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 :)

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

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

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

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

\(pt\Leftrightarrow 2015\sqrt{2015x-2014}-2015+\sqrt{2016x-2015}-1=0\)

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

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

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

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

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

dệ mà m :v bình phương đi :v

Câu 3:

$\Delta=2015^2-4.2013.2=2011^2$

Do đó pt có 2 nghiệm:

$x_1=\frac{2015+2011}{2.2013}=1$

$x_2=\frac{2015-2011}{2.2013}=\frac{2}{2013}$

Đáp án B.

Câu 4:

Theo định lý Viet, tổng các nghiệm của pt là:

$S=\frac{-b}{a}=\frac{-3}{1}=-3$

Đáp án B.

Câu 5:

PT (C) có $\Delta'=1-4.3.2<0$ nên PT này vô nghiệm

Đáp án C.

Nhận xét: Tổng các hệ số của phương trình bằng 0 => phương trình có 1 nghiệm là 1

=> vế trái có nhân tử (x - 1)

pt <=> (x⁴ - 1 ) + (2015x³ - 2015x²) - (2015x - 2015)  = 0

<=> (x-1)(x+1).(x² + 1) + 2015x²(x - 1) - 2015.(x - 1) = 0

<=> (x - 1).[(x+1).(x² + 1) + 2015x² - 2015] = 0

<=> (x -1). [(x+1).(x² + 1) + 2015(x² - 1)] = 0

<=> (x -1). [(x+1).(x² + 1) + 2015(x - 1)(x+1)] = 0

<=> (x -1).(x+1).(x² + 1 + 2015x - 2015 ) = 0  

<=> x - 1 = 0 hoặc  x+ 1 = 0 hoặc x² + 1 + 2015x - 2015  = 0

+) x - 1 = 0 <=> x = 1

+) x + 1 = 0 <=> x = -1

+) x² + 1 + 2015x - 2015 = 0 <=> x² + 2015x - 2014 = 0 

<=> x² +2.x. \(\frac{2015}{2}\) + \(\left(\frac{2015}{2}\right)^2\) - \(\left(\frac{2015}{2}\right)^2\)   - 2015 = 0

<=> \(\left(x-\frac{2015}{2}\right)^2=\frac{2015^2+4030}{2}\)

<=>  \(x-\frac{2015}{2}=\sqrt{\frac{2015^2+4030}{2}}\) hoặc \(x-\frac{2015}{2}=-\sqrt{\frac{2015^2+4030}{2}}\)

<=> \(x=\frac{2015}{2}+\sqrt{\frac{2015^2+4030}{2}}\)hoặc \(x=\frac{2015}{2}-\sqrt{\frac{2015^2+4030}{2}}\)

Vậy pt có 4 nghiệm...

chính xác nè bạn nhớ sai ruj:

x⁴+2015x²+2014x+2015=0

<=>x⁴-x+2015x²+2015x+2015=0

<=>x(x³-1)+2015(x²+x+1)=0

<=>x(x-1)(x²+x+1)+2015(x²+x+1)=0

<=>(x²+x+1)[x(x-1)-2015]=0

<=>(x²+x+1)(x²-x-2015)=0

<=>x²+x+1=0 hoặc x²-x-2015=0

*x²+\(2x.\frac{1}{2}\)+\(\frac{1}{4}+\frac{3}{4}\)=0 

<=>(x+1/2)²+3/4=0(vô lí)

*x²-\(2x.\frac{1}{2}+\frac{1}{4}-\frac{8061}{4}\)

<=>(x-1/2)²-8061/4=0

<=>(x-1/2)²           =8061/4

<=>x-1/2              =\(\sqrt{\frac{8061}{4}}\)

<=>x                    =\(\sqrt{\frac{8061}{4}+}\frac{1}{2}\)

\(a,m=3=>x^2+3x-2=0\)

\(\Delta=3^2-4\left(-2\right)=17>0\)

pt có 2 nghiệm pb \(\left[{}\begin{matrix}x1=\dfrac{-3+\sqrt{17}}{2}\\x2=\dfrac{-3-\sqrt{17}}{2}\end{matrix}\right.\)

b,\(\Delta=m^2-4\left(-2\right)=m^2+8>0\)

=> pt đã cho luôn có 2 nghiệm phân biệt x1,x2 với mọi m

theo vi ét \(\left\{{}\begin{matrix}x1+x2=-m\\x1x2=-2\end{matrix}\right.\)

có \(x1^2x2+x2^2x1=2014< =>x1x2\left(x1+x2\right)=2014\)

\(< =>-2\left(-m\right)=2014< =>m=1007\)

a) Thay m=3 vào phương trình, ta được:

\(x^2+3x-2=0\)

\(\Delta=3^2-4\cdot1\cdot\left(-2\right)=9+8=17\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-3-\sqrt{17}}{2}\\x_2=\dfrac{-3+\sqrt{17}}{2}\end{matrix}\right.\)