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 :)

Theo em những bài toán khó lớp 9 hay lớp 10 mik mang lên H.VN nhé OLM ít ng trl lắm ạ

ĐKXĐ: sin x<>0

=>x<>kΩ

Đặt \(t=\sin x+\frac{1}{\sin x}\)

=>\(\left|t\right|\ge2\cdot\sqrt{\sin x\cdot\frac{1}{\sin x}}=2\)

\(t^2=\left(\sin x+\frac{1}{\sin x}\right)^2=\sin^2x+\frac{1}{\sin^2 x}+2\)

\(\Rightarrow\sin^2x+\frac{1}{\sin^2x}=t^2-2\)

Phương trình trở thành: \((t^2-2)+4t-7=0\)

=>\(t^2+4t-9=0\)

=>\(t^2+4t+4=13\)

=>\(\left(t+2\right)^2=13\)

=>\(\left[\begin{array}{l}t+2=\sqrt{13}\\ t+2=-\sqrt{13}\end{array}\right.\Rightarrow\left[\begin{array}{l}t=-2+\sqrt{13}\left(loại\right)\\ t=-2-\sqrt{13}\left(nhận\right)\end{array}\right.\)

\(t=-2-\sqrt{13}\)

=>\(\sin x + \frac{1}{\sin x} = -2 - \sqrt{13}\)

=>\(\sin^2 x + (2 + \sqrt{13})\sin x + 1 = 0\) (1)

\(\Delta = (2 + \sqrt{13})^2 - 4 = 4 + 4\sqrt{13} + 13 - 4 = 13 + 4\sqrt{13}\)

Do đó: Phương trình (1)có hai nghiệm phân biệt là:

\(\left[\begin{array}{l} \sin x = \frac{-(2+\sqrt{13}) + \sqrt{13+4\sqrt{13}}}{2} \approx -0,192 \quad (\text{Thỏa mãn } -1 \le \sin x \le 1) \\ \sin x = \frac{-(2+\sqrt{13}) - \sqrt{13+4\sqrt{13}}}{2} \approx -5,413 \quad (\text{Loại do } \sin x < -1) \end{array}\right.\)

=>\(\left[\begin{array}{l} x = \arcsin\left(\frac{-(2+\sqrt{13}) + \sqrt{13+4\sqrt{13}}}{2}\right) + k2\pi \\ x = \pi - \arcsin\left(\frac{-(2+\sqrt{13}) + \sqrt{13+4\sqrt{13}}}{2}\right) + k2\pi \end{array}\right. \quad (k \in \mathbb{Z})\)

Ta có:

Chọn đáp án B

giải hộ nha

\(2016x^{2017}+2017y^{2016}=2015\left(1\right)\)

Có 2016x²⁰¹⁷ là số chẵn, 2015 là số lẻ 

=> 2017y²⁰¹⁶ là số lẻ => y²⁰¹⁶ là số lẻ

Đặt y¹⁰⁰⁸ = 2k+1 \(\left(k\in Z\right)\)

Có y²⁰¹⁶ = (2k+1)² = 4k²+4k+1

=> 2017y²⁰¹⁶ = 2017 (4k²+4k+1) = 2017.4.(k²+k)+2017

Lại có: \(2017.4.\left(k^2+k\right)\equiv0\left(mod4\right)\)

           \(2017\equiv1\left(mod4\right)\)

suy ra: \(2017y^{2016}\equiv1\left(mod4\right)\)

mà   \(2016x^{2017}\equiv0\left(mod4\right)\)

\(\Rightarrow2016x^{2017}+2017y^{2016}\equiv1\left(mod4\right)\left(2\right)\)

Lại có: \(2015\equiv3\left(mod4\right)\left(3\right)\)

Từ (1), (2) và (3) => PT vô nghiệm