\(2015\sqrt{2015x-2014}+\sqrt{2016x-2015}=2016\)

ĐK:\(x\ge\frac{2015}{2016}\)

\(\Leftrightarrow2015\left(\sqrt{2015x-2014}-1\right)+\sqrt{2016x-2015}-1=0\)

\(\Leftrightarrow2015\frac{2015x-2014-1}{\sqrt{2015x-2014}+1}+\frac{2016x-2015-1}{\sqrt{2016x-2015}+1}=0\)

\(\Leftrightarrow2015\frac{2015x-2015}{\sqrt{2015x-2014}+1}+\frac{2016x-2016}{\sqrt{2016x-2015}+1}=0\)

\(\Leftrightarrow2015\frac{2015\left(x-1\right)}{\sqrt{2015x-2014}+1}+\frac{2016\left(x-1\right)}{\sqrt{2016x-2015}+1}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{2015^2}{\sqrt{2015x-2014}+1}+\frac{2016}{\sqrt{2016x-2015}+1}\right)=0\)

Dễ thấy: \(\frac{2015^2}{\sqrt{2015x-2014}+1}+\frac{2016}{\sqrt{2016x-2015}+1}>0\)

\(\Rightarrow x-1=0\Rightarrow x=1\)

ĐKXĐ: \(x\notin\left\{-\dfrac{1}{2014};-\dfrac{2}{2015};-\dfrac{3}{2016};-\dfrac{4}{2017}\right\}\)

Ta có: \(\dfrac{1}{2014x+1}-\dfrac{1}{2015x+2}=\dfrac{1}{2016x+3}-\dfrac{1}{2017x+4}\)

\(\Leftrightarrow\dfrac{2015x+2-2014x-1}{\left(2014x+1\right)\left(2015x+2\right)}=\dfrac{2017x+4-2016x-3}{\left(2016x+3\right)\left(2017x+4\right)}\)

\(\Leftrightarrow\dfrac{x+1}{\left(2014x+1\right)\left(2015x+2\right)}-\dfrac{x+1}{\left(2016x+3\right)\left(2017x+4\right)}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{\left(2014x+1\right)\left(2015x+2\right)}-\dfrac{1}{\left(2016x+3\right)\left(2017x+4\right)}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\\dfrac{1}{\left(2014x+1\right)\left(2015x+2\right)}=\dfrac{1}{\left(2016x+3\right)\left(2017x+4\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\4058210x^2+6043x+2=4066272x^2+14115x+12\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\8062x^2+8072x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\8062x^2+8062x+10x+10=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\8062x\left(x+1\right)+10\left(x+1\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\\left(x+1\right)\left(8062x+10\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x+1=0\\8062x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-1\\8062x=-10\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(nhận\right)\\x=\dfrac{-5}{4031}\left(nhận\right)\end{matrix}\right.\)

Vậy: \(S=\left\{-1;\dfrac{-5}{4031}\right\}\)

câu này làm ntn :)

bạn có: x^4 + 2016x^2 + 2015x + 2016
= x^4 + x^3 + x^2 - x^3 - x^2 - x + 2016x^2 + 2016x + 2016
= x^2(x^2 + x + 1) - x(x^2 + x + 1) + 2016(x^2 + x + 1)
= (x^2 + x + 1)(x^2 - x + 2016)

  \(x^4+2016x^2+2015x+2016\)

=\(x^4+x^3+x^2+2015x^2+2015x+2015+1-x^3\)

=\(x^2\left(x^2+x+1\right)+2015\left(x^2+x+1\right)+\left(1-x\right)\left(x^2+x+1\right)\)

=\(\left(x^2+x+1\right)\left(x^2+2015+1-x\right)\)

=\(\left(x^2+x+1\right)\left(x^2-x+2016\right)\)

tui ko bít bạn học lớp mí

lớp999999

Vào câu hỏi tương tự đi

\(=x^4-x+2016x^2+2016x+2016.\)

\(=x\left(x^3-1\right)+2016\left(x^2+x+1\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)+2016\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2016\right)\)

2015x⁴ + 2016x² + x + 2016

= (2015x⁴ + 2015x³ + 2015x²) + (- 2015x³ - 2015x² - 2015x) + (2016x² + 2016x + 2016)

= (x² + x + 1)(2015x² - 2015x + 2016)

Vào câu trả lời tương tự đi có đáp án đó