\(\frac{2016+x}{x}+\frac{2520+x}{x}+\frac{3024+x}{x}=2523\)

\(\Leftrightarrow\frac{2016+x+2520+x+3024+x}{x}=2523\)

\(\Leftrightarrow\frac{7560+3x}{x}=\frac{2523x}{x}\)Khử mẫu : \(7560+3x=2523x\)

\(\Leftrightarrow7560=2520x\Leftrightarrow x=3\)

\(\Rightarrow\frac{2016}{x}+\frac{x}{x}+\frac{2520}{x}+\frac{x}{x}+\frac{3024}{x}+\frac{x}{x}=2523\)

\(\Rightarrow\frac{2016}{x}+\frac{2520}{x}+\frac{3024}{x}=2520\)

\(\Rightarrow\frac{1}{x}\left(2016+2520+3024\right)=2520\)

\(\Rightarrow x=3\)

bài này đơn giản thôi!! x=3 . Vì không biết viết phân số nên tớ ko trả lời kèm lời giải được !! muốn biết thì hỏi lại tớ nhá !!

Cảm ơn các bạn nha

\(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+3+...+2016}\right)\)

\(A=\left(1-\frac{1}{\frac{2\left(2+1\right)}{2}}\right)\left(1-\frac{1}{\frac{3\left(3+1\right)}{2}}\right)...\left(1-\frac{1}{\frac{2016\left(2016+1\right)}{2}}\right)\)

\(A=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}...\frac{2016.2017-2}{2016.2017}\)(1)

Mà \(2016.2017-2=2016\left(2018-1\right)+2016-2018\)

\(=2016\left(2018-1+1\right)-2018=2016.2018-2018=2018.2015\)(2)

Từ (1) và (2), ta có:

\(A=\frac{4.1}{2.3}.\frac{5.2}{3.4}.\frac{6.3}{4.5}...\frac{2018.2015}{2016.2017}=\frac{\left(4.5.6...2018\right)\left(1.2.3...2015\right)}{\left(2.3.4...2016\right)\left(3.4.5...2017\right)}=\frac{1009}{3024}\)

vô tcn của PTD/KM ?, https://olm.vn/thanhvien/kimmai123az, toàn câu tl copy, con giẻ rách này ko nên sông nx

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Còn rất rất nhìu nx, ko có t/g

\(\frac{x-1}{2016}+\frac{x-2}{2015}+\frac{x-3}{2014}+...+\frac{x-2016}{1}=2016\)

\(\Rightarrow\frac{x-1}{2016}-1+\frac{x-2}{2015}-1+\frac{x-3}{2014}-1+...+\frac{x-2016}{1}-1=2016-2016\)

\(\Rightarrow\frac{x-2017}{2016}+\frac{x-2017}{2015}+\frac{x-2017}{2014}+...+\frac{x-2017}{1}=0\)

\(\Rightarrow\left(x-2017\right).\left(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+...+1\right)=0\)

Mà \(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+...+1\ne0\Rightarrow x-2017=0\)

=> x = 2017

\(\frac{x-1}{2016}+\frac{x-2}{2015}+\frac{x-3}{2014}+...+\frac{x-2016}{1}=2016\)

\(\Leftrightarrow\frac{x-1}{2016}-1+\frac{x-2}{2015}-1+\frac{x-3}{2014}-1+...+\frac{x-2016}{1}-1=0\)

\(\Leftrightarrow\frac{x-2017}{2016}+\frac{x-2017}{2015}+\frac{x-2017}{2014}+...+\frac{x-2017}{1}=0\)

\(\Leftrightarrow\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}+...+1\right)=0\)

Có: \(\frac{1}{2016}+\frac{1}{2015}+...+1\ne0\)

\(\Rightarrow x-2017=0\)

\(\Rightarrow x=2017\)

<=> \(\frac{x-1}{2016}+\frac{x-2}{2015}+\frac{x-3}{2014}+....+\frac{x-2016}{1}-2016=0\)\(=0\)

<=> \(\left(\frac{x-1}{2016}-1\right)+\left(\frac{x-2}{2015}-1\right)+...+\left(\frac{x-2016}{1}-1\right)=0\)

<=> \(\frac{x-2017}{2016}+\frac{x-2017}{2015}+...+\frac{x-2017}{1}=0\)

<=> \(\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}+...+\frac{1}{1}\right)=0\)

<=> \(x-2017=0\)\(\left(do\frac{1}{2016}+\frac{1}{2015}+...+\frac{1}{1}>0\right)\)

<=> \(x=2017\)

Vậy x = 2017

đúng thì

a) \(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)

\(\Leftrightarrow\frac{x+2015}{5}+\frac{5}{5}+\frac{x+2016}{4}+\frac{4}{4}=\frac{x+2017}{3}+\frac{3}{3}+\frac{x+2018}{2}+\frac{2}{2}\)

\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2002}{2}\)

\(\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)

\(\Leftrightarrow\left(x+2020\right).\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)

