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\(\frac{x-3}{2017}+\frac{x-2}{2018}=\frac{x-2018}{2}+\frac{x-2017}{3}\\\Leftrightarrow \left(\frac{x-3}{2017}-1\right)+\left(\frac{x-2}{2018}-1\right)=\left(\frac{x-2018}{2}-1\right)+\left(\frac{x-2017}{3}-1\right)\\\Leftrightarrow \frac{x-2020}{2017}+\frac{x-2020}{2018}=\frac{x-2020}{2}+\frac{x-2020}{3}\\ \Leftrightarrow\frac{x-2020}{2017}+\frac{x-2020}{2018}-\frac{x-2020}{2}-\frac{x-2020}{3}=0\\ \Leftrightarrow\left(x-2020\right)\left(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2}-\frac{1}{3}\right)=0\\ \Leftrightarrow x-2020=0\left(Vi\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2}-\frac{1}{3}\ne0\right)\\ \Leftrightarrow x=2020\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{2020\right\}\)
\(\frac{x-3}{2017}+\frac{x-2}{2018}=\frac{x-2018}{2}+\frac{x-2017}{3}\)
\(\Leftrightarrow\) \(\frac{x-3}{2017}-1+\frac{x-2}{2018}-1=\frac{x-2018}{2}-1+\frac{x-2017}{3}-1\)
\(\Leftrightarrow\) \(\frac{x-2020}{2017}+\frac{x-2020}{2018}=\frac{x-2020}{2}+\frac{x-2020}{3}\)
\(\Leftrightarrow\) \(\frac{x-2020}{2017}+\frac{x-2020}{2018}-\frac{x-2020}{2}-\frac{x-2020}{3}=0\)
\(\Leftrightarrow\) (x - 2020)(\(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2}-\frac{1}{3}\)) = 0
\(\Leftrightarrow\) x - 2020 = 0
\(\Leftrightarrow\) x = 2020
Vậy S = {2020}
Chúc bn học tốt!!
\(\frac{x-3}{2017}-\frac{x-2}{2018}=\frac{x-2018}{2}+\frac{x-2017}{3}\)
\(\Leftrightarrow\frac{x-3}{2017}-1-\frac{x-2}{2018}-1=\frac{x-2018}{2}-1+\frac{x-2017}{3}-1\)
\(\Leftrightarrow\frac{x-2020}{2017}-\frac{x-2020}{2018}=\frac{x-2020}{2}+\frac{x-2020}{3}\)
\(\Leftrightarrow\frac{x-2020}{2017}-\frac{x-2020}{2018}-\frac{x-2020}{2}-\frac{x-2020}{3}=0\)
\(\Leftrightarrow\left(x-2020\right)\left(\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2}-\frac{1}{3}\right)=0\)
\(\Leftrightarrow x-2020=0\Leftrightarrow x=2020\)
Ta có \(\frac{2015}{2016}.x+\frac{2016}{2017}.x+\frac{2017}{2018}.x=\frac{2018}{2019}.x\)
<=>\(\frac{2015}{2016}.x+\frac{2016}{2017}.x+\frac{2017}{2018}x-\frac{2018}{2019}x=0\)
<=>x\(\left(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}-\frac{2018}{2019}\right)=0\)
Vì \(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}-\frac{2018}{2019}\) không thể bằng 0
Vậy x=0
Ta có 1 nghiệm thỏa mãn S=\(\left\{0\right\}\)
À khác cái dấu nhưng đề phải là giải phương trình chứ
Đặt 2017-x=a => x-2018=-a-1 phương trình trở thành:
\(\frac{a^2+a\left(-a-1\right)+\left(a-1\right)^2}{a^2-a\left(-a-1\right)+\left(a-1\right)^2}=\frac{19}{49}\)
\(\Leftrightarrow\frac{a^2+a+1}{3a^2+3a+1}=\frac{19}{49}\)
\(\Leftrightarrow49\left(a^2+a+1\right)=19\left(3a^2+3a+1\right)\)
\(\Leftrightarrow49a^2+49a+49=57a^2+57a+19\)
\(\Leftrightarrow8a^2+8a-30=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=\frac{3}{2}\\a=-\frac{5}{2}\end{cases}\Rightarrow\orbr{\begin{cases}x=2015,5\\x=2019,5\end{cases}}}\)
Vậy......................
