+2}\right)=\dfrac{1}{3}\left(\dfrac{3n+2-2}{2.\left(3n+2\right)}\right)=\dfrac{n}{6n+4}=VP\rightarrow dpcm\)
B) tương tự
Gọi \(ƯC\left(12n+1;30n+2\right)=d\)
\(\Rightarrow12n+1⋮d\Rightarrow60n+5⋮d\)
và \(30n+2⋮d\Rightarrow60n+ 4⋮d\)
Do đó \(60n+5-60n-4⋮d\)
\(\Rightarrow1⋮d\)
\(\Rightarrow d=1\)
Vậy \(\dfrac{12n+1}{30n+2}\) là phân số tối giản.
Gọi (12n+1),(30n+2) là d (1)
=>30n+2 \(⋮\) d
=> 2(30n + 2) \(⋮\) d hay 60n +4 \(⋮\) d
Tương tự ta chưng minh:
12n + 1 \(⋮\)d (2)
=> 5(12n+1) \(⋮\) d hay 60n +5 \(⋮\)d
Do đó (60n + 5) - ( 60n +4 ) \(⋮\)d hay 1 \(⋮\) d
=> d = 1 hoặc -1
Từ (1) và(2) ta có( 12n+1 ;30n+2) =1
=> P/s 12n + 1 /30n+2 là ps tối giản