\(\Leftrightarrow x+2020=0\)

\(\Leftrightarrow x=-2020\)

Vậy : \(x=-2020\)

Chúc bạn học tốt !!

a) \(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\\ \left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\\ \frac{x+2020}{5}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2020}{2}\\ \frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\\ \left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\\ \Rightarrow x+2020=0\\ \Rightarrow x=-2020\)

Vậy x = -2020

b) \(\frac{x+2015}{5}+\frac{x+2016}{6}=\frac{x+2017}{7}+\frac{x+2018}{8}\\ \left(\frac{x+2015}{5}-1\right)+\left(\frac{x+2016}{6}-1\right)=\left(\frac{x+2017}{7}-1\right)+\left(\frac{x+2018}{8}-1\right)\\ \frac{x+2010}{5}+\frac{x+2010}{6}=\frac{x+2010}{7}+\frac{x+2010}{8}\\ \frac{x+2010}{5}+\frac{x+2010}{6}-\frac{x+2010}{7}-\frac{x+2010}{8}=0\\ \left(x+2010\right)\left(\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}\right)=0\\ \Rightarrow x+2010=0\\ \Rightarrow x=-2010\)

Vậy x = -2010

\(\sqrt{x-2016}+\sqrt{y-2017}+\sqrt{z-2018}+3024=\frac{1}{2}\left(x+y+z\right)\)

\(\Leftrightarrow2\left(\sqrt{x-2016}+\sqrt{y-2017}+\sqrt{z-2018}+3024\right)=x+y+z\)

\(\Leftrightarrow2\sqrt{x-2016}+2\sqrt{y-2017}+2\sqrt{z-2018}+6048=x+y+z\)

\(\Leftrightarrow x-2\sqrt{x-2016}+y-2\sqrt{y-2017}+z-2\sqrt{z-2018}+6048=0\)

\(\Leftrightarrow x-2016-2\sqrt{x-2016}+1+y-2017+2\sqrt{y-2017}+1+z-2018-2\sqrt{z-2018}+1=0\)

ĐK : \(x\ge2016;y\ge2017;z\ge2018\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2016}-1=0\\\sqrt{y-2017}-1=0\\\sqrt{z-2018}-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2016}=1\\\sqrt{y-2017}=1\\\sqrt{z-2018}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2017\\y=2018\\z=2019\end{matrix}\right.\)

Lời giải:

Áp dụng BĐT AM-GM:

\(\sqrt{x-2016}\leq \frac{1+(x-2016)}{2}=\frac{x-2015}{2}\)

\(\sqrt{y-2017}\leq \frac{1+(y-2017)}{2}=\frac{y-2016}{2}\)

\(\sqrt{z-2018}\leq \frac{1+(z-2018)}{2}=\frac{z-2017}{2}\)

Cộng theo vế thu được:

\(\sqrt{x-2016}+\sqrt{y-2017}+\sqrt{z-2018}+3024\leq \frac{x-2015}{2}+\frac{y-2016}{2}+\frac{z-2017}{2}+3024=\frac{x+y+z}{2}\)

Dấu "=" xảy ra khi \(\left\{\begin{matrix} x-2016=1\\ y-2017=1\\ z-2018=1\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=2017\\ y=2018\\ z=2019\end{matrix}\right.\)

\(\sqrt{x-2016}+\sqrt{y-2017}+\sqrt{z-2018}+3024=\frac{1}{2}\left(x+y+z\right)\)

\(\Leftrightarrow2\left(\sqrt{x-2016}+\sqrt{y-2017}+\sqrt{z-2018}+3024\right)=x+y+z\)

\(\Leftrightarrow2\sqrt{x-2016}+2\sqrt{y-2017}+2\sqrt{z-2018}+6048=x+y+z\)

\(\Leftrightarrow x-2\sqrt{x-2016}+y-2\sqrt{y-2017}+z-2\sqrt{z-2018}+6048=0\)

\(\Leftrightarrow x-2016-2\sqrt{x-2016}+1+y-2017+2\sqrt{y-2017}+1+z-2018-2\sqrt{z-2018}+1=0\)

\(\Leftrightarrow\left(\sqrt{x-2016}-1\right)^2+\left(\sqrt{y-2017}-1\right)^2+\left(\sqrt{z-2018}-1\right)^2=0\)

\(ĐK:x\ge2016;y\ge2017;z\ge2018\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-2016}-1=0\\\sqrt{y-2017}-1=0\\\sqrt{z-2018}-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}\sqrt{x-2016}=1\\\sqrt{y-2017}=1\\\sqrt{z-2018}=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2017\\y=2018\\z=2019\end{cases}}}\)

nhân đôi 2 vế rồi chuyển vế trái sang vế phải, ta có:

\(\left(\sqrt{x-2016}-1\right)^2\) + \(\left(\sqrt{y-2017}-1\right)^2\)

+ \(\left(\sqrt{z-2018}-1\right)^2\)

= 0