Ta có:\(\frac{x-2}{2017}+1+\frac{x-3}{2018}+1=\frac{x-4}{2019}+1+\frac{x-5}{2020}+1\)
\(\Rightarrow\frac{x+2015}{2017}+\frac{x+2015}{2018}-\frac{x+2015}{2019}-\frac{x+2015}{2020}=0\)
\(\Rightarrow\left(x+2015\right)\left(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)
Mà \(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}>0\)
\(\Rightarrow x+2015=0\Rightarrow x=-2015\)
\(S=\left\{-2015\right\}\)
Chứng minh Nesbit 4 số rồi áp dụng nhé
\(\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}=\frac{a^2}{a\left(b+c\right)}+\frac{b^2}{b\left(c+d\right)}+\frac{c^2}{c\left(d+a\right)}+\frac{d^2}{d\left(a+b\right)}\) (*)
Theo Cauchy - Schwarz dạng engel , ta có
(*) \(\ge\frac{\left(a+b+c+d\right)^2}{a\left(b+c\right)+b\left(c+d\right)+c\left(d+a\right)+d\left(a+b\right)}\)
\(=\frac{2\left(a+c\right)\left(b+d\right)+\left(a+c\right)^2+\left(b+d\right)^2}{\left(a+c\right)\left(b+d\right)+2ac+2bd}\ge\frac{2\left(a+c\right)\left(b+d\right)+4ac+4bd}{\left(a+c\right)\left(b+d\right)+2ac+2bd}=2\)
Đẳng thức xảy ra <=> a = c và b = d
Áp dụng bất đẳng thức Nesbit cho 4 số ,ta có
\(\frac{2018}{x+y}+\frac{x}{y+2017}+\frac{y}{2017+2018}+\frac{2017}{x+2018}\ge2\)
Đẳng thức xảy ra <=> y = 2018 , x = 2017
\(\frac{2-x}{2017}-1=\frac{1-x}{2018}-\frac{x}{2019}\)
\(\Leftrightarrow\) \(\frac{2-x}{2017}+1=\frac{1-x}{2018}+1-\frac{x}{2019}+1\)
\(\Leftrightarrow\) \(\frac{2019-x}{2017}=\frac{2019-x}{2018}-\frac{2019-x}{2019}\)
\(\Leftrightarrow\) \(\frac{2019-x}{2017}-\frac{2019-x}{2018}+\frac{2019-x}{2019}=0\)
\(\Leftrightarrow\) \(\left(2019-x\right)\left(\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}\right)=0\)
Mà \(\left(\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}\right)\ne0\)
\(\Rightarrow\) \(2019-x=0\) \(\Leftrightarrow\) \(x=2019\)
\(\Rightarrow\) \(S=\left\{2019\right\}\)
\(\frac{x-2017}{2018}+\frac{x-2018}{2017}=\frac{2017}{x-2018}+\frac{2018}{x-2017}\)
\(\Leftrightarrow\frac{2017.\left(x-2017\right)+2018.\left(x-2018\right)}{2018.2017}=\frac{2017.\left(x-2017\right)+2018.\left(x-2018\right)}{\left(x-2018\right).\left(x-2017\right)}\)
\(\Leftrightarrow2018.2017=\left(x-2018\right).\left(x-2017\right)\)
\(\Leftrightarrow2018.2017=x^2-4035x+2018.2017\)
\(\Leftrightarrow x^2-4035x=2018.2017-2018.2017\)
\(\Leftrightarrow x.\left(x-4035\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-4035=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=4035\end{cases}}\)
Vậy tập nghiệm của phương trình là \(S=\left\{0;4035\right\}